Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Students can download 12th Business Maths Chapter 7 Probability Distributions Ex 7.1 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 1.
Define Binomial distribution.
Solution:
A random variable X is said to follow a binomial distribution with parameter ‘n’ and ‘p’ if it assumes only non-negative value and its probability mass function is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q1

Question 2.
Define Bernoulli trials.
Solution:
A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q, is called a Bernoulli trial.
Example 1, Tossing of a coin (Head or Tail)
Example 2, Writing an exam (Pass or Fail)

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 3.
Derive the mean and variance of binomial distribution.
Solution:
Let X be a random variable with the Binomial distribution.
The probability function of X is
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.2

Question 4.
Write down the conditions for which the binomial distribution can be used.
Solution:
The binomial distribution can be used under the following conditions:

  • The number of trials (or) observations ‘n’ is fixed (finite).
  • Each observation is independent of each other.
  • In every trial, there are only two possible outcomes – success or failure.
  • The probability of success ‘p’ is the same for each outcome.

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 5.
Mention the properties of the binomial distribution.
Solution:
Property 1:
The binomial distribution is symmetrical when the probability of success ‘p’ is 0.5 (or) when a number of trials ‘n’ is very large. In other words, if p = q = 1/2, the distribution is symmetric about the median. If p ≠ q, then it is skewed distribution, (p < 0.5 → positively skewed, p > 0.5 → negatively skewed)

Property 2:
The variance is less than mean (i,e,) npq < np

Question 6.
If 5% of the items produced turn out to be defective, then find out the probability that out of 10 items selected at random there are
(i) exactly three defectives
(ii) at least two defectives
(iii) exactly 4 defectives
(iv) find the mean and variance
Solution:
Let p be the probability of a defective item.
Given that, p = 5% = \(\frac{5}{100}\) = 0.05
So q = 1 – p = 1 – 0.05 = 0.95. Also n = 10.
Let X be the random variable which follows the binomial distribution. Then X ~ B (10, 0.05)
(i) P(exactly three defectives) = P(X = 3)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q6
(ii) P(atleast two defectives) = P (X ≥ 2) = 1 – P (X < 2)
= 1 – [P(X = 1) + P(X = 0)]
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q6.1
(iii) P(exactly 4 defectives) = P(X = 4)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q6.2
(iv) We know that mean = np = (10) (0.05) = 0.5
Variance = npq = (10) (0.05) (0.95) = 0.475

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 7.
In a particular university, 40% of the students are having newspaper reading habit. Nine university students are selected to find their views on reading habit. Find the probability that
(i) none of those selected has newspaper reading habit
(ii) all those selected have newspaper reading habit
(iii) at least two-third have newspaper reading habit.
Solution:
Let X be the binomial random variable which denotes the number of students having newspaper reading habit.
It is given that 40% of students have reading habit.
p = \(\frac{40}{100}\) = 0.4 and q = 1 – 0.4 = 0.6
(i) P(none of selected have newspaper reading habit) = P(X = 0)
Now X ~ B (9, 0.4)
The p.m.f is given by P (X = x) = p (x) = \(^{9} \mathrm{C}_{x}(0.4)^{x}(0.6)^{9-x}\)
P(X = 0) = \(^{9} \mathrm{C}_{0}(0.4)^{0}(0.6)^{9}\) = (0.6)9 = 0.01008 (using calculator)
(ii) P (all selected have newspaper reading habit)
= P (X = 9)
= \(^{9} \mathrm{C}_{9}(0.4)^{9}(0.6)^{0}\)
= (0.4)9
= 0.000262 (using calculator)
(iii) P (at least two third have newspaper reading habit) = P (X ≥ 6)
{9 students are selected. Two third of them means \(\frac{2}{3}\) (9) = 6}
Now P (X ≥ 6) = P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q7
= (84) (0.004096) (0.216) + 36 (0.0016384) (0.36) + 9 (0.00065536) (0.6) + 0.000262
= 0.074318 + 0.021234 + 0.003539 + 0.000262
= 0.099353

Question 8.
In a family of 3 children, what is the probability that there will be exactly 2 girls?
Solution:
Let X denote the binomial variable which denotes the number of girls.
Given that n = 3 and p = q = \(\frac {1}{2}\)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q8
Hence the probability that there will be exactly 2 girls is. 0.375.

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 9.
Defects in yarn manufactured by a local mill can be approximated by a distribution with a mean of 1.2 defects for every 6 metres of length. If lengths of 6 metres are to be inspected, find the probability of fewer than 2 defects.
Solution:
Given mean np = 1.2 and n = 6
p = \(\frac{1.2}{6}\) = 0.2, q = 1 – 0.2 = 0.8
Let X be a binomial variable denoting the number of defects, (i.e,) X ~ B (6, 0.2)
p.m.f is given by P (X = x) = \(^{6} \mathrm{C}_{x}(0.2)^{x}(0.8)^{6-x}\)
We want P(X < 2) = P(X = 0) + P (X = 1)
= \(^{6} \mathrm{C}_{0}(0.2)^{0}(0.8)^{6}+^{6} \mathrm{C}_{1}(0.2)^{1}(0.8)^{5}\)
= (0.8)6 + 6 (0.2) (0.8)5
= 0.262144 + 0.393216
= 0.65536
Thus if lengths of 6 metres are to be inspected, the probability of less than 2 defects is 0.65536.

Question 10.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
(i) exactly one will be defective
(ii) none will be defective
(iii) at most 2 will be defective
Solution:
Let X be the random variable denoting the number of defective bolts.
The probability of defective bolts p = \(\frac{18}{100}\) = 0.18 ⇒ q = 0.82.
Also n = 4
The p.m.f is P (X = x ) = \(^{4} \mathrm{C}_{x}(0.18)^{x}(0.82)^{4-x}\)
(i) P (exactly one defective) = P(X = 1)
= \(^{4} \mathrm{C}_{1}(0.18)^{1}(0.82)^{3}\)
= 4 (0.18) (0.82)3
= 0.3969
(ii) P (no defective) = P(X = 0)
= \(^{4} \mathrm{C}_{0}(0.18)^{0}(0.82)^{4}\)
= (0.82)4
= 0.45212
(iii) P (atmost 2 defective) = P(X ≤ 2)
= P(X = 2) + P(X = 1) + P(X = 0)
= \(^{4} \mathrm{C}_{2}\) (0.18)2 (0.82)2 + 0.3969 + 0.45212
= 0.1307 + 0.3969 + 0.45212
= 0.97972

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 11.
If the probability of success is 0.09, how many trials are needed to have a probability of at least one success as 1/3 or more?
Solution:
Given p = 0.09 (success)
q = 0.91 (failure)
We have to find number of trials ‘n.’
According to the problem,
P(X ≥ 1 ) > \(\frac{1}{3}\)
(We must have atleast one success)
1 – P(X < 1) > \(\frac{1}{3}\)
1 – P(X = 0) > \(\frac{1}{3}\)
(or) P(X = 0) < \(\frac{2}{3}\)
Using p.m.f, we have,
\(^{n} \mathrm{C}_{0}(0.09)^{0}(0.91)^{n}<\frac{2}{3}\)
(0.91)n < \(\frac{2}{3}\)
we can use log tables to calculate (or) by trial method try for n = 1, 2,…… using calculator.
We observe that (0.91)5 < \(\frac{2}{3}\). Thus we need minimum 5 trial or more.

Question 12.
Among 28 professors of a certain department, 18 drive foreign cars and 10 drive locally made cars. If 5 of these professors are selected at random, what is the probability that at least 3 of them drive foreign cars?
Solution:
Here n = 5, p = \(\frac{18}{28}=\frac{9}{14}\), q = \(\frac{10}{28}=\frac{5}{14}\)
(i.e.) the probability of professors driving foreign cars p = \(\frac{9}{14}\), and those who drive local cars q = \(\frac{5}{14}\).
Let X be the Binomial random variable denoting persons who drive foreign cars.
Then the p.m.f of X is given by P (X = x) = \(^{5} \mathrm{C}_{x}\left(\frac{9}{14}\right)^{x}\left(\frac{5}{14}\right)^{5-x}\)
We want P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X = 5)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q12

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 13.
Out of 750 families with 4 children each, how many families would be expected to have
(i) at least one boy
(ii) at most 2 girls
(iii) and children of both sexes?
Assume equal probabilities for boys and girls.
Solution:
Given that 750 families are considered each with 4 children. We will find the probabilities for one particular family and then multiply by 750.
In other words, n = 4, p = q = \(\frac{1}{2}\) (since boy and girl child have equal probability).
Let X denote the binomial random variable which denotes the number of boys in the family.
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q13
So out of 750 families the number of families would be expected to have atleast one boy is \(\frac{15}{16}\) × 750 = 703
(ii) P(atmost 2 girls) = P(2G, 2B) + P(1G, 3B) + P(0G, 4B)
= P(X = 2) + P(X = 3) + P(X = 4)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q13.1
Thus out of 750 families, 516 families would be expected to have atmost 2 girls.
(iii) P(children of both sexes) = P(both boys and girls)
Out of 4 children the sample space is given by {BGGG, BBGG, BBBG}and each case in any order.
So we require P(1B, 3G) + P(2B, 2G) + P(3B, 1G)
(i.e,) P(X = 1) + P(X = 2) + P(X = 3)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q13.2
Thus out of 750 families, 656 families would be expected to have children of both sexes.

