Category: Class 12

Students can download 12th Business Maths Chapter 7 Probability Distributions Ex 7.4 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 7.4

Choose the correct answer:

Question 1.
Normal distribution was invented by ______
(a) Laplace
(b) De-Moivre
(c) Gauss
(d) all the above
Answer:
(b) De-Moivre

Question 2.
If X ~ N (9, 81) the standard normal variate Z will be ______
(a) Z = \(\frac{x-81}{9}\)
(b) Z = \(\frac{X-9}{81}\)
(c) Z = \(\frac{X-9}{9}\)
(d) Z = \(\frac{9-x}{9}\)
Answer:
(c) Z = \(\frac{X-9}{9}\)
Hint:
µ = 9, σ = 9
Z = \(\frac{X-9}{9}\)

Question 3.
If Z is a standard normal variate, the proportion of items lying between Z = -0.5 and Z = -3.0 is ________
(a) 0.4987
(b) 0.1915
(c) 0.3072
(d) 0.3098
Answer:
(c) 0.3072
Hint:

P(-0.5 < Z < -3)
By symmetry we want P (0.5 < Z < 3)
= P(0 < Z < 3) – P(0 < Z < 0.5)
= 0.49865 – 0.1915
= 0.30715 ~ 0.3072

Question 4.
If X ~ N(µ, σ²), the maximum probability at the point of inflexion of normal distribution is ______
(a) \(\left(\frac{1}{\sqrt{2 \pi}}\right) e^{\frac{1}{2}}\)
(b) \(\left(\frac{1}{\sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)}\)
(c) \(\left(\frac{1}{\sigma \sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)}\)
(d) \(\left(\frac{1}{\sqrt{2 \pi}}\right)\)
Answer:
(c) \(\left(\frac{1}{\sigma \sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)}\)
Hint:

Question 5.
In a parametric distribution the mean is equal to variance is ________
(a) binomial
(b) normal
(c) Poisson
(d) all of the above
Answer:
(c) Poisson

Question 6.
In turning out certain toys in a manufacturing company, the average number of defectives is 1%. The probability that the sample of 100 toys there will be 3 defectives is __________
(a) 0.0613
(b) 0.613
(c) 0.00613
(d) 0.3913
Answer:
(a) 0.0613
Hint:

Question 7.
The parameters of the normal distribution f(x) = \(\frac{1}{\sqrt{72 \pi}} \frac{e^{(-x-10)^{2}}}{72}-\infty<x<\infty\)
(a) (10, 6)
(b) (10, 36)
(c) (6, 10)
(d) (36, 10)
Answer:
(b) (10, 36)
Hint:
Comparing f(x) with p.d.f of normal distribution, µ = 10,
\(\sigma \sqrt{2 \pi}=\sqrt{72 \pi}=\sqrt{36 \times 2 \pi}=6 \sqrt{2 \pi}\)
σ = 6

Question 8.
A manufacturer produces switches and experiences that 2 per cent switches are defective. The probability that in a box of 50 switches, there are at most two defective is:
(a) 2.5 e^-1
(b) e^-1
(c) 2e^-1
(d) none of the above
Answer:
(a) 2.5 e^-1
Hint:

Question 9.
An experiment succeeds twice as often as it fails. The chance that in the next six trials, there shall be at least four successes is _______
(a) \(\frac {240}{729}\)
(b) \(\frac {489}{729}\)
(c) \(\frac {496}{729}\)
(d) \(\frac {251}{729}\)
Answer:
(c) \(\frac {496}{729}\)
Hint:
Let X be the binomial random variable. Given p = 2q
From p + q = 1 ,we get p = \(\frac{2}{3}\), q = \(\frac{1}{3}\)
We want P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)

Question 10.
If for a binomial distribution b(n, p) mean = 4 and variance = \(\frac{4}{3}\), the probability, P(X ≥ 5) is equal to _______
(a) (2/3)⁶
(b) (2/3)⁵ (1/3)
(c) (1/3)⁶
(d) 4(2/3)⁶
Answer:
(d) 4(2/3)⁶
Hint:

Question 11.
The average percentage of failure in a certain examination is 40. The probability that out of a group of 6 candidates atleast 4 passed in the examination are _______
(a) 0.5443
(b) 0.4543
(c) 0.5543
(d) 0.4573
Answer:
(a) 0.5443
Hint:
The percentage of success p = 0.6 ⇒ q = 0.4
P (X ≥ 4) = P (X = 4) + P (X = 5) + P (X = 6)

= 0.31104 + 0.186624 + 0.046656
= 0.54432

Question 12.
Forty per cent of the passengers who fly on a certain route do not check in any luggage. The planes on this route seat 15 passengers. For a full flight, what is the mean of the number of passengers who do not check in any luggage?
(a) 6.00
(b) 6.45
(c) 7.20
(d) 7.50
Answer:
(a) 6.00
Hint:
n = 15, p = 0.4 ⇒ mean (np) = 6

Question 13.
Which of the following statements is/are true regarding the normal distribution curve?
(а) it is a symmetrical and bell-shaped curve
(b) it is asymptotic in that each end approaches the horizontal axis but never reaches it
(c) its mean, median and mode are located at the same point
(d) all of the above statements are true
Answer:
(d) all of the above statements are true

Question 14.
Which of the following cannot generate a Poisson distribution?
(а) The number of telephone calls received in a ten-minute interval
(b) The number of customers arriving at a petrol station
(c) The number of bacteria found in a cubic foot of soil
(d) The number of misprints per page
Answer:
(b) The number of customers arriving at a petrol station

Question 15.
The random variable X is normally distributed with a mean of 70 and a standard deviation of 10. What is the probability that X is between 72 and 84?
(a) 0.683
(b) 0.954
(c) 0.271
(d) 0.340
Answer:
(d) 0.340
Hint:
µ = 70, σ = 10
P(72 < X < 84) = P(\(\frac{72-70}{10}\) < Z < \(\frac{84-70}{10}\))
= P(0.2 < Z < 1.4)
= P(0 < Z < 1.4) – P(0 < Z < 0.2)
= 0.4192 – 0.0793
= 0.3399
= 0.340

Question 16.
The starting annual salaries of newly qualified chartered accountants (CA’s) in South Africa follow a normal distribution with a mean of ₹ 180,000 and a standard deviation of ₹ 10,000. What is the probability that a randomly selected newly qualified CA will earn between ₹ 165,000 and ₹ 175,000.
(a) 0.819
(b) 0.242
(c) 0.286
(d) 0.533
Answer:
(b) 0.242
Hint:

µ = 180,000, σ = 10,000
P(165,000 < X < 175,000)
= P(\(\frac{165,000-180,000}{10,000}\) < Z < \(\frac{175,000-180,000}{10,000}\))
= P(-1.5 < Z < -0.5)
By symmetry of the normal curve,
= P (0.5 < Z < 1.5)
= P (0 < Z < 1.5) – P(0 < Z < 0.5)
= 0.4332 – 0.1915
= 0.2417
= 0.242

Question 17.
In a large statistics class the heights of the students are normally distributed with a mean of 172 cm and a variance of 25 cm. What proportion of students are between 165cm and 181 cm in height?
(a) 0.954
(b) 0.601
(c) 0.718
(d) 0.883
Answer:
(d) 0.883
Hint:

µ = 172, σ² = 25 ⇒ σ = 5
P (165 < X < 181)
= P(\(\frac{165-172}{5}\) < Z < \(\frac{181-172}{5}\))
= P(-1.4 < Z < 1.8)
= P (-1.4 < Z < 0) + P(0 < Z < 1.8)
= P(0 < Z < 1.4) + P(0 < Z < 1.8)
= 0.4192 + 0.4641
= 0.8833

Question 18.
A statistical analysis of long-distance telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What proportion of calls lasts less than 180 seconds?
(a) 0.214
(b) 0.094
(c) 0.933
(d) 0.067
Answer:
(d) 0.067
Hint:

µ = 240, σ = 40
P(X < 180)
= P(Z < \(\frac{180-240}{40}\))
= P(Z < -1.5) = P(Z > 1.5)
= 0.5 – P (0 < Z < 1.5)
= 0.5 – 0.4332
= 0.0668
= 0.067

Question 19.
Cape town is estimated to have 21% of homes whose owners subscribe to the satellite service, DSTV. If a random sample of your home in taken, what is the probability that all four home subscribe to DSTV?
(a) 0.2100
(b) 0.5000
(c) 0.8791
(d) 0.0019
Answer:
(d) 0.0019
Hint:

Question 20.
Using the standard normal table, the sum of the probabilities to the right of z = 2.18 and to the left of z = -1.75 is:
(a) 0.4854
(b) 0.4599
(c) 0.0146
(d) 0.0547
Answer:
(d) 0.0547
Hint:

P(Z > 2.18) + P(Z < -1.75)
= 0.5 – P(0 < Z < 2.18) + P (Z > 1.75)
= 0.5 – P(0 < Z < 2.18) + 0.5 – P(0 < Z < 1.75)
= 0.5 – 0.4854 + 0.5 – 0.4599
= 0.0547

Question 21.
The time until first failure of a brand of inkjet printers is normally distributed with a mean of 1,500 hours and a standard deviation of 200 hours. What proportion of printers fails before 1000 hours?
(a) 0.0062
(b) 0.0668
(c) 0.8413
(d) 0.0228
Answer:
(a) 0.0062
Hint:
µ = 1500, σ = 200
P(X < 1000)
= P(Z < \(\frac{1000-1500}{200}\))
= P(Z < -2.5) = P (Z > 2.5)
= 0.5 – P (0 < Z < 2.5)
= 0.5 – 0.4938
= 0.0062

Question 22.
The weights of newborn human babies are normally distributed with a mean of 3.2 kg and a standard deviation of 1.1 kg. What is the probability that a randomly selected newborn baby weight less than 2.0 kg?
(a) 0.138
(b) 0.428
(c) 0.766
(d) 0.262
Answer:
(a) 0.138
Hint:
µ = 3.2, σ = 1.1
P (X < 2)
= P( Z < \(\frac{2-3.2}{1.1}\))
= P(Z < -1.09) = P(Z > 1.09)
= 0.5 – P(0 < Z < 1.09)
= 0.5 – 0.3621
= 0.138

Question 23.
Monthly expenditure on their credit cards, by credit card holders from a certain bank, follows a normal distribution with a mean of ₹ 1,295.00 and a standard deviation of ₹ 750.00. What proportion of credit card holders spend more than ₹ 1,500.00 on their credit cards per month?
(a) 0.487
(b) 0.394
(c) 0.500
(d) 0.791
Answer:
(b) 0.394
Hint:
µ = 1295, σ = 750
P(X > 1500)
= P(Z > \(\frac{1500-1295}{750}\))
= P (Z > 0.27)
= 0.5 – P (0 < Z < 0.27)
= 0.5 – 0.1064
= 0.3936 ~ 0.394

Question 24.
Let z be a standard normal variable. If the area to the right of z is 0.8413, then the value of z must be:
(a) 1.00
(b) -1.00
(c) 0.00
(d) -0.41
Answer:
(b) -1.00
Hint:

P(Z > -Z) = 0.8413
⇒ P(-z < Z < 0) + 0.5 = 0.8413
P(0 < Z < z) = 0.8413 – 0.5 = 0.3413
from normal tables, z = 1.
The required value is -z = -1

Question 25.
If the area to the left of a value of z (z has a standard normal distribution) is 0.0793, what is the value of z?
(a) -1.41
(b) 1.41.
(c) -2.25
(d) 2.25
Answer:
(a) -1.41
Hint:

P(Z < -z) = 0.0793 By symmetry, P (Z > z) = 0.0793
(i.e) 0.5 – P (0 < Z < z) = 0.0793
P(0 < Z < z) = 0.4207 from normal tables, z = 1.41
Thus the required value is -1.41 (since it is left of Z = 0)

Question 26.
If P(Z > z) = 0.8508 what is the value of z (z has a standard normal distribution)?
(a) -0.48
(b) 0.48
(c) -1.04
(d) -0.21
Answer:
(c) -1.04
Hint:

P (Z > z) = 0.8508
Since the given value is more than we take z to the left of Z = 0 axis.
P (Z > -z) = 0.8508
P(-z < Z < 0) + 0.5 = 0.8508
P(0 < Z < z) = 0.3508
From normal tables, z = 1.04
So required value is -1.04

Question 27.
If P(Z > z) = 0.5832 what is the value of z (z has a standard normal distribution)?
(a) -0.48
(b) 0.48
(c) 1.04
(d) -0.21
Answer:
(d) -0.21
Hint:

P(0 < Z < z) = 0.5832 – 0.5 = 0.0832
From tables, z = 0.21
Since z is to the left of Z = 0, the required value is -0.21

Question 28.
In a binomial distribution, the probability of success is twice as that of failure. Then out of 4 trials, the probability of no success is _______
(a) \(\frac {16}{81}\)
(b) \(\frac {1}{16}\)
(c) \(\frac {2}{27}\)
(d) \(\frac {1}{81}\)
Answer:
(d) \(\frac {1}{81}\)
Hint:

Students can download 12th Business Maths Chapter 8 Sampling Techniques and Statistical Inference 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 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

Question 1.
Explain the types of sampling.
Solution:
The different types of sampling are

• simple random sampling
• Stratified random sampling and
• Systematic sampling

(i) In simple random sampling, every item of the population has an equal chance for being selected. The sampling can be done with replacement (or) without replacement. A random sampling from a finite population with replacement is equivalent to sampling from an infinite population without replacement. This technique will give useful results only if the population is homogeneous. The following are some of the methods of selecting a random sample.

(a) Use of an unbiased die or coin: If we have to choose between two alternatives, a coin is tossed and depending on the head or tail course of action is taken. A die can be employed if there are six different alternatives.

(b) Lottery sampling: Here a random sample is selected by identifying each element of the population by means of a card of a pack of uniform cards or (by writing the number on pieces of paper) and to select a required number of cards after thorough mixing of the cards.

(c) Random numbers: Random numbers are formed of ‘random digits’ and arranged in the form of a table having a number of rows and columns. Tippett’s numbers form one such table wherein 40,000 digits were selected at random from census reports and combined by groups of four into 10,000 numbers.

(ii) In stratified random sampling, a population of units is divided into L sub-populations of N₁, N₂, …… N_L. The sub-populations being non-overlapping and mutually exhaustive so that N = N₁ + N₂ + …….. + N_L. Each sub-populations is known as a stratum. If we select n₁, n₂, ……. n_l items, respectively, from these strata, we get a stratified sample. If a simple random sample is taken from each stratum, the whole procedure is referred to as stratified random sampling.