Question 14.
Forty percent of business travellers carry a laptop. In a sample of 15 business travelers
(i) what is the probability that 3 will have a laptop?
(ii) what is the probability that 12 of the travelers will not have a laptop?
(iii) what is the probability that atleast three of the travelers have a laptop?
Solution:
Let X be the binomial variables which denotes the number of business travellers having a laptop.
Given that n = 15 and P = 40% = 0.4. So q = 1 – 0.4 = 0.6. Thus X ~ B (15, 0.4).
The p.m.f of X is given by P (X = x) = \(^{15} \mathrm{C}_{x}(0.4)^{x}(0.6)^{15-x}\)
(i) P(3 travellers will have a laptop) = P (X = 3)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q14
Note: The calculation can be done by method of logarithms also.
P(X = 3) = 455 (0.064) (0.002177) = 0.0634
(ii) P(12 of the travellers will not have a laptop)
= P(15 – 12 = 3 will have a laptop)
= P(X = 3) = 0.0634 (from the previous subdivision)
(iii) P(atleast three of the travellers have a laptop)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q14.1
Using (0.6)12 = 0.002177 from the previous subdivision, we have
= 1 – (0.002177) [10.08 + 2.16 + 0.216]
= 1 – (0.002177) (12.456)
= 1 – 0.02712
= 0.9729

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 15.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 successes.
Solution:
Let p be the probability of getting a doublet, (i.e,) probability of success. When we throw a pair of dice there are 36 possibilities. The number of doublets is 6 [(1, 1) (2, 2), (3, 3) (4, 4) (5, 5) (6, 6)].
So p = \(\frac{6}{36}=\frac{1}{6}\)
q = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Let X be the random variable denoting the number of doublet in 4 throws.
Then X ~ B (4, \(\frac{1}{6}[/latex)]
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q15
Hence the probability of 2 successes is [latex]\frac{25}{216}\)

Question 16.
The mean of a binomial distribution is 5 and the standard deviation is 2. Determine the distribution.
Solution:
Given mean = 5 and standard deviation = 2
(i.e,) np = 5 and √npq = 2 ⇒ npq = 4
5q = 4 ⇒ q = \(\frac{4}{5}\), p = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Again np = 5 gives \(\frac{n}{5}\) = 5 ⇒ n = 25
So the p.m.f of the distribution is given by P (X = x) = \(\left(\begin{array}{c}
25 \\
x
\end{array}\right)\left(\frac{1}{5}\right)^{x}\left(\frac{4}{5}\right)^{25-x}\)

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 17.
Determine the binomial distribution for which the mean is 4 and variance 3. Also find P(X = 15)
Solution:
Given mean = 4 and variance is 3.
(i.e,) np = 4 and npq = 3
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q17

Question 18.
Assume that a drug causes a serious side effect at a rate of three patients per one hundred. What is the probability that at least one person will have side effects in a random sample of ten patients taking the drug?
Solution:
According to the problem, n = 10, p = \(\frac{3}{100}\) = 0.03 where p is the probability that a drug causes side effect. Now X ~ B (10, 0.03). The p.m.f is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q18
Thus the probability that at least one person will have side effects is 0.2626.

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 19.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease. Assume that the probability of recovery is 0.73. What is the probability that at least 3 of the 5 mice recover?
Solution:
Given n = 5 and the probability of recovery p = 0.73.
So q = 1 – 0.73 = 0.27. X ~ B (5, 0.73).
The p.m.f of X is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q19
Thus the probability that at least 3 of the 5 mice recover is 0.8743.

Question 20.
An experiment succeeds twice as often as it fails, what is the probability that in the next five trials there will be
(i) three successes and
(ii) at least three successes.
Solution:
Given a number of trials n = 5.
Let P be the probability of success and q be the probability of failure. It is given that p = 2q.
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q20
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q20.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Students can download 12th Business Maths Chapter 7 Probability Distributions Additional Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Choose the correct answer:

Question 1.
If X is a poisson variate with P (X = 1) = P (X = 2), the mean of the poisson variate is equal to _____
(a) 1
(b) 2
(c) -2
(d) 3
Answer:
(b) 2
Hint:
P(X = 1) = P(X = 2)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems I Q1

Question 2.
If ‘λ’ is the mean of the Poisson distributions, then P (X = 0) is given by ________
(a) \(e^{-\lambda}\)
(b) \(e^{\lambda}\)
(c) e
(d) \(\lambda^{-e}\)
Answer:
(a) \(e^{-\lambda}\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems I Q2

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 3.
A larger standard deviation for a normal distribution with an unchanged mean indicates that the curve becomes, ____
(a) narrower and more peaked
(b) flatter and wider
(c) more skewed to the right
(d) more skewed to the left
Answer:
(b) flatter and wider

Question 4.
If X ~ N (0, 4), the value of P(|X| ≥ 2.2) is ______
(a) 0.2321
(b) 0.8438
(c) 0.2527
(d) 0.2714
Answer:
(d) 0.2714
Hint:
P (|X| ≥ 2.2)
= P (X > 2.2) + P (X < -2.2)
= P(Z > \(\frac{2.2-0}{2}\)) + P(Z < \(\frac{-2.2-0}{2}\))
= P(Z > 1.1) + P(Z < -1.1)
= 2 P(Z > 1.1) (by symmetry)
= 2[0.5 – P(0 < Z < 1.1)]
= 2 [0.5 – 0.3643]
= 0.2714

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 5.
For a binomial distribution.

  1. If n = 1, then E(X) is ________
  2. The variance is always ________
  3. Successive trials are ______
  4. Negatively skewed when _______
  5. Number of parameters is _______
  6. n = 10, p = 0.3, variance is _______
  7. Is symmetrical when ______
  8. n = 6, p = 0.9, P (X = 7) is _______
  9. Mean, median and mode will be equal when ______

Answer:

  1. p
  2. less than mean
  3. independent
  4. p > \(\frac {1}{2}\)
  5. two
  6. 2.1
  7. p = q
  8. zero
  9. p = 0.5

Question 6.
For a Normal distribution

  1. The parameters which controls the flatness of the curve is _____ & ______
  2. If Y = 5X + 10 and X ~ N (10, 25), then mean of Y is _______
  3. Normal curve is asymptotic to the ________
  4. The median corresponds to the value of Z = _______
  5. The area under the normal curve on either side of mean is _______

Answer:

  1. µ, σ
  2. 60
  3. X-axis
  4. µ
  5. 0.5

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 7.
Heights of college girls follows a normal distribution with mean 65 inches and a standard deviation of 3 inches. About what proportion of girls are between 65 and 67 inches tall?
(a) 0.75
(b) 0.5
(c) 0.25
(d) 0.17
Answer:
(c) 0.25
Hint:
P (65 < X < 67)
= P(\(\frac{65-65}{3}\) < Z < \(\frac{67-65}{3}\))
= P(0 < Z < 67)
= 0.2486 ~ 0.25

Question 8.
Which one of these is a binomial random variable?
(a) time is taken by a randomly chosen student to complete an exam
(b) number of books bought by a randomly chosen student
(c) number of women taller than 150 cm in a random sample of 10 women
(d) number of CD’s a randomly selected person owns
Answer:
(c) number of women taller than 150 cm in a random sample of 10 women

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 9.
Pulse rates of adult men follows a normal distribution with a mean of 70 and a standard deviation of 8. Which choice tells how to find the proportion of men that have a pulse rate greater than 78?
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems I Q9
(a) find the area to the left of Z = 1 under the normal curve
(b) find the area between Z = -1 and Z = 1 under the standard normal curve
(c) find the area to the right of Z = 1 under the standard normal curve
(d) find the area to the right of Z = -1 under the standard normal curve
Answer:
(c) find the area to the right of Z = 1 under the standard normal curve
Hint:
P(X > 78) = P(Z > \(\frac{78-70}{8}\)) = P(Z > 1)