(iii) Systematic sampling is a form of restricted random selection which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in serial order and every ith element, starting from any of the first items is chosen. For example, suppose we require a 5% sample of students from a college where there are 2000 students, we select a random number from 1 to 20. If it is 12, then our sample consists of students with numbers 12, 32, 52, 72, …… 1992.

Question 2.
Write a short note on sampling distribution and standard error.
Solution:
Sampling distribution: Sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.
For instance if we draw a sample of size n from a given finite population of size N, then the total number of possible samples is \(^{\mathrm{N}} C_{n}=\frac{\mathrm{N} !}{n !(N-n) !}=k\) (say). For each of these k samples we can compute some statistic, t = t(x₁, x₂, x₃ ,… x_n), in particular the mean \(\bar{X}\) the variance S², etc., is given below

The set of the values of the statistic so obtained, one for each sample constitutes the sampling distribution of the statistic.

Standard Error:
The standard deviation of the sampling distribution of a statistic is known as its Standard Error abbreviated as S.E. The Standard Errors (S.E.) of some of the well-known statistics, for large samples, are given below, where n is the sample size, σ² is the population variance.

Question 3.
Explain the procedures of testing of hypothesis.
Solution:
Hypothesis testing addresses the important question of how to choose among alternative propositions while controlling and minimizing the risk of wrong decisions. A hypothesis which is tested for possible rejection is called null hypothesis H₀ and the hypothesis which is opposite to this is the alternative hypothesis H₁. There are two basic types of decision problems that can be considered in a hypothesis testing procedure.
(a) whether a population parameter has changed from or differs from a particular value.
(b) whether the sample has come from the population that has a parameter value less than or more than the hypothesized value.

The set of all possible values of the sample statistic is referred to as the sample space. The test procedure divides the sample space into two parts called the acceptance region and rejection region (critical region). In the case of the two-tailed test, we find two values C₁ and C₂ which set the limits on the amount of sampling variation consistent with the null hypothesis H₀. For the one-tailed test, we find only one value C₁

When the hypothesis H₀ is rejected when it is true the error is Type I error. When H0 is accepted when it is false it is called Type II error. When the calculated value of the test statistic is less than the table value, we accept the null hypothesis H₀; otherwise, accept alternative hypothesis H₁.

Question 4.
Explain in detail about the test of significance of a single mean.
Solution:
A random sample of size n (n ≥ 30) is drawn from a population. We want to test the population mean has a specified value µ₀.
Procedure for testing: (For two-tail test)
The null hypothesis is H₀ : µ = µ₀.
The alternative hypothesis is H₁ : µ ≠ µ₂
Since n is large the sampling distribution of \(\bar{x}\) (the sample mean) is approximately normal.
The test statistic

For a significance level α = 0.05 (5% level)
If |Z| < 1.96, H0 is accepted at 5% level. If |Z| > 1.96, H₀ is rejected at 5% level
For α = 0.01 (1% level), if |Z| < 2.58, H₀ is accepted. If |Z| > 2.58, H₀ is rejected.
Procedure for one tail test: (left tail)
H₀ : µ ≥ µ₀
H₁ : µ < µ₀
At α = 0.05, |Z| = 1.645
If Z < -1.645, H₀ is rejected If Z > -1.645, H₀ is accepted
One tail test: (right tail)
If Z < 1.645, H₀ is accepted If Z > 1.645, H₀ is rejected at 5% level of significance.

Question 5.
Determine the standard error of proportion for a random sample of 500 pineapples was taken from a large consignment and 65 were found to be bad.
Solution:
The standard error of proportion is S.E. = \(\sqrt{\mathrm{PQ} / \mathrm{n}}\)
Given sample size n = 500 and P = \(\frac{65}{500}\)

Question 6.
A sample of 100 students is drawn from a school. The mean weight and variance of the sample are 67.45 kg and 9 kg. respectively. Find (a) 95% (b) 99% confidence intervals for estimating the mean weight of the students.
Solution:
Given Sample size n = 100
Sample mean \(\bar{x}\) = 67.45 kg
Sample S.D s = √9 = 3 kg
(a) 95% confidence interval for population mean µ is,


Therefore 99% confidence intervals for estimating the mean weight of the students is (66.68, 68.22)

Question 7.
The mean I.Q of a sample of 1600 children was 99. Is it likely that this was a random sample from a population with mean I.Q 100 and standard deviation 15? (Test at 5% level of significance)
Solution:
Given
n (Sample size) = 1600 children
\(\bar{x}\) (Sample mean) = 99
µ (Population mean) = 100
σ (Population SD) = 15
α (Level of significance) = 5%
Null hypothesis H₀ : µ = 100
(Sample has been drawn from a population with mean 100 and S.D 15)
Alternative hypothesis H₁ : µ ≠ 100
(Sample has not been taken from a population with mean 100 and S.D 15)
Test statistic

The significant value or table value \(Z_{\alpha / 2}\) = 1.96. Comparing the values, we see that 2.67 > 1.96 (or) Z > \(Z_{\alpha / 2}\) at 5% level of significance. So the null hypothesis is rejected. Hence we conclude that the random sample is not taken from a population with mean I.Q = 100 and S.D = 15.

Students can download 12th Business Maths Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3 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 8 Sampling Techniques and Statistical Inference Ex 8.3

Choose the correct answer.

Question 1.
A ______ may be finite or infinite according to as the number of observations or items in it is finite or infinite.
(a) Population
(b) census
(c) parameter
(d) none of these
Answer:
(a) Population

Question 2.
A _______ of statistical individuals in a population is called a sample.
(a) Infinite set
(b) finite subset
(c) finite set
(d) entire set
Answer:
(b) finite subset

Question 3.
A finite subset of statistical individuals in a population is called ________
(a) a sample
(b) a population
(c) universe
(d) census
Answer:
(a) a sample

Question 4.
Any statistical measure computed from sample data is known as _______
(a) parameter
(b) statistic
(c) infinite measure
(d) uncountable measure
Answer:
(b) statistic

Question 5.
A ______ is one where each item in the universe has an equal chance of known opportunity of being selected.
(a) Parameter
(b) random sample
(c) statistic
(d) entire data
Answer:
(b) random sample

Question 6.
A random sample is a sample selected in such a way that every item in the population has an equal chance of being included ______
(a) Harper
(b) Fisher
(c) Karl Pearson
(d) Dr. Yates
Answer:
(a) Harper

Question 7.
Which one of the following is probability sampling?
(a) purposive sampling
(b) judgement sampling
(c) simple random sampling
(d) Convenience sampling
Answer:
(c) simple random sampling

Question 8.
In simple random sampling from a population of units, the probability of drawing any unit at the first draw is ______
(a) \(\frac{n}{\mathrm{N}}\)
(b) \(\frac{1}{\mathrm{N}}\)
(c) \(\frac{N}{\mathrm{n}}\)
(d) 1
Answer:
(b) \(\frac{1}{\mathrm{N}}\)

Question 9.
In _______ the heterogeneous groups are divided into homogeneous groups.
(a) Non-probability sample
(b) a simple random sample
(c) a stratified random sample
(d) systematic random sample
Answer:
(c) a stratified random sample

Question 10.
Errors in sampling are of ______
(a) Two types
(b) three types
(c) four types
(d) five types
Answer:
(a) Two types

Question 11.
The method of obtaining the most likely value of the population parameter using statistic is called ________
(a) estimation
(b) estimator
(c) biased estimate
(d) standard error
Answer:
(a) estimation

Question 12.
An estimator is a sample statistic used to estimate a ______
(a) population parameter
(b) biased estimate
(c) sample size
(d) census
Answer:
(a) population parameter

Question 13.
________ is a relative property, which states that one estimator is efficient relative to another.
(a) efficiency
(b) sufficiency
(c) unbiased
(d) consistency
Answer:
(a) efficiency

Question 14.
If probability P[|\(\bar{\theta}\) – θ| < ε] → 1µ as n → ∞ for any positive ε then \(\bar{\theta}\) is said to ______ estimator of θ.
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Answer:
(d) consistent

Question 15.
An estimator is said to be _______ if it contains all the information in the data about the
parameter it estimates.
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Answer:
(b) sufficient

Question 16.
An estimate of a population parameter given by two numbers between which the parameter would be expected to lie is called an _______ interval estimate of the parameter.
(a) point estimate
(b) interval estimation
(c) standard error
(d) confidence
Answer:
(b) interval estimation

Question 17.
A ________ is a statement or an assertion about the population parameter.
(a) hypothesis
(b) statistic
(c) sample
(d) census
Answer:
(a) hypothesis

Question 18.
Type I error is ______
(a) Accept H₀ when it is true
(b) Accept H₀ when it is false
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Answer:
(c) Reject H₀ when it is true

Question 19.
Type II error is ______
(a) Accept H₀ when it is wrong
(b) Accept H₀ when it is true
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Answer:
(a) Accept H₀ when it is wrong

Question 20.
The standard error of sample mean is ________
(a) \(\frac{\sigma}{\sqrt{2 n}}\)
(b) \(\frac{\sigma}{n}\)
(c) \(\frac{\sigma}{\sqrt{n}}\)
(d) \(\frac{\sigma^{2}}{\sqrt{n}}\)
Answer:
(c) \(\frac{\sigma}{\sqrt{n}}\)

Students can download 12th Business Maths Chapter 7 Probability Distributions Ex 7.3 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 7 Probability Distributions Ex 7.3

Question 1.
Define Normal distribution.
Solution:
A random variable X is said to follow a normal distribution with parameters µ (mean) and σ² (variance) if its probability density function is given by

Question 2.
Define Standard normal variate.
Solution:
A random variable Z = \(\frac{X-\mu}{\sigma}\) is called a standard normal variate with mean 0 and standard deviation 1 (i.e.) Z ~ N (0, 1). Its probability density function is given by

Question 3.
Write down the conditions in which the Normal distribution is a limiting case of the binomial distribution.
Solution:
The Normal distribution is a limiting case of Binomial distribution under the following conditions:

• n, the number of trials is infinitely large, i.e. n → ∞
• neither p (or q ) is very small.

Question 4.
Write down any five chief characteristics of Normal probability curve.
Solution:
Chief Characteristics of the Normal Probability Curve are as follows:

• The curve is bell-shaped and symmetrical about the line x = µ.
• Mean, median and mode of the distribution coincide.
• The total area under the normal curve is equal to unity.
• For a given µ and σ, there is only one normal distribution.
• The Points of inflexion are given by x = µ ± σ

Question 5.
In a test on 2,000 electric bulbs, it was found that bulbs of a particular make, was normally distributed with an average life of 2,040 hours and a standard deviation of 60 hours. Estimate the number of bulbs likely to burn for
(i) more than 2,150 hours
(ii) less than 1,950 hours
(iii) more 1,920 hours but less than 2,100 hours.
Solution:
Let X be the numbers of hours for which the bulbs are in use. It is given that X is normally distributed with mean 2040 hours and a standard deviation of 60 hours, (i.e) X ~ N (2040, 60²)
(i) P (X > 2150)
We change to the standard normal variate.

The total area to the right of Z = 0 is 0.5.
The area between Z = 0 and 1.833 is 0.4664 (from tables)
So P (Z > 1.833) = 0.5 – 0.4664 = 0.0336
The number of bulbs likely to bum for more than 2150 hours is 2000 × 0.0336 = 67.2 ~ 67

(ii) We want P (X < 1950)
= P(\(\frac{\mathrm{X}-\mu}{\sigma}<\frac{1950-2040}{60}\))
= P(Z < -1.5)
The area between Z = -1.5 and Z = 0 is same as area between Z = 0 and Z = 1.5.
From the tables, area between Z = 0 and Z = 1.5 is 0.4332
P(Z < -1.5) = 0.5 – 0.4332 = 0.0668
Hence the number of bulbs likely to bum for less than 1950 hours is 2000 × 0.0668 = 133.6 ~ 134

(iii) We want P(1920 < X < 2100)
When X = 1920,
\(Z=\frac{X-\mu}{\sigma}=\frac{1920-2040}{60}=-2\)
When X = 2100,
\(\mathrm{Z}=\frac{2100-2040}{60}=\frac{60}{60}=1\)

So P(1920 < X < 2100) = P(-2 < Z < 1)
= P(-2 < Z < 0) + P(0 < Z < 1)
= P (0 < Z < 2) + P (0 < Z < 1)
= 0.4772 + 0.3413
= 0.8185
Hence the number of bulbs likely to bum for more than 1920 hours but less than 2100 hours is 2000 × 0.8185 = 1637.

Question 6.
In a distribution, 30% of the items are under 50 and 10% are over 86. Find the mean and standard deviation of the distribution.
Solution:
Let X be the normal random variable denoting the number of items in the distribution.
It is given that 30% of items are under 50
⇒ P (X < 50) = 30% = 0.3.
Also, 10% are over 86
⇒ P (X > 86) = 10% = 0.1.
We have to find µ and σ.
Representing the given data diagrammatically,
Where Z₁ = \(\frac{50-\mu}{\sigma}\) and Z₂ = \(\frac{86-\mu}{\sigma}\)

From the diagram,
P (-Z₁ < Z < 0) = 0.2
By symmetry P (0 < Z < Z₁) = 0.2
Z₁ = 0.525 (from the normal table)
Hence -0.525 = \(\frac{50-\mu}{\sigma}\)
(i.e.) 50 – µ = -0.525σ …….. (1)
Again P(0 < Z < Z₂) = 0.4
Z₂ = 1.28
Hence \(\frac{86-\mu}{\sigma}\) = 1.28
(or) 86 – µ = 1.28σ ……. (2)
Solving (1) and (2)
50 – µ = -0.525σ
86 – µ = 1.28σ
Subtracting,
36 = 1.28σ + 0.525σ
36 = 1.805σ
σ = 19.94
Using this in (2),
86 – µ = 1.28 (19.94)
86 – µ = 25.52
µ = 60.48
Hence the mean of the distribution is 60.48 and standard deviation is 19.94.