2 Mark Questions

Question 1.
The probability that a radio manufactured by a company will be defective is \(\frac{1}{10}\). If 15 such radios are inspected, find the probability that exactly 3 will be defective.
Solution:
Given n = 15, p = \(\frac{1}{10}\) = 0.1, q = 0.9 10
Let X be the binomial variable, denoting the number of radios.
We want probability of exactly 3 defectives, (i.e) P (X = 3)
Now P (X = 3) = \(^{15} \mathrm{C}_{3}(0.1)^{3}(0.9)^{12}\)
= 455 (0.001) [(0.9)4]3
= (455) (0.001) (0.6561)3
= 0.1285

Question 2.
The probability that a bulb produced in a factory will fuse after 10 days is 0.05. Find the probability that out of 5 such bulbs, not more than 1 will fuse after 400 days of use.
Solution:
Given that X ~ B (5, 0.05)
(i.e.) n = 5, p = 0.05, q = 0.95
P(X ≤ 1) = P(X = 0) + P(X = 1)
= \(^{5} \mathrm{C}_{0}(0.05)^{0}(0.95)^{5}+^{5} \mathrm{C}_{1}(0.05)^{1}(0.95)^{4}\)
= (0.95)5 + 5 (0.05) (0.95)4
= (0.95)4 + [0.95 + 0.25]
= 0.9774

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 3.
The number of traffic accidents that occur in a particular road follows a Poisson distribution with a mean of 9.4. Find the probability that less than two accidents will occur on this road during a randomly selected month.
Solution:
Let X be the Poisson variate with mean λ = 9.4.
Now P(X < 2) = P (X = 0) + P (X = 1)
= \(e^{-9.4}\) [1 + 9.4]
= \(e^{-9.4}\) (10.4)
= 0.00086

Question 4.
What is the probability of getting 2 Sundays out of 15 days selected at random?
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems II Q4

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 5.
Find the mean and standard deviation of a Poisson variate X which satisfies the condition P(X = 2 ) = P(X = 3)
Solution:
P(X = 2 ) = P(X = 3)
\(\frac{e^{-\lambda} \lambda^{2}}{2 !}=\frac{e^{-\lambda} \lambda^{3}}{3 !}\)
1 = \(\frac{\lambda}{3}\) (or) λ = 3
Variance for Poisson distribution is λ = 3. Hence s.d = √3

Question 6.
Between 5 to 6 p.m., the average number of phone calls per minute is 4. What is the probability that there is no phone call during a minute?
Solution:
Let X be the Poisson variate denoting the number of phone calls per minute.
Given that mean λ = 4. We want P(X = 0)
Now P(X = 0) = \(\frac{e^{-4} \lambda^{0}}{0 !}\) = e-4 = 0.0183

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 7.
2% cars are defective. What is the probability that one of 150 cars, there is exactly one defective car?
Solution:
Let X be the Poisson variate denoting the number of defective cars. The mean λ is given by
λ = \(\frac {2}{100}\) × 150 = 3
So P(X = 1) = \(\frac{e^{-3}(3)^{1}}{1 !}\) = 3e-3 = 0.1494

Question 8.
What is the standard deviation of the number of recoveries among 48 patients, the probability of recovering is 0.75.
Solution:
Given n = 48, p = 0.75, q = 0.25
This is a binomial distribution.
The variance is given by Var (X) = npq
= (48) (0.75) (0.25)
= 9 .
So the standard deviation is √9 = 3

3 and 5 Mark Questions

Question 1.
What is the probability of success of the binomial distribution satisfying the following condition: 4P(X = 4) = P(X = 2) and having other parameter as 6?
Solution:
Given n = 6, 4P(X = 4) = P(X = 2).
We have to find p.
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q1
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q1.1

Question 2.
Wool fibre breaking strengths are normally distributed with mean µ = 23.56 and S.D σ = 4.55. What proportion of fibres would have a breaking strength of 14.45 or less?
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q2
Let X be the normal variable denoting the breaking strengths.
We want to find
P(X ≤ 14.45) = P(Z ≤ \(\frac{14.45-23.56}{4.55}\)) = P(Z ≤ -2.00)
By Symmetry,
P(Z ≤ -2) = P(Z ≥ 2)
= 0.5 – P(0 ≤ Z ≤ 2)
= 0.5 – 0.4772
= 0.0228
Hence the proportion of fibres with a breaking strength of 14.45 or less is 2.28%

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 3.
The finish times for marathon runners during a race are normally distributed with a mean of 195 minutes and a standard deviation of 25 minutes.
(a) What is the probability that a runner will complete the marathon within 3 hours?
(b) Calculate to the nearest minute, the time by which the first 8% runners have completed the marathon?
(c) What proportion of the runners will complete the marathon between 3 hours and 4 hours?
Solution:
Let X be the normal variate denoting the finish time of the marathon runners.
Given the mean µ = 195 and s.d σ = 25
(a) P(X ≤ 3 hours)
= P(X ≤ 180 minutes)
= P(Z ≤ \(\frac{180-195}{25}\))
= P(Z ≤ -0.6)
By symmetry
P(Z ≤ -0.6)
= P(Z ≤ 0.6)
= 0.5 – P(0 ≤ Z ≤ 0.6)
= 0.5 – 0.2258
= 0.2742
(i.e) Probability of a runner taking less than 3 hours is 0.2742
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q3
(b) Given the probability is 8% = 0.08
P (Z ≥ z) = 0.08
0.5 – P(0 ≤ Z ≤ z) = 0.08
P(0 ≤ Z ≤ z) = 0.42
z = 1.41 (from normal tables)
Hence \(\frac{X-195}{25}\) = -1.41
X = 25 (-1.41) + 195 = 159.75 = 160 minutes
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q3.1
(c) P(3 hours < X < 4 hours)
= P (180 min < X < 240 min)
= P(\(\frac{180-195}{25}\) < Z < \(\frac{240-195}{25}\))
= P(-0.6 < Z < 1.8)
= P (-0.6 < Z < 0) + P (0 < Z < 1.8)
= P(0 < Z < 0.6) + P(0 < Z < 1.8)
= 0.2258 + 0.4641
= 0.6899
Hence the proportion of runners taking between 3 hours and 4 hours is 68.99%
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q3.2

Question 4.
The probability that a driver must stop at any one traffic light is 0.2. There are 15 sets of traffic lights on the journey.
(а) What is the probability that a student must stop at exactly 2 of the 15 sets of traffic lights?
(b) What is the probability that a student will be stopped at 1 or more of the 15 sets of traffic lights?
Solution:
Let X be the binomial random variable denoting the number of traffic lights.
Given n = 15, p =0.2, q = 0.8
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q4

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 5.
A radioactive source emits 4 particles on average during a five-second period.
(а) Calculate the probability that it emits 3 particles during a five-second period.
(b) Find the probability that it emits at least one particle during a 5 second period.
(c) During a 10 second period, what is the probability that 6 particles are emitted?
Solution:
Let X be the Poisson variable denoting the number of particles emitted.
Given the mean λ = 4
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q5

Question 6.
Given that X ~ B (n, p) and E(X) = 24, Var(X) = 8, find the values of n and p.
Solution:
We know E(X) = np = 24 and Var(X) = npq = 8
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q6

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems

Question 7.
Given that X ~ N (6, 4), find the values of ‘a’ and ‘b’ such that P (X ≤ a) = 0.6500 and P(X ≤ b) = 0.8200
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q7
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q7.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q7.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Additional Problems III Q7.3

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Students can download 12th Business Maths Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 1.
The probability function of a random variable X is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q1
Evaluate the following probabilities.
(i) P(X ≤ 0)
(ii) P(X < 0)
(iii) P(|X| ≤ 2)
(iv) P(0 ≤ X ≤ 10)
Solution:
(i) P(X ≤ 0) = P (X = 0) + P (X = -2)
\(=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
(ii) P(X < 0) = P (X = – 2) = \(\frac{1}{4}\)
(iii) P(|X| ≤ 2) = P(-2 ≤ X ≤ 2)
= P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{1}{4}\) + 0 + \(\frac{1}{4}\) + 0 + 0
= \(\frac{1}{2}\)
(iv) P(0 ≤ X ≤ 10) = P(X = 0) + P(X = 10) + 0
\(=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

Question 2.
Let X be a random variable with cumulative distribution function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q2
(a) Compute: (i) P(1 ≤ X ≤ 2) and (ii) P(X = 3).
(b) Is X a discrete random variable? Justify your answer.
Solution:
(a) (i) P(1 ≤ X ≤ 2) = F(2) – F(1)
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q2.1
(a) (ii) P(X = 3) = 0. The given random variable is continuous r.v.
Hence the probability for a particular value of X is zero.
(b) X is not discrete since the cumulative distribution function is a continuous function. It is not a step function.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 3.
The p.d.f. of X is defined as
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q3
Find the value of k and also find P(2 ≤ X ≤ 4).
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q3.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q3.2

Question 4.
The probability distribution function of a discrete random variable X is
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q4
where k is some constant.
Find (a) k and (b) P(X > 2).
Solution:
(a) Given X is a discrete random variable.
The probability distribution can be written as
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q4.1
We know that Σp(x) = 1
⇒ 2k + 3k + 4k = 1
⇒ 9k = 1
⇒ k = 1/9
(b) P(X > 2) = P(X = 3) + P(X = 5)
= 3k + 4k
= 7k
= \(\frac{7}{9}\)

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 5.
The probability density function of a continuous random variable X is
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q5
where a and b are some constants.
Find (i) a and b if E(X) = \(\frac{3}{5}\)
(ii) Var(X)
Solution:
Given that X is a continuous random variable and f(x) is density function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q5.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q5.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q5.3
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q5.4

Question 6.
Prove that if E(X) = 0, then V(X) = E(X2).
Solution:
Given E(X) = 0. To show V(X) = E (X2)
We know that Var (X) = E(X2) – [E(X)]2
So if E(X) = 0, Var (X) = E(X2)
From the definition of the variance of X also we can see the result.
Var(X) = Σ[x – E(x)]2 p(x)
If E (X) = 0, then V(X) = Σ x2 p(x) = E(X2)

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 7.
What is the expected value of a game that works as follows: I flip a coin and if tails pay you ₹ 2; if heads pay you ₹ 1. In either case, I also pay you ₹ 50.
Solution:
Let X be the expected value of the game.
The probability distribution is given by,
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q7

Question 8.
Prove that
(i) V(aX) = a2 V(X)
(ii) V(X + b) = V(X)
Solution:
(i) To show V(aX) = a2 V(X)
We know V(X) = E(X2) – [E(X)]2
So V(aX) = [E(a2 X2)] – [E(aX)]2
= a2 E(X2) – [aE(X)]2
= a2 E(X2) – a2 [E(X)]2
= a2 {{E(X2) – [E(X)]2}
= a2 V(X)

(ii) V(X + b) = V(X)
LHS = V(X + b) = E[(X + b)2] – {E(X + b)}2
= E [X2 + 2bX + b2] – [E(X) + b]2
= E(X2) + 2bE(X) + b2 – [(E(X))2 + b2 + 2bE(X)]
= E(X2) + 2bE(X) + b2 – [E(X)]2 – b2 – 2bE(X)
= E(X2) – [E(X)]2
= V(X)
= RHS

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 9.
Consider a random variable X with p.d.f
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q9
Find E(X) and V(3X – 2).
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q9.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q9.2

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 10.
The time to failure in thousands of hours of an important piece of electronic equipment used in a manufactured DVD player has the density function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q10
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable denoting the life of the piece of equipment.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Q10.1
Thus the expected life of the piece of equipment is \(\frac{1}{2}\) hrs (in thousands).

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Students can download 12th Business Maths Chapter 5 Numerical Methods Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 1.
If f(x) = eax then show that f(0), ∆f(0), ∆2 f(0) are in G.P
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q1

Question 2.
Prove that
(i) (1 + ∆) (1 – ∇) = 1
(ii) ∆∇ = ∆ – ∇
(iii) E∇ = ∆ – ∇E
Solution:
(i) To show (1 + ∆) (1 – ∇) = 1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q2.1
Hence proved.

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 3.
A second degree polynomial passes through the point (1, -1) (2, -1) (3, 1) (4, 5). Find the polynomial.
Solution:
Given values can be tabulated as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q3
We have to find a second-degree polynomial.
We use Newton’s forward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q3.1
The difference table is given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q3.2
(or) y = x2 – 3x + 1 is the required second-degree polynomial which passes through the given points

Question 4.
Find the missing figures in the following table
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q4
Solution:
Since only four values of y are given, the polynomial which fits the data is of degree three.
Hence fourth differences are zero.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q4.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q4.2
Thus the missing figures are 14.25 and 23.5

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 5.
Find f(0.5) if f(-1) = 202, f(0) = 175, f(1) = 82 and f(2) = 55
Solution:
The given data can be written as
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5
We have to find y when x = 0.5.
We use Newton’s forward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5.1
The difference table is
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5.2
Using these values we get,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5.3
Thus the value of f(0.5) = 128.5

Question 6.
From the following data find y at x = 43 and x = 84
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6
Solution:
We have to find the value of y at (a) x = 43 and (b) x = 84
(a) x = 43.
The value of y is required at the beginning of the table.
So we use Newton’s forward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.1
Here x = 43 , x0 = 40, h = 10
So 43 = 40 + 10n
n = 0.3
The difference table is given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.2
y = 184 + 6 – 0.21
y = 189.79

(b) x = 84.
The value of y is required at the end of the table.
So we use Newton’s backward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.3
We use the back difference values from the table
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.4
Hence the value of y at x = 43 is 189.75 and the value of y at x = 84 is 286.96

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 7.
The area A of a circle of diameter ‘d’ is given for the following values
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7
Find the approximate values for the areas of circles of diameter 82 and 91 respectively.
Solution:
Let diameter be x and area be y
We have to find value of y when (a) x = 82 and (b) x = 91
We first find the difference as given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7.3
Hence the area of a circle when the diameter is 82 is 5281
area of a circle when the diameter is 91 is 6504.

Question 8.
If u0 = 560, u1 = 556, u2 = 520, u4 = 385, show that u3 = 465
Solution:
Given u0 = 560, u1 = 556, u2 = 520, u4 = 385
Since only four values are given,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q8

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 9.
From the following table obtain a polynomial of degree y in x
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9
Solution:
To find a polynomial y = f(x)
Here x0 = 1, h = 1
x = x0 + nh
x = 1 + n(1)
n = x – 1
We find the forward differences as below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9.1
Using Newton’s forward interpolation formula,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9.3

Question 10.
Using Lagrange’s interpolation formula find a polynomial which passes through the points (0, -12), (1, 0), (3, 6) and (4, 12).
Solution:
The given values are
x0 = 0, y0 = -12
x1 = 1, y1 = 0
x2 = 3, y2 = 6
x3 = 4, y3 = 12
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q10
By Lagrange’s interpolaiton formula,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q10.1
⇒ y = (x – 1) (x – 3) (x – 4) – x (x – 1) (x – 4) + x (x – 1) (x – 3)
⇒ y = (x – 1) (x – 4)[(x – 3) – x] + x (x – 1) (x – 3)
⇒ y = (x – 1) (x – 4) (-3) + x (x – 1) (x – 3)
⇒ y = (x – 1) [-3x + 12 + x2 – 3x]
⇒ y = (x – 1) (x2 – 6x + 12)
⇒ y = x3 – 6x2 + 12x – x2 + 6x – 12
y = x3 – 7x2 + 18x – 12 is the required polynomial which passes through the given points

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Students can download 12th Business Maths Chapter 6 Random Variable and Mathematical Expectation Ex 6.3 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Choose the correct answer:

Question 1.
The value which is obtained by multiplying possible values of a random variable with a probability of occurrence and is equal to the weighted average is called ______
(a) Discrete value
(b) Weighted value
(c) Expected value
(d) Cumulative value
Answer:
(c) Expected value

Question 2.
Demand of products per day for three days are 21, 19, 22 units and their respective probabilities are 0.29, 0.40, 0.35. Profit per unit is 0.50 paisa then expected profits for three days are _______
(a) 21, 19, 22
(b) 21.5, 19.5, 22.5
(c) 0.29, 0.40, 0.35
(d) 3.045, 3.8, 3.85
Answer:
(d) 3.045, 3.8, 3.85
Hint:
The expected profit for three days are as follows:
For day 1 ⇒ 21 × 0.29 × 0.5 = 3.045
For day 2 ⇒ 19 × 0.4 × 0.5 = 3.8
For day 3 ⇒ 22 × 0.35 × 0.5 = 3.85

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 3.
Probability which explains x is equal to or less than particular value is classified as _______
(a) discrete probability
(b) cumulative probability
(c) marginal probability
(d) continuous probability
Answer:
(b) cumulative probability

Question 4.
Given E(X) = 5 and E(Y) = -2, then E(X – Y) is _______
(a) 3
(b) 5
(c) 7
(d) -2
Answer:
(c) 7
Hint:
E(X – Y) = E(X) – E (Y) = 5 – (-2) = 7 .