Question 7.
X is normally distributed with mean 12 and SD 4. Find P(X ≤ 20) and P(0 ≤ X ≤ 12).
Solution:
Given X ~ N (12, 42), (i.e) mean (µ) = 12 and s.d (σ) = 4.
P(X ≤ 20)
= P(\(\frac{X-\mu}{\sigma} \leq \frac{20-12}{4}\))
= P(X ≤ 2)
Now P(Z ≤ 2) = P(Z ≤ 0) + P(0 ≤ Z ≤ 2)
= 0.5 + 0.4772
= 0.9772

P(0 ≤ X ≤ 12)
= P(\(\frac{0-12}{4} \leq \frac{X-\mu}{\sigma} \leq \frac{12-12}{4}\))
= P (-3 ≤ Z ≤ 0)
= P(0 ≤ Z ≤ 3)
P(0 ≤ Z ≤ 3) = 0.49865 (from normal tables)

Question 8.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation of 3.0 inches, how many students have height
(a) greater than 72 inches
(b) less than or equal to 64 inches
(c) between 65 and 71 inches.
Solution:
Given X is the normal random variables denoting the height of the students with mean µ = 68 and s.d (σ) = 3.
(a) To find P(X > 72)
= P(Z > \(\frac{72-68}{3}\))
= P(Z > \(\frac{4}{3}\))
= P(Z > 1.33)
Now P(Z > 1.33) = 0.5 – P(0 < Z < 1.33)
= 0.5 – 0.4082
= 0.0918
Hence number of students whose height is greater than 72 inches is 500 × 0.0918 = 45.9 ~ 46

(b) To find P(X ≤ 64)
= P(Z ≤ \(\frac{64-68}{3}\))
= P(Z ≤ -1.33)
By Symmetry,
P(Z ≤ -1.33) = P(Z ≥ 1.33) = 0.0918 (see before section)
Hence number of students whose heights are less than or equal to 64 inches is 0.0918 × 500 = 46

(c) To find P (65 < X < 71)
= P(\(\frac{65-68}{3}\) < Z < \(\frac{71-68}{3}\))
= P(-1 < Z < 1)
= P(-1 < Z < 0) + P(0 < Z < 1)
= 2P(0 < Z < 1) {By symmetry}
= 2 (0.3413)
= 0.6826
Hence number of students with height between 65 and 71 inches is (500) (0.6826) = 341.3 ~ 342.

Question 9.
In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second. Find the probability that it will take less than 16.35 seconds to develop prints.
Solution:
Let X be the normal random variable denoting the developing time of the prints.
Given that mean µ = 16.28 and s.d σ = 0.12
To find P(X < 16.35)
When X = 16.35, \(\mathrm{Z}=\frac{\mathrm{X}-\mu}{\sigma}=\frac{16.35-16.28}{0.12}\)
Z = 0.58
So P(X < 16.35) = P(Z < 0.58)
Now P(Z < 0.58) = 0.5 + P(0 < Z < 0.58)
= 0.5 + 0.2190
= 0.7190
Thus the probability that it will take less than 16.35 seconds to develop prints is 0.719.

Question 10.
Time taken by a construction company to construct a flyover is a normal variate with mean 400 labour days and a standard deviation of 100 labour days. If the company promises to construct the flyover in 450 days or less and agree to pay a penalty of ₹ 10,000 for each labour day spent in excess of 450. What is the probability that
(i) the company pays a penalty of at least ₹ 2,00,000?
(ii) the company takes at most 500 days to complete the flyover?
Solution:
Let X be the normal variate denoting the number of labour days.
Given mean µ = 400 and s.d σ = 100
(i) The company pays penalty of ₹ 2,00,000 at the rate of ₹ 10,000 per each extra labour day.
No. of extra days = \(\frac{2,00,000}{10,000}\) = 20
So the probability that company pays a penalty of atleast ₹ 2,00,000 is probability that labour days should be atleast 450 + 20 = 470 days, (i.e.) P (X ≥ 470)
Now P(X ≥ 470) = P(Z ≥ \(\frac{470-400}{100}\)) = P (Z ≥ 0.7)
P(Z ≥ 0.7) = 0.5 – P(0 ≤ Z ≤ 0.7) = 0.5 – 0.258 = 0.242

(ii) P (X ≤ 500) = P(Z ≤ \(\frac{500-40}{100}\)) = P (Z ≤ 1)
P(Z ≤ 1) = 0.5 + P (0 ≤ Z ≤ 1) = 0.5 + 0.3413 = 0.8413
Thus the probability that the company takes at most 500 days to complete the flyover is 0.8413.

Students can download 12th Business Maths Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2 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 8 Sampling Techniques and Statistical Inference Ex 8.2

Question 1.
Mention two branches of statistical inference?
Solution:
The two branches of statistical inference are estimation and testing of hypothesis.

Question 2.
What is an estimator?
Answer:
An estimator is a statistic that is used to infer the value of an unknown population parameter in a statistical model. The estimator is a function of the data arid so it is also a random variable.

Question 3.
What is an estimate?
Solution:
Any specific numerical value of the estimator is called an estimate. For example, sample means are used to estimate population means.

Question 4.
What is point estimation?
Solution:
Point estimation involves the use of sample data to calculate a single value which is to serve as a best estimate of an unknown population parameter. For example the mean height of 145 cm from a sample of 15 students is‘a point estimate for the mean height of the class of 100 students.

Question 5.
What is interval estimation?
Solution:
Interval estimation is the use of sample data to calculate an interval of possible values of an unknown population parameter. For example the interval estimate for the population mean is (101.01, 102.63).This gives a range within which the population mean is most likely to be located.

Question 6.
What is confidence interval?
Solution:
A confidence interval L a type of interval estimate, computed from the statistics of the observed data, that might contain the true value of an unknown population parameter. The numbers at the upper and lower end of a confidence interval are called confidence limits. For example, if mean is 7.4 with confidence interval (5.4, 9.4), then the numbers 5.4 and 9.4 are the confidence limits.

Question 7.
What is null hypothesis? Give an example.
Solution:
A null hypothesis is a type of hypothesis, that proposes that no statistical significance exists in a set of given observations. For example, let the average time to cook a specific dish is 15 minutes. The null hypothesis would be stated as “The population mean is equal to 15 minutes”, (i.e) H₀ : µ = 15

Question 8.
Define the alternative hypothesis.
Solution:
The alternative hypothesis is the hypothesis that is contrary to the null hypothesis and it is denoted by H₁.
For example if H₁ : µ = 15, then the alternative hypothesis will be : H₁ : µ ≠ 15, (or) H₁ : µ < 15 (or) H₁ : µ > 15.

Question 9.
Define the critical region.
Solution:
The critical region is the region of values that corresponds to the rejection of the null hypothesis at some chosen probability level. For the two-tailed test, the critical region is given below.

where α is the level of significance.

Question 10.
Define critical value.
Solution:
A critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis. It depends on the level of significance. For example, if the confidence level is 90% then the critical value is 1.645.

Question 11.
Define the level of significance
Solution:
The level of significance is defined as the probability of rejecting a null hypothesis by the test when it is really true, which is denoted as α. That is P(Type 1 error) = α. For example, the level of significance 0.1 is related to the 90% confidence level.

Question 12.
What is a type I error?
Solution:
In statistical hypothesis testing, a Type f error is the rejection of a true null hypothesis. Example of Type I errors includes a test that shows a patient to have a disease when he does not have the disease, a fire alarm going on indicating a fire when there is no fire (or) an experiment indicating that medical treatment should cure a disease when in fact it does not.

Question 13.
What is the single-tailed test?
Solution:
A single-tailed test or a one-tailed test is a statistical test in which the critical area of a distribution is one-sided so that it is either greater than or less than a certain value, but not both. For the null hypothesis H₀ : µ = 16.91, the alternative hypothesis H₁ : µ > 16.91 or H₁ : µ < 16.91 are one-tailed tests.