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 5.
A variable that can assume any possible value between two points is called _______
(a) discrete random variable
(b) continuous random variable
(c) discrete sample space
(d) random variable
Answer:
(b) continuous random variable

Question 6.
A formula or equation used to represent the probability distribution of a continuous random variable is called ______
(a) probability distribution
(b) distribution function
(c) probability density function
(d) mathematical expectation
Answer:
(c) probability density function

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 7.
If X is a discrete random variable and p(x) is the probability of X, then the expected value of this random variable is equal to ________
(a) Σ f(x)
(b) Σ[x + f(x)]
(c) Σ f(x) + x
(d) ΣxP(x)
Answer:
(d) ΣxP(x)

Question 8.
Which of the following is not possible in probability distribution?
(a) Σ p(x) ≥ 0
(b) Σ p(x) = 1
(c) Σ xp(x) = 2
(d) p(x) = -0.5
Answer:
(d) p(x) = -0.5
Hint:
p(x) = -0.5 is not possible since the probability cannot be negative.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 9.
If c is a constant, then E(c) is ________
(a) 0
(b) 1
(c) c f(c)
(d) c
Answer:
(d) c

Question 10.
A discrete probability distribution may be represented by ______
(a) table
(b) graph
(c) mathematical equation
(d) all of these
Answer:
(d) all of these

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 11.
A probability density function may be represented by ________
(a) table
(b) graph
(c) mathematical equation
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 12.
If c is a constant in a continuous probability distribution, then p(x = c) is always equal to ______
(a) zero
(b) one
(c) negative
(d) does not exist
Answer:
(a) zero

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 13.
E[X – E(X)] is equal to ______
(a) E(X)
(b) V[X]
(c) 0
(d) E(X) – X
Answer:
(c) 0
Hint:
E[X – E(X)] = E(X) – E [E(X)] = E(X) – E(X) = 0

Question 14.
E[X – E(X)]2 is ______
(a) E(X)
(b) E(X2)
(c) V(X)
(d) S.D (X)
Answer:
(c) V(X)
Hint:
E[X – E(X)]2 = E[X2 – 2XE(X) + E(X)2]
= E[X2] – 2[E(X)]2 + [E(X)]2
= E[X2] – [E(X)]2
= Var (X)

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 15.
If the random variable takes negative values, then the negative values will have ________
(a) positive probabilities
(b) negative probabilities
(c) constant probabilities
(d) difficult to tell
Answer:
(a) positive probabilities

Question 16.
If we have f(x) = 2x, 0 ≤ x ≤ 1, then f(x) is a ________
(a) probability distribution
(b) probability density function
(c) distribution function
(d) continuous random variable
Answer:
(b) probability density function

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 17.
A discrete probability function p(x) is always ________
(a) non-negative
(b) negative
(c) one
(d) zero
Answer:
(a) non-negative

Question 18.
In a discrete probability distribution, the sum of all the probabilities is always equal to ______
(a) zero
(b) one
(c) minimum
(d) maximum
Answer:
(b) one

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 19.
The expected value of a random variable is equal to its _______
(a) variance
(b) standard deviation
(c) mean
(d) covariance
Answer:
(c) mean

Question 20.
A discrete probability function p(x) is always non-negative and always lies between ________
(a) 0 and ∞
(b) 0 and 1
(c) -1 and +1
(d) -∞ and +∞
Answer:
(b) 0 and 1

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 21.
The probability density function p(x) cannot exceed _______
(a) zero
(b) one
(c) mean
(d) infinity
Answer:
(b) one

Question 22.
The height of persons in a country is a random variable of the type ________
(a) discrete random variable
(b) continuous random variable
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(b) continuous random variable

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 23.
The distribution function F(x) is equal to ______
(a) P(X = x)
(b) P(X ≤ x)
(c) P(X ≥ x)
(d) all of these
Answer:
(b) P(X ≤ x)

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Students can download 12th Business Maths Chapter 5 Numerical Methods Ex 5.3 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Choose the correct answer.

Question 1.
2 y0 = ______
(a) y2 – 2y1 + y0
(b) y2 + 2y1 – y0
(c) y2 + 2y1 + y0
(d) y2 – y1 + 2y0
Answer:
(a) y2 – 2y1 + y0
Hint:
2 y0 = ∆(∆y0) = ∆(y1 – y0) = ∆y1 – ∆y0
= (y2 – y1) – (y1 – y0)
= y2 – 2y1 + y0

Question 2.
∆f(x) = _______
(a) f(x + h)
(b) f(x) – f(x + h)
(c) f(x + h) – f(x)
(d) f(x) – f(x – h)
Answer:
(c) f(x + h) – f(x)
Hint:
∆f(x) = f(x + h) – f(x)

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Question 3.
E = ______
(a) 1 + ∆
(b) 1 – ∆
(c) 1 + ∇
(d) 1 – ∇
Answer:
(a) 1 + ∆
Hint:
E = 1 + ∆

Question 4.
If h = 1, then ∆(x2) = ________
(a) 2x
(b) 2x – 1
(c) 2x + 1
(d) 1
Answer:
(c) 2x + 1
Hint:
∆(x2) = (x + h)2 – x2 = (x + 1)2 – x2 = 2x + 1

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Question 5.
If c is a constant then ∆c = ______
(a) c
(b) ∆
(c) ∆2
(d) 0
Answer:
(d) 0

Question 6.
If m and n are positive integers then ∆mn f(x) = _______
(a) ∆m+n f(x)
(b) ∆m f(x)
(c) ∆n f(x)
(d) ∆m-n f(x)
Answer:
(a) ∆m+n f(x)

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Question 7.
If ‘n’ is a positive integer ∆n [∆-n f(x)] _______
(a) f(2x)
(b) f(x + h)
(c) f(x)
(d) ∆ f(2x)
Answer:
(c) f(x)

Question 8.
E f(x) = _______
(a) f(x – h)
(b) f(x)
(c) f(x + h)
(d) f(x + 2h)
Answer:
(c) f(x + h)

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Question 9.
∇ = _______
(a) 1 + E
(b) 1 – E
(c) 1 – E-1
(d) 1 + E-1
Answer:
(c) 1 – E-1

Question 10.
∇ f(a) = ______
(a) f(a) + f(a – h)
(b) f(a) – f(a + h)
(c) f(a) – f(a – h)
(d) f(a)
Answer:
(c) f(a) – f(a – h)

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Question 11.
For the given points (x0 , y0) and (x1, y1) the Lagrange’s formula is ______
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3 Q11
Answer:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3 Q11.1

Question 12.
Lagrange’s interpolation formula can be used for ________
(a) equal intervals only
(b) unequal intervals only
(c) both equal and unequal intervals
(d) none of these
Answer:
(c) both equal and unequal intervals

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3

Question 13.
If f(x) = x2 + 2x + 2 and the interval of differencing is unity then ∆ f(x) _______
(a) 2x – 3
(b) 2x + 3
(c) x + 3
(d) x – 3
Answer:
(b) 2x + 3
Hint:
f(x) = 2x2 + 2x + 2
h = 1
∆f(x) = (x + 1)2 + 2(x + 1) + 2 – x2 – 2x – 2
= x2 + 2x + 1 +2x + 2 + 2 – x2 – 2x – 2
= 2x + 3

Question 14.
For the given data find the value of ∆3 y0 is _________
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3 Q14
(a) 1
(b) 0
(c) 2
(d) -1
Answer:
(b) 0
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.3 Q14.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Students can download 12th Business Maths Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 1.
Find the expected value for the random variable of an unbiased die.
Solution:
Let X denote the number on the top side of the unbiased die.
The probability mass function is given by the following table.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q1
The expected value for the random variable X is E(X) = \(\sum_{x} x P_{x}(x)\)
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q1.1

Question 2.
Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q2
Solution:
Expected value of X, E(X) = \(\sum_{x} x P_{x}(x)\)
E(X) = (0 × 0.2) + (1 × 0.1) + (2 × 0.4) + (3 × 0.3)
= 0 + 0. 1 + 0.8 + 0.9
= 1.8