Question 14.
A sample of 100 items, draw from a universe with mean value 4 and S.D 3, has a mean value 3.5. Is the difference in the mean significant?
Solution:
Given Sample size n = 100
Sample mean \(\bar{x}\) = 3.5
Population mean µ = 4
Population SD σ = 3
Now, null hypothesis H₀ : µ = 4
Alternative hypothesis H₁ : µ ≠ 4 (Two tail)
We take level of significance α = 5% = 0.05
The table value \(\mathrm{Z}_{\alpha / 2}\) = 1.96
Test statistic

Since the alternative hypothesis is of the two-tailed test we can take |Z| = 1.667. We observe that 1.667 < 1.96 (i.e) |Z| < \(\mathrm{Z}_{\alpha / 2}\). So at 5% level of significance, the null hypothesis H0 is accepted. Therefore, we conclude that there is no significant difference between the sample mean and the population mean.

Question 15.
A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be reasonably regarded as a sample from a large population with a mean height of 67.39 inches and standard deviation of 1.30 inches?
Solution:
Given Sample size n = 400
Sample mean \(\bar{x}\) = 67.47
Population mean µ = 67.39
Population SD σ = 1.3
Null hypothesis H₀ : µ = 67.39 inches
(the sample has been drawn from the population with mean heights 67.39 inches)
Alternative hypothesis H₁ : µ ≠ 67.39 inches
(the sample has not been drawn from the population with mean height 67.39 inches)
The level of significance α = 5% = 0.05
Test statistic

The significant value or table value \(\mathrm{Z}_{\alpha / 2}\) = 1.96. We see that 1.2308 < 1.96 (i.e) Z < \(\mathrm{Z}_{\alpha / 2}\). Since the calculated value is less than the table value at 5% level of significance, the null hypothesis is accepted. Hence we conclude that the data does not provide us with any evidence against the null hypothesis. Thus, the sample has been drawn from a large population with a mean height of 67.39 inches and S.D 1.3 inches.

Question 16.
The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state’s education system, the scores of 100 of the state’s students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores.
Solution:
Given Population mean µ = 76
Population SD σ = 8
Sample size n = 100
Sample mean \(\bar{x}\) = 72
Significance level α = 0.05
Null hypothesis H0 : µ = 76
(i.e) there is no difference between the state scores and the national scores.
Alternative hypothesis H₁ : µ ≠ 76
(i.e) there is a significant difference between the state scores and the nationals scores of the aptitude test.
Test statistic

The significant value or table value \(\mathrm{Z}_{\alpha / 2}\) = 1.96. Comparing the calculated value and table value, we find that |Z| > \(\mathrm{Z}_{\alpha / 2}\) (i.e) 5 > 1.96. So the null hypothesis is rejected and we accept the alternative hypothesis. So we conclude that at the significance level of 5%, there is a difference between the state scores and the national scores of the nationally administered amplitude test.

Question 17.
The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation of 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance?
Solution:
Given Population mean µ = 1800
Population SD σ = 100
Sample size n = 50
Sample mean \(\bar{x}\) = 1850
Significance level α = 0.01
Null hypothesis H₀ : µ = 1800
(i.e) the breaking strength of the cables has not increased, after the new technique in the manufacturing process.
Alternative hypothesis H₁ : µ > 1800 (i.e) the new technique was successful.
Test statistic

The table value for the one-tailed test is Zα = 2.33.
Comparing the calculated value and table value, we find that Z > Zα (i.e.) 3.536 > 2.33.

Inference: Since the calculated value is greater than the table value at 1 % level of significance, the null hypothesis is rejected and we accept the alternative hypothesis. We conclude that by the new technique in the manufacturing process the breaking strength of the cables is increased. So the claim is supported at 0.01 level of significance.

Students can download 12th Business Maths Chapter 7 Probability Distributions Ex 7.2 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 7 Probability Distributions Ex 7.2

Question 1.
Define Poisson distribution.
Solution:
Poisson distribution is a discrete frequency distribution which gives the probability of a number of independent events occurring in a fixed time. It is useful for characterizing events with very low probabilities of occurrence within some definite time or space.

Question 2.
Write any 2 examples for Poisson distribution.
Solution:
Examples of Poisson distribution are given by

• The number of printing mistakes per page in a textbook.
• A number of lightning per second.
• The number of bacteria in one cubic centimetre.

Question 3.
Write the conditions for which the Poisson distribution is a limiting case of the binomial distribution.
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:

• the number of trials ‘n’ is indefinitely large i.e, → ∞
• the probability of success ‘p’ in each trial is very small, i.e, p → 0
• np = λ is finite. Thus p = \(\frac{\lambda}{n}\) and q = 1 – \(\frac{\lambda}{n}\), λ > 0

Question 4.
Derive the mean and variance of the Poisson distribution.
Solution:
Let X be a Poisson random variable with parameter λ. The p.m.f is given by



Thus the mean and variance of Poisson distribution are both equal to λ.

Question 5.
Mention the properties of Poisson distribution.
Solution:

1. Poisson distribution is the only distribution in which the mean and variance are equal.
2. The probability that an event occurs in a given time, distance, area or volume is the same.

Question 6.
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? [Given e^-2.8 = 0.06]
Solution:
Let X denote the number of deaths due to the disease
P(death) = \(\frac{7}{1000}\) = 0.007 ⇒ p = 0.007 and n = 400
The value of mean λ = np = (0.007) (400) = 2.8
Hence X follows a Poisson distribution with

So the probability of just 2 deaths on account of this disease in a group of 400 is 0.2352.

Question 7.
It is given that 5% of the electric bulbs manufactured by a company are defective. Using Poisson distribution find the probability that a sample of 120 bulbs will contain no defective bulb.
Solution:
Given p = \(\frac{5}{100}\) = 0.05 and n = 120
⇒ λ = np = (0.05) (120) = 6
Thus X is a Poisson random variable with P (X = x) = \(\frac{e^{-6} 6^{x}}{x !}\)
We want P (no defective bulb) = P (X = 0)
= \(\frac{e^{-6} 6^{0}}{0 !}\)
= e^-6
= 0.0025 (Using exponent table)
Thus the probability that a sample of 120 bulbs will not contain any defective bulb is 0.0025.

Question 8.
A car hiring firm has two cars. The demand for cars on each day is distributed as a Poisson variate, with mean 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused.
Solution:
Let X be the Poisson variable denoting the demand for the cars.
It is given that mean is 1.5 ⇒ λ = 1.5
(i) P (Neither car is used) = P (X = 0) = \(\frac{e^{-1.5}(1.5)^{0}}{0 !}=e^{-1.5}=0.2231\)
(ii) Some demand is refused when demand is more than 2 since the firm has only 2 cars. So we want P (X > 2)
Now P (X > 2) = 1 – P (X ≤ 2)
= 1 – [P(X = 2) + P(X = 1) + P(X = 0)]

Question 9.
The average number of phone calls per minute into the switchboard of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be
(i) no phone at all
(ii) exactly 3 calls
(iii) at least 5 calls.
Solution:
Let X be the Poisson variable denoting the number of phone calls per minute.

P (X = 1) = \(\frac{e^{-2.5}(2.5)}{1 !}\) = (0.08208) (2.5) = 0.2052
Using the above values and P (X = 0) and P (X = 3) from the previous subdivisions in (A) we get,
P(X ≥ 5) = 1 – [0.1336 + 0.2138 + 0.2565 + 0.2052 + 0.08208]
= 1 – 0.89118
= 0.10882

Question 10.
The distribution of the number of road accidents per day in a city is Poisson with mean 4. Find the number of days out of 100 days when there will be
(i) no accident
(ii) at least 2 accidents and
(iii) at most 3 accidents.
Solution:
Let X be the Poisson variable denoting the number of accidents per day.
Given that mean is 4 (i.e,) λ = 4. The p.m.f is given by P(X = x) = \(\frac{e^{-4} 4^{x}}{x !}\)
(i) P (no accident) = P(X = 0) = e^-4 = 0.0183
For 100 days we have 100 × 0.0183 = 1.83 ~ 2
Hence out of 100 days there will be no accident for 2 days.

(ii) P (atleast 2 accidents) = P (X ≥ 2)
= 1 – P (X < 2)
= 1 – [P(X = 1) + P(X = 0)]
= 1 – [e^-4 (4) + e^-4]
= 1 – (0.0183) (5)
= 1 – 0.0915
= 0.9085
For 100 days we have 100 × 0.9085 ~ 91
Hence out of 100 days there will be at least 2 accidents for 91 days.

(iii) P (atmost 3 accidents) = P (X ≤ 3)
= P (X = 0 ) + P (X = 1 ) + P (X = 2) + P (X = 3)
= \(e^{-4}\left[1+\frac{4}{1}+\frac{16}{2}+\frac{64}{6}\right]\)
= (0.0183) [23.6667]
= 0.4331
For 100 days we have 100 × 0.4331 ~ 43
Hence out of 100 days, there will be atmost 3 accidents for 43 days.

Question 11.
Assuming that a fatal accident in a factory during the year is 1/1200, calculate the probability that in a factory employing 300 workers there will be at least two fatal accidents in a year, (given e^-0.25 = 0.7788).
Solution:
Let X denote the number of accidents.
Given that probability of accidents ‘p’ is 1/1200 and n = 300

= 1 – [e^-0.25 + e^-0.25 (0.25)]
= 1 – e^-0.25 (1.25)
= 1 – (0.7788) (1.25)
= 0.0265
Thus the probability that there will be atleast two fatal accidents in a year is 0.0265.

Question 12.
The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute
(i) No customer appears
(ii) three or more customers appear.
Solution:
Let X denote the number of customers.
Given λ = 2
(i) P (no customer) = P (X = 0)
= \(\frac{e^{-2}(2)^{0}}{0 !}\)
= e^-2
= 0.1353
(ii) P (3 or more customers) = P (X ≥ 3)

Thus during a given minute, the probability that three or more customers appear is 0.3235.

Students can download 12th Business Maths Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1 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 8 Sampling Techniques and Statistical Inference Ex 8.1

Question 1.
What is the population?
Solution:
A population is a set of similar items or events which is of interest for some question or experiment. A population can be specific or vague. Examples of population defined vaguely include the number of newborn babies in Tamil Nadu, a total number of tech startups in India, the average height of all exam candidates, mean weight of taxpayers in Chennai etc. Examples of population defined specifically include a number of fans produced in a particular factory, the number of students in a class, the number of boys and girls in a tuition centre etc.

Question 2.
What is the sample?
Solution:
A sample is a set of data collected from a statistical population by a defined procedure. The elements of a sample are called sample size or sample points. Samples are collected and statistics are calculated from the samples, so that one can make inferences from the sample to the population.

Question 3.
What is statistic?
Solution:
A statistic is used to estimate the value of a population parameter. For instance, we selected a random sample of 100 students from a school with 1000 students. The average height of the sampled students would be an example of a statistic. Examples, sample variance, sample quartiles, sample percentiles, sample moments etc.

Question 4.
Define parameter.
Solution:
A parameter is any numerical quantity that characterizes a given population or some aspect of it. This means the parameter tells us something about the whole population. For example, the population mean µ, variance σ2, population proportion P, population correlation ρ.

Question 5.
What is the sampling distribution of a statistic?
Solution:
Sampling distribution of a statistic is the probability distribution of a given random sample based statistic. It may be considered as the distribution of the statistic for all possible samples from the same population of a given sample size.
Example, the sampling distribution of the mean for n = 2 is given below:

Question 6.
What is the standard error?
Solution:
The standard error (S.E) of a statistic is the standard deviation of its sampling distribution. If the parameter or the statistic is the mean, it is called the standard error of the mean (SEM). The standard error provides a rough estimate of the interval in which the population parameters is likely to fall.

Question 7.
Explain in detail about simple random sampling with a suitable example.
Solution:
(i) Simple random sampling:
In this technique, the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. Simple random sampling may be done, with or without replacement of the samples selected. In a simple random sampling with replacement, there is a possibility of selecting the same sample any number of times. So, simple random sampling without replacement is followed.
Thus in simple random sampling from a population of N units, the probability of drawing any unit at the first draw is \(\frac{1}{N}\), the probability of drawing any unit in the second draw from among the available (N – 1) units is \(\frac{1}{(N-1)}\), and so on. Several methods have been adopted for random selection of the samples from the population. Of those, the following two methods are generally used and which are described below.

1. Lottery method
This is the most popular and simplest method when the population is finite. In this method, all the items of the population are numbered on separate slips of paper of the same size, shape and colour. They are folded and placed in a container and shuffled thoroughly. Then the required numbers of slips are selected for the desired sample size. The selection of items thus depends on chance.

For example, if we want to select 10 students, out of 100 students, then we must write the names/roll number of all the 100 students on slips of the same size and mix them, then we make a blindfold selection of 10 students. This method is called unrestricted random sampling because units are selected from the population without any restriction. This method is mostly used in lottery draws. If the population or universe is infinite, this method is inapplicable.

2. Table of Random number
When the population size is large, it is difficult to number all the items on separate slips of paper of same size, shape and colour. The alternative method is that of using the table of random numbers. The most practical, easy and inexpensive method of selecting a random sample can be done through “Random Number Table”. The random number table has been so constructed that each of the digits 0, 1, 2,…, 9 will appear approximately with the same frequency and independently of each other.

The various random number tables available are

• L.H.C. Tippett random number series
• Fisher and Yates random number series
• Kendall and Smith random number series
• Rand Corporation random number series.

Tippett’s table of random numbers is most popularly used in practice.

An example to illustrate how Tippett’s table of random numbers may be used is given below. Suppose we have to select 20 items out of 6,000. The procedure is to number all the 6,000 items from 1 to 6,000. A page from Tippett’s table may be selected and the first twenty numbers ranging from 1 to 6,000 are noted down. If the numbers are above 6000, choose the next number ranging from 1 to 6000. Items bearing those numbers will be selected as samples from the population. Making use of the portion of the random number table given, the required random samples are shaded. Here, we consider row-wise selection of random numbers.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
Stratified Random Sampling
In stratified random sampling, first divide the population into subpopulations, which are called strata. Then, the samples are selected from each of the strata through random techniques. The collection of all the samples from all strata gives the stratified random samples.

When the population is heterogeneous or different segments or groups with respect to the variable or characteristic under study, then the Stratified Random Sampling method is studied.. First, the population is divided into the homogeneous number of sub-groups or strata before the sample is drawn. A sample is drawn from each stratum at random. Following steps are involved in selecting a random sample in a stratified random sampling method.

(a) The population is divided into different classes so that each stratum will consist of more or less homogeneous elements. The strata are so designed that they do not overlap each other.
(b) After the population is stratified, a sample of a specified size is drawn at random from each stratum using Lottery Method or Table of Random Number Method.