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 3.
The following table is describing the probability mass function of the random variable X.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q3
Find the standard deviation of x.
Solution:
The standard deviation of X, σx is given by σx = √Var[X]
Now Var(X) = E(X2) – [E(X)]2
From the given table,
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q3.1
Hence the standard deviation of X is 2.15

Question 4.
Let X be a continuous random variable with probability density function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q4
Find the expected value of X.
Solution:
The expected value of the random variable is given by E(X) = \(\int_{-\infty}^{\infty} x f(x) d x\)
According to the problem we have,
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q4.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 5.
Let X be a continuous random variable with probability density function
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q5
Find the mean and variance of X.
Solution:
Given that X is a continuous random variable.
The mean of X is the expected value of X.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q5.1

Question 6.
In investment, a man can make a profit of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38. Find the expected gain.
Solution:
Let X be the random variable which denotes the gain in the investment. It is given that X takes the value 5000 with probability 0.62 and -8000 with a probability 0.38.
(Note that we take -8000 since it is a loss)
The probability distribution is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q6
E(X) = (0.38) (-8000) + (0.62) (5000)
= -3040 + 3100
= 60
Hence the expected gain is ₹ 60

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 7.
What are the properties of Mathematical expectation?
Solution:
The properties of Mathematical expectation are as follows:
(i) E(a) = a, where ‘a’ is a constant
(ii) Addition theorem: For two r.v’s X and Y, E(X + Y) = E(X) + E(Y)
(iii) Multiplication theorem: E(XY) = E(X) E(Y)
(iv) E(aX) = aE(X), where ‘a’ is a constant
(v) For constants a and b, E(aX + b) = a E(X) + b

Question 8.
What do you understand by Mathematical expectation?
Solution:
The expected value of a random variable gives a measure of the center of the distribution of the variable. In other words, E(X) is the long-term average value of the variable. The expected value is calculated as a weighted average of the values of a random variable in a particular experiment. The weights are the probabilities. The mean of the random variable X is µX = E(X).

Question 9.
How do you define variance in terms of Mathematical expectation?
Solution:
Let X be a random variable. Let E(X) denote the expectation of X.
Then the variance is defined in terms of the mathematical expectation as follows.
(a) X is discrete r.v with p.m.f p(x). Then Var(X) = \(\sum[x-\mathrm{E}(\mathrm{X})]^{2} p(x)\)
(b) X is continuous r.v with p.d.f fx(x). Then Var(X) = \(\int_{-\infty}^{\infty}[X-E(X)]^{2} f_{X}(x) d x\)

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 10.
Define Mathematical expectation in terms of a discrete random variable.
Solution:
Let X be a discrete random variable with probability mass function (p.m.f) P(x). Then, its expected value is defined by E(X) = \(\sum_{x} x p(x)\)
In other words, if x1, x2, x3,…… xn are the different values of X, and p(x1), p(x2) …..p(xn) are the corresponding probabilities, then E(X) = x1 p(x1) + x2 p(x2) + x3 p(x3) +… xn p(xn)

Question 11.
State the definition of Mathematical expectation using a continuous random variable.
Solution:
Let X be a continuous random variable with probability density function f(x). Then the expected value of X is
\(\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x\)
If the integral exists, E(X) is the mean of the values, otherwise, we say that the mean does not exist.

Question 12.
In a business venture, a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What are his expected, variance and standard deviation of profit?
Solution:
Let X be the random variable denoting the profit of the business venture.
The probability distribution of X is given as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q12
E(X) = (-1000) (0.6) + (2000) (0.4)
= – 600 + 800
= 200
E(X2) = (-1000)2 (0.6) + (2000)2 (0.4)
= 6,00,000 + 16,00,000
= 22,00,000
V(X) = E(X2) – [E(X)]2
= 22,00,000 – 40000
= 21,60,000
Standard deviation = √Var[X]
= \(\sqrt{2160000}\)
= 1469.69
Thus the expected value of profit is ₹ 200. The variance of profit is ₹ 21,60,000 and the standard deviation of profit is ₹ 1469.69.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 13.
The number of miles an automobile tyre lasts before it reaches a critical point in tread wear can be represented by a p.d.f.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q13
Find the expected number of miles (in thousands) a tyre would last until it reaches the critical tread wear point.
Solution:
Let the continuous random variable X denote the number of miles (in thousands) till an automobile tyre lasts.
The expected value is E(X) = \(\int_{-\infty}^{\infty} x f(x) d x\)
From the problem we have,
\(E(X)=\int_{0}^{\infty}(x) \frac{1}{30} e^{\frac{-x}{30}} d x\)
We use integration by parts to evaluate the integral
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q13.1
Hence the expected number of miles is 30,000.

Question 14.
A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains.
Solution:
Let X be the discrete random variable which denotes the gain of the person.
The probability distribution of X is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q14
(Here, since a coin is tossed the probability is equal for the outcomes head or tail)
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Q14.1
Thus the expectation of his gains is 1 and the variance of his gains is 9.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 15.
Let X be a random variable and Y = 2X + 1. What is the variance of Y if the variance of X is 5?
Solution:
Given X is a random variable and Y = 2X + 1 and Var(X ) = 5
Var (Y) = Var (2X + 1) = (2)2 = 4
Var X = 4(5) = 20

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2

Students can download 12th Business Maths Chapter 5 Numerical Methods Ex 5.2 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2

Question 1.
Using graphic method, find the value of y when x = 48 from the following data:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q1
Solution:
The given points are (40, 6.2), (50, 7.2) (60, 9.1) and (70, 12).
We plot the points on a graph with suitable scale
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q1.1
The value of y when x = 48 is 6.8

Question 2.
The following data relates to indirect labour expenses and the level of output
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q2
Estimate the expenses at a level of output of 350 units, by using the graphic method.
Solution:
Take the units of output along the x-axis, labour expenses along the y-axis.
The points to be plotted are (200, 2500), (300, 2800) (400, 3100), (640, 3820), (540, 3220), (580, 3640)
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q2.1
From the graph, the expenses at a level of output of 350 units are ₹ 2940.

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2

Question 3.
Using Newton’s forward interpolation formula find the cubic polynomial.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q3
Solution:
Newton’s forward interpolation formula is
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q3.1
y(x) = 1 + x – (x2 – x) + 2[x3 – 3x2 + 2x]
y(x) = 1 + x – x2 + x + 2x3 – 6x2 + 4x
f(x) = y = 2x3 – 7x2 + 6x + 1 is the required cubic polynomial

Question 4.
The population of a city in a census taken once in 10 years is given below.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q4
Estimate the population in the year 1955.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q4.1
Let ‘x’ denote the year of the census. Let ‘y’ represent the population in lakhs. We have to find the population in the year 1955 (i.e) the value of y when x = 1955. Since the value of y is required near the beginning of the table, we use Newton’s forward interpolation formula.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q4.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q4.3
Thus the estimated population in the year 1955 is 36.784 lakhs

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2

Question 5.
In an examination the number of candidates who secured marks between certain intervals was as follows:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q5
Estimate the number of candidates whose marks are less than 70.
Solution:
Let x be the marks and y be the number of candidates. The given class intervals are not continuous. So we make them continuous and find the cumulative frequency.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q5.1
The difference table is as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q5.2
Since the required value of y is near the end of the table, we use Newton’s backward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q5.3
Hence the estimated value of the number of candidates whose marks are less than 70 is 197

Question 6.
Find the value of f(x) when x = 32 from the following table
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q6
Solution:
To find y = f(x) when x = 32 from the given table. Since the required value of y is near the beginning of the table, we use Newton’s forward interpolation formula.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q6.1
The difference table is as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q6.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q6.3
Hence the value of f(x) when x = 32 is 15.45

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2

Question 7.
The following data gives the melting point of an alloy of lead and zinc where ‘t’ is the temperature in degree c and P is the percentage of lead in the alloy
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q7
Find the melting point of the alloy containing 84 per cent lead.
Solution:
To find T when P = 84. The required value of T is near the end of the table. So we use Newton’s backward interpolation formula.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q7.1
⇒ 90 + n(10) = 84
⇒ n = \(\frac{84-90}{10}=\frac{-6}{10}=-0.6\)
The difference table is given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q7.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q7.3
Hence the melting point of the alloy containing 84 per cent lead is 286.9°C

Question 8.
Find f(2.8) from the following table.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q8
Solution:
To find y = f(x) at x = 2.8 from the given table. We use Newton’s backward interpolation formula since the required value is near the end of the table.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q8.1
2.8 = 3 + n(1)
n = 2.8 – 3 = -0.2
The difference table given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q8.2
y = 34 – 4.6 – 1.12 – 0.288
y = 27.992
Hence the value of f(x) at x = 2.8 is 27.992

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2

Question 9.
Using interpolation estimate the output of a factory in 1986 from the following data
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q9
Solution:
Let x denote the year and y represent the output. The x values are not equidistant. So we use Lagrange’s formula
x0 = 1974, y0 = 25
x1 = 1978, y1 = 60
x2 = 1982, y2 = 80
x3 = 1990, y3 = 170
For x = 1986 we have to find y value
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q9.1
We find the different values separately and substitute in the formula.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q9.2
y = 6.25 – 60 + 120 + 42.5
y = 108.75
The output of the factory in 1986 is 109 (thousand tonnes)

Question 10.
Use Lagrange’s formula and estimate from the following data the number of workers getting income not exceeding Rs. 26 per month.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q10
Solution:
Let x represent the income per month and y denote the number of workers.
From the given data,
x0 = 15, y0 = 36
x1 = 25, y1 = 40
x2 = 30, y2 = 45
x3 = 35, y3 = 48
We have to find the value of y at x = 26
By Lagrange’s interpolation formula,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q10.1
The different values are given in the table below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q10.2
y = -0.432 + 31.68 + 11.88 – 2.112
y = 41.016
Thus the number of workers getting income not exceeding Rs.26 per month is 42

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2

Question 11.
Using interpolation estimate the business done in 1985 from the following data.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q11
Solution:
Let x denote the year of business and y (in lakhs) denote the amount of business.
From the given data,
x0 = 1982, y0 = 150
x1 = 1983, y1 = 235
x2 = 1984, y2 = 365
x3 = 1986, y3 = 525
We have to find the value of y when x = 1985.
By Lagrange’s interpolation formula,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q11.1
We form a table for the different values
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q11.2
y = 481.25
Thus the business done in the year 1985 is estimated as 481.25 lakhs

Question 12.
Using interpolation, find the value of f(x) when x = 15
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q12
Solution:
We have to find the value of y when x = 15.
From the given data,
x0 = 3, y0 = 42
x1 = 7, y1 = 43
x2 = 11, y2 = 47
x3 = 19, y3 = 60
Since the intervals are unequal, we use the Lagrange’s interpolation formula,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q12.1
The different values are given in the table below.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.2 Q12.2
y = 10.5 – 43 + 70.5 + 15
y = 53
Hence the value of f(x) when x = 15 is 53

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Students can download 12th Business Maths Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 1.
Construct cumulative distribution function for the given probability distribution.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 1
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 2
Thus the cumulative distribution function is given by,
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 3

Question 2.
Let X be a discrete random variable with the following p.m.f.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 4
Find and plot the c.d.f. of X.
Solution:
The probability distribution function can be written as follows:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 5
We obtain
F(3) = P(x ≤ 3) = P(3) = 0.3
F(5) = P(x ≤ 5) = P(3) + P(5) = 0.3 + 0.2 = 0.5
F(8) = P(x ≤ 8) = P(3) + P(5) + P(8) = 0.5 + 0.3 = 0.8
F(10) = P(x ≤ 10) = P(3) + P(5) + P(8) + P(10) = 0.8 + 0.2 = 1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 6
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 7

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 3.
The discrete random variable X has the following probability function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 8
where k is a constant. Show that k = \(\frac{1}{18}\)
Solution:
The given probability function can be written as follows:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 9
Since the condition of probability mass function is \(\sum_{i=1}^{\infty} p\left(x_{i}\right)=1\), we have
2k + 4k + 6k + 6k = 1
⇒ 18k = 1
⇒ k = \(\frac{1}{18}\)

Question 4.
The discrete random variable X has the probability function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 10
Show that k = 0.1
Solution:
\(\sum_{i=1}^{\infty} p\left(x_{i}\right)=1\)
gives k + 2k + 3k + 4k = 1
⇒ 10k = 1
⇒ k = \(\frac{1}{10}\) = 0.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 5.
Two coins are tossed simultaneously. Getting a head is termed as a success. Find the probability distribution of the number of successes.
Solution:
Let X be the number of observed heads. The sample space S = {(H H), (H T), (T H), (T T)}
X takes the values 2, 1, 1, 0. Hence the probability distribution of the number of successes is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 11

Question 6.
A random variable X has the following probability function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 12
(i) Find k
(ii) Evaluate P( X < 6), P(X ≥ 6) and P(0 < X < 5)
(iii) If P(X ≤ x) > \(\frac{1}{2}\), then find the minimum value of x.
Solution:
(i) Since the condition of probability mass function
\(\sum_{i=1}^{\infty} p\left(x_{i}\right)=1\)
\(\sum_{i=0}^{7} p\left(x_{i}\right)=1\)
0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
⇒ 10k2 + 9k – 1 = 0
⇒ (10k – 1) (k + 1) = 0
⇒ k = \(\frac{1}{10}\) and k = -1
Since p(x) cannot be negative, k = -1 is not applicable. Hence k = \(\frac{1}{10}\)

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

(ii) P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P (X = 3) + P(X = 4) + P (X = 5)
= 0 + k + 2k + 2k + 3k + k2
= 8k + k2
= 8(\(\frac{1}{10}\)) + (\(\frac{1}{10}\))2 (∵ k = \(\frac{1}{10}\))
= \(\frac{8}{10}+\frac{1}{100}=\frac{81}{100}\)
Method 2:
P(X < 6) = 1 – P(X ≥ 6)
= 1 – [P(X = 6) + P (X = 7)]
= 1 – [2k2 + 7k2 + k]
= 1 – 9k2 – k
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 13
P(0 < X < 5) = P(X = 1) + P(X = 2) + P (X = 3) + P(X = 4)
= k + 2k + 2k + 3k = 8k
= \(\frac{8}{10}\)

(iii) P(X ≤ x) > \(\frac{1}{2}\)
To find the minimum value of x, let us construct the cumulative distribution function of X.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 14
From the table we see that
P(X ≤ 0) = 0
P(X ≤ 1) = \(\frac{1}{10}\) = 0.1
P(X ≤ 2) = \(\frac{3}{10}\) = 0.3
P(X ≤ 3) = \(\frac{5}{10}\) = 0.5
P(X ≤ 4) = \(\frac{8}{10}\) = 0.8
So x = 4

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 7.
The distribution of a continuous random variable X in range (-3, 3) is given by p.d.f.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 15
Verify that the area under the curve is unity.
Solution:
We know that area under a curve f(x) between x = a and x = b is \(\int_{a}^{b} f(x) d x\)
Here the area is given by \(\int_{-3}^{3} f(x) d x\)
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 16
Thus the area under the curve is unity.

Question 8.
A continuous random variable X has the following distribution function:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 17
Find (i) k and (ii) the probability density function.
Solution:
We have \(\frac{d}{d x}\)F(x) = f(x) ≥ 0, where F(x) is the distribution function and f(x) is the probability density function.
Here F(x) = 0 for x ≤ 1
f(x) = 0 for x ≤ 1
Again F(x) = 1 for x > 3
f(x) = \(\frac{d}{d x}\) (1) = 0 for x > 3
In 1 < x ≤ 3, F(x) = k(x – 1)4
f(x) = \(\frac{d}{d x}\) (k(x – 1)4) = 4k(x – 1)3
(i) We know that \(\int_{-\infty}^{\infty} f(x) d x\) = 1
This gives \(\int_{1}^{3} 4 k(x-1)^{3} d x=1\)
\(\left[k(x-1)^{4}\right]_{1}^{3}=1\)
k[16 – 0] = 1
k = \(\frac{1}{16}\)
(ii) The probability density function is
f(x) = \(\frac{1}{4}(x-1)^{3}\), 1 < x ≤ 3
= 0, otherwise

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 9.
The length of time (in minutes) that a certain person speaks on the telephone is found to be a random phenomenon, with a probability function specified by the probability density function f(x) as
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 18
(a) Find the value of A that makes f(x) a p.d.f.
(b) What is the probability that the number of minutes that person will talk over the phone is (i) more than 10 minutes, (ii) less than 5 minutes and (iii) between 5 and 10 minutes.
Solution:
(a) For f(x) to be a p.d.f we must have \(\int_{-\infty}^{\infty} f(x) d x=1\)
According to the problem,
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 19
(b) (i) The probability that the number of minutes that person will talk is more than 10 minutes is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 20
(b) (ii) The probability that the number of minutes is less than 5 is given by 5
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 21
(b) (iii) The probability that the number of minutes that person will talk is between 5 and 10 minutes is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 22

Question 10.
Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 23
(a) Is the distribution function continuous? If so, give its probability density function?
(b) What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes and (iii) between 1 and 3 minutes?
Solution:
(a) Yes. The distribution function is continuous. It is defined for all values of -∞ < x < ∞.
We know that \(\frac{d}{d x}\) F(x) = f(x). So we differentiate the given distribution function in the intervals where it is defined to find the density function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 24
Thus the probability density function is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 25
(b) (i) The probability that a person will have to wait for more than 3 minutes is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 26
Method 2:
P (X ≥ 3) = 1 – P(X ≤ 3) = 1 – F(3)
From the question F (3) = \(\frac{3}{4}\)
So P (X ≥ 3) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
(b) (ii) The probability that a person will have to Wait for less than 3 minutes is
P(X ≤ 3) = F(3) = \(\frac{3}{4}\)
(b) (iii) Probability that the person will have to wait between 1 and 3 minutes is
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 27

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 11.
Define the random variable.
Solution:
A Random variable is a set of possible values from a random experiment. The set of possible values is called the sample space. A random variable (r.v) is denoted by a capital letter such as X, Y and Z etc. If X and Y are r.v’s then X + Y is also an r.v.