Stratified random sampling is applied in the field of the different legislative areas as strata in election polling, division of districts (strata) in a state etc…

Ex: From the following data, select 68 random samples from the population of the heterogeneous group with a size of 500 through stratified random sampling, considering the following categories as strata.

• Category 1: Lower income class – 39%
• Category 2: Middle income class – 38%
• Category 3: Upper income class – 23%

Solution:

Question 9.
Explain in detail about systematic random sampling with example.
Solution:
Systematic sampling:
In systematic sampling, randomly select the first sample from the first k units. Then every kth member, starting with the first selected sample, is included in the sample.

Systematic sampling is a commonly used technique if the complete and up-to-date list of the sampling units is available. We can arrange the items in numerical, alphabetical, geographical or in any other order. The procedure of selecting the samples starts with selecting the first sample at random, the rest being automatically selected according to some pre-determined ( pattern. A systematic sample is formed by selecting every item from the population, where k refers to the sample interval. The sampling interval can be determined by dividing the size of the population by the size of the sample to be chosen.
That is k = \(\frac{\mathrm{N}}{n}\), where k is an integer.
k = Sampling interval, N = Size of the population, n = Sample size.

Procedure for selection of samples by systematic sampling method
(i) If we want to select a sample of 10 students from a class of 100 students, the sampling interval is calculated as \(k=\frac{N}{n}=\frac{100}{10}=10\)
Thus sampling interval = 10 denotes that for every 10 samples one sample has to be selected.
(ii) The first sample is selected from the first 10 (sampling interval) samples through random selection procedures.
(iii) If the selected first random sample is 5, then the rest of the samples are automatically selected by incrementing the value of the sampling interval (k = 10) i.e., 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.

Ex: Suppose we have to select 20 items out of 6,000. The procedure is to number all the 6,000 items from 1 to 6,000. The sampling interval is calculated as k = \(\frac{N}{n}=\frac{6000}{20}\) = 300. Thus sampling interval = 300 denotes that for every 300 samples one sample has to be selected. The first sample is selected from the first 300 (sampling interval) samples through random selection procedures. If the selected first random sample is 50, then the rest of the samples are automatically selected by incrementing the value of the sampling interval (k = 300) ie,50, 350, 650, 950, 1250, 1550, 1850, 2150, 2450, 2750, 3050, 3350, 3650, 3950, 4250, 4550, 4850, 5150, 5450, 5750. Items bearing those numbers will be selected as samples from the population.

Question 10.
Explain in detail about sampling error.
Solution:
Sampling Errors: Errors, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors. Sampling errors are inherent in the method of sampling. They may arise accidentally without any bias or prejudice.

Sampling Errors arise primarily due to the following reasons:

• Faulty selection of the sample instead of the correct sample by defective sampling technique.
• The investigator substitutes a convenient sample if the original sample is not available while investigation.
• In area surveys, while dealing with borderlines it depends upon the investigator whether to include them in the sample or not. This is known as Faulty demarcation of sampling units.

Question 11.
Explain in detail about the non-sampling error.
Solution:
Non-Sampling Errors:
The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments( tape, scale) are called Non-Sampling errors.
It may arise in the following ways:

• Due to negligence and carelessness of the part of either investigator or respondents.
• Due to the lack of trained and qualified investigators.
• Due to the framing of a wrong questionnaire.
• Due to applying the wrong statistical measure
• Due to incomplete investigation and sample survey.

Question 12.
State any two merits of simple random sampling.
Solution:

• In simple random sampling personal bias is completely eliminated.
• This method is economical as it saves time, money and labour.

Question 13.
State any three merits of stratified random sampling.
Solution:

• A random stratified sample is superior to a simple random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled.
• A stratified random sample can be kept small in size without losing its accuracy.
• It is easy to administer if the population under study is sub-divided.

Question 14.
State any two demerits of systematic random sampling.
Solution:

• Systematic samples are not random samples.
• If N is not a multiple of n, then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

Question 15.
State any two merits for systematic random sampling.
Solution:
Merits of systematic sampling are given below:

• This method distributes the sample more evenly over the entire listed population.
• The time and work are reduced much.

Question 16.
Using the following Tippet’s random number table.

Draw a sample of 10 three-digit numbers which are even numbers.
Solution:
There are many ways to select a sample of 10 3-digit even numbers. From the table, start from the first number and move along the column. Select the first three digits as the number. If it is an odd number, move to the next number. The selected sample is 416, 664, 952, 748, 524, 914, 154, 340, 140, 276.

Question 17.
A wholesaler in apples claims that only 4 % of the apples supplied by him are defective. A random sample of 600 apples contained 36 defective apples. Calculate the standard error concerning good apples.
Solution:
Sample size = 600
No. of defective apples = 36
Sample proportion p = \(\frac{36}{600}\) = 0.06
Population proportion P = probability of defective apples = 4% = 0.04
Q = 1 – P = 1 – 0.04 = 0.96
The S.E for sample proportion is given by S.E

Question 18.
A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Tamil Nadu State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 lbs. Calculate the standard error of the mean.
Solution:
Given n = 1000, \(\bar{X}\) = 119, σ = 30

Question 19.
A random sample of 60 observations was drawn from a large population and its standard deviation was found to be 2.5. Calculate the suitable standard error that this sample is taken from a population with standard deviation 3?
Solution:
Given sample size n = 60
Sample standard deviation = 2.5
Population standard deviation σ = 3
The S.E is given by

Question 20.
In a sample of 400 population from a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village?
Solution:
Given sample size 400 and 230 are vegetarian eaters.
So sample proportionp = \(\frac{230}{400}\) = 0.575
Population proportion P = Prob (vegetarian eaters from the village) = \(\frac{1}{2}\)
(Since vegetarian and non-vegetarian foods are equally popular)
Q = 1 – P = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)

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

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)

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

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.

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)

(ii) P(atleast two defectives) = P (X ≥ 2) = 1 – P (X < 2)
= 1 – [P(X = 1) + P(X = 0)]

(iii) P(exactly 4 defectives) = P(X = 4)

(iv) We know that mean = np = (10) (0.05) = 0.5
Variance = npq = (10) (0.05) (0.95) = 0.475

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)

= (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 = \(\frac {1}{2}\)

Hence the probability that there will be exactly 2 girls is. 0.375.

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

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)

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.

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)

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)

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)

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)

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

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)]

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) = \(\left(\begin{array}{c}
25 \\
x
\end{array}\right)\left(\frac{1}{5}\right)^{x}\left(\frac{4}{5}\right)^{25-x}\)

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

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

Thus the probability that at least one person will have side effects is 0.2626.

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

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.

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)

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:

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

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

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

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?

(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

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:

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

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.

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:

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%

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

(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

(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%

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

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

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

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:

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

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.

(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)

(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.

Question 3.
The p.d.f. of X is defined as

Find the value of k and also find P(2 ≤ X ≤ 4).
Solution:

Question 4.
The probability distribution function of a discrete random variable X is

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

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
= \(\frac{7}{9}\)

Question 5.
The probability density function of a continuous random variable X is

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.

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)

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,

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

Question 9.
Consider a random variable X with p.d.f

Find E(X) and V(3X – 2).
Solution:

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.

Find the expected life of this piece of equipment.
Solution:
Let X be the random variable denoting the life of the piece of equipment.

Thus the expected life of the piece of equipment is \(\frac{1}{2}\) hrs (in thousands).