Question 12.
Explain what are the types of a random variable?
Solution:
Random variables are classified into two types namely discrete and continuous random variables.
Discrete random variable: A discrete random variable has a finite number of possible values or an infinite sequence of countable real numbers.
Examples:

  1. X denotes the number of hits when trying 20 free throws.
  2. X denotes the number of customers who arrive at the bank from 8.30 – 9.30 AM Mon – Fri.
  3. X denotes the number of balls sold during a week in a particular shop.

Continuous random variable:
A continuous random variable takes all values in an interval of real numbers.
Examples:

  1. X denotes the time it takes for a bulb to bum out.
  2. X denotes the weight of a truck in a truck – weighing station.
  3. X denotes the amount of water in a bottle.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 13.
Define discrete random variable.
Solution:
A variable which can assume a finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable.
Examples:

  1. Marks obtained in an exam.
  2. Number of chocolates in a box.
  3. Number of phone calls during a day.
  4. Number of TV sets sold during a month by a dealer.

Question 14.
What do you understand by continuous random variable?
Solution:
A random variable which can take on any value (integral as well as fraction) in the interval is called a continuous random variable.
Examples:

  1. The speed of a train.
  2. Electricity consumption in kilowatt-hours.
  3. Height of people in a population
  4. Weight of students in a class.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 15.
Describe what is meant by a random variable.
Solution:
A Random variable is a set of possible values from a random experiment. The set of possible values is called the sample space. A random variable (r.v) is denoted by a capital letter such as X, Y and Z etc. If X and Y are r.v’s then X + Y is also an r.v.

Question 16.
Distinguish between discrete and continuous random variable.
Solution:

Discrete Variable Continuous Variable
1. A variable which can take only certain values. 1. A variable which can take any value in a particular limit.
2. The value of the variables can increase incomplete numbers. 2. Its value increases infractions but not in jumps.
3. Example: Number of students who opt for commerce in class 11, say 30, 35, 40, 45, and 50. 3. Example: Height, Weight and age of family members: 50.5 kg, 30 kg, 42.8 kg and 18.6 kg.
4. Binomial, Poisson, Hypergeometric probability distributions come under this category. 4. Normal, student’s t and chi-square distribution come under this category.

Question 17.
Explain the distribution function of a random variable.
Solution:
The discrete cumulative distribution function or distribution function of a real-valued discrete random variable X, which takes the countable number of points x1, x2,….. with corresponding probabilities p(x1), p(x2),…. is defined by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 28
If X is a continuous random variable with probability density function fx(x), then the function FX(x) is defined by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 29
is called the distribution function of the continuous random variable.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 18.
Explain the terms
(i) Probability mass function
(ii) Probability density function and
(iii) Probability distribution function.
Solution:
(i) Probability mass function: A probability mass function (p.m.f) is a function that gives the probability that a discrete random variable is exactly equal to some value. This is the means of defining a discrete probability distribution. Let X be a random variable with values x1, x2,…. xn….. Then the p.m.f is defined by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 30
(ii) Probability density function: A probability density function (p.d.f) or density of a continuous random variable, is a function whose value at any given sample (or point) in the sample space gives the probability of the r.v. falling within the range of values. This probability is given by the area under the density function. The probability that an r.v. X takes a value in the interval [a, b] is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 31
(iii) Probability distribution function: A probability distribution is a list of all of the possible outcomes of a random variable along with their corresponding probability values. When the r.v is discrete, the distribution function is called a probability mass function (p.m.f), and when the r.v is continuous, the distribution function is called a probability density function (p.d.f)
Examples:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 32

Question 19.
What are the properties of
(i) discrete random variable and
(ii) continuous random variable?
Solution:
Discrete Random Variable:

  • A variable which can take only certain values.
  • The value of the variables can increase incomplete numbers
  • Binomial, Poisson, Hypergeometric probability distributions come under this category
  • Example: Number of students who opt for commerce in class 11, say 30, 35, 40, 45, and 50.

Continuous random variable:

  • A variable which can take any value in a particular limit.
  • Its value increases infractions but not in jumps.
  • Normal, student’s t and chi-square distribution come under this category.
  • Example: Height, Weight and age of family members: 50.5 kg, 30 kg, 42.8 kg and 18.6 kg.

Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 20.
State the properties of the distribution function.
Solution:

  • Property 1: The distribution function F is increasing, (i.e) if x ≤ y, then F(x) ≤ F(y)
  • Property 2: F(x) is continuous from right, (i.e) for each x ∈ R, F (x+) = F (x)
  • Property 3: F (∞) = 1
  • Property 4: F (-∞) = 0
  • Property 5: F'(x) = f(x)
  • Property 6: P(a ≤ X ≤ b) = F(b) – F(a)

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Students can download 12th Business Maths Chapter 5 Numerical Methods Ex 5.1 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 1.
Evaluate (log ax)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q1

Question 2.
If y = x3 – x2 + x – 1 calculate the values of y for x = 0, 1, 2, 3, 4, 5 and form the forward differences table.
Solution:
Given y = x3 – x2 + x – 1
When
x = 0, y = 0 – 0 + 0 – 1 = -1
x = 1, y = 1 – 1 + 1 – 1 = 0
x = 2, y = 8 – 4 + 2 – 1 = 5
x = 3, y = 27 – 9 + 3 – 1 = 20
x = 4, y = 64 – 16 + 4 – 1 = 51
x = 5, y = 125 – 25 + 5 – 1 = 104
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q2

Question 3.
If h = 1 then prove that (E-1 ∆) x3 = 3x2 – 3x + 1.
Solution:
h = 1 To prove (E-1 ∆) x3 = 3×2 – 3x + 1
L.H.S = (E-1 ∆) x3 = E-1 (∆x3)
= E-1[(x + h)3 – x3]
= E-1( x + h)3 – E-1(x3)
= (x – h + h)3 – (x – h)3
= x3 – (x – h)3
But given h = 1
So(E-1 ∆) x3 = x3 – (x – 1)3
= x3 – [x3 – 3x2 + 3x – 1]
= 3x2 – 3x + 1
= RHS

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 4.
If f(x) = x2 + 3x then show that ∆f(x) = 2x + 4
Solution:
f(x) = x2 + 3 x
∆f(x) = f(x + h) – f(x)
= (x + h)2 + 3(x + h) – x2 – 3x
= x2 + 2xh + h2 + 3x + 3h – x2 – 3x
= 2xh + 3h + h2
Put h = 1, ∆f(x) = 2x + 4

Question 5.
Evaluate \(\Delta\left[\frac{1}{(x+1)(x+2)}\right]\) by taking ‘1’ as the interval of differencing.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q5
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q5.1

Question 6.
Find the missing entry in the following table
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q6
Solution:
Since only four values of y are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q6.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 7.
Following are the population of a district
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q7
Find the population of the year 1911?
Solution:
Since five values are given, the polynomial which fits the data is of degree four.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q7.1
From the given table
y0 = 363, y1 = 391, y2 = 421, y4 = 467 and y5 = 501
501 – 5(467) + 10y3 – 10(421) + 5(391) – 363 = 0
501 – 2335 + 10y3 – 4210 + 1955 – 363 = 0
-501 + 2335 + 4210 – 1955 + 363 = 10y3
10y3 = 4452
y3 = 445.2
Hence the population of the year 1911 is 445 thousand

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 8.
Find the missing entries from the following.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q8
Solution:
Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q8.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q8.2