{"id":16464,"date":"2024-03-11T10:09:49","date_gmt":"2024-03-11T04:39:49","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=16464"},"modified":"2024-03-12T15:40:37","modified_gmt":"2024-03-12T10:10:37","slug":"samacheer-kalvi-12th-business-maths-solutions-chapter-7-ex-7-3","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-business-maths-solutions-chapter-7-ex-7-3\/","title":{"rendered":"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3"},"content":{"rendered":"<p>Students can download 12th Business Maths Chapter 7 Probability Distributions Ex 7.3 Questions and Answers, <a href=\"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-business-maths-book-solutions\/\">Samacheer Kalvi 12th Business Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n<h2>Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3<\/h2>\n<p>Question 1.<br \/>\nDefine Normal distribution.<br \/>\nSolution:<br \/>\nA random variable X is said to follow a normal distribution with parameters \u00b5 (mean) and \u03c3<sup>2<\/sup> (variance) if its probability density function is given by<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41298\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q1.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q1\" width=\"386\" height=\"74\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q1.png 386w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q1-300x58.png 300w\" sizes=\"(max-width: 386px) 100vw, 386px\" \/><\/p>\n<p>Question 2.<br \/>\nDefine Standard normal variate.<br \/>\nSolution:<br \/>\nA 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<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41299\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q2.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q2\" width=\"290\" height=\"62\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheerkalvi.Guru_.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3\" width=\"195\" height=\"28\" \/><\/p>\n<p>Question 3.<br \/>\nWrite down the conditions in which the Normal distribution is a limiting case of the binomial distribution.<br \/>\nSolution:<br \/>\nThe Normal distribution is a limiting case of Binomial distribution under the following conditions:<\/p>\n<ul>\n<li>n, the number of trials is infinitely large, i.e. n \u2192 \u221e<\/li>\n<li>neither p (or q ) is very small.<\/li>\n<\/ul>\n<p>Question 4.<br \/>\nWrite down any five chief characteristics of Normal probability curve.<br \/>\nSolution:<br \/>\nChief Characteristics of the Normal Probability Curve are as follows:<\/p>\n<ul>\n<li>The curve is bell-shaped and symmetrical about the line x = \u00b5.<\/li>\n<li>Mean, median and mode of the distribution coincide.<\/li>\n<li>The total area under the normal curve is equal to unity.<\/li>\n<li>For a given \u00b5 and \u03c3, there is only one normal distribution.<\/li>\n<li>The Points of inflexion are given by x = \u00b5 \u00b1 \u03c3<\/li>\n<\/ul>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheerkalvi.Guru_.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3\" width=\"195\" height=\"28\" \/><\/p>\n<p>Question 5.<br \/>\nIn 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<br \/>\n(i) more than 2,150 hours<br \/>\n(ii) less than 1,950 hours<br \/>\n(iii) more 1,920 hours but less than 2,100 hours.<br \/>\nSolution:<br \/>\nLet 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, 60<sup>2<\/sup>)<br \/>\n(i) P (X &gt; 2150)<br \/>\nWe change to the standard normal variate.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41302\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q5\" width=\"381\" height=\"146\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.png 381w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5-300x115.png 300w\" sizes=\"(max-width: 381px) 100vw, 381px\" \/><br \/>\nThe total area to the right of Z = 0 is 0.5.<br \/>\nThe area between Z = 0 and 1.833 is 0.4664 (from tables)<br \/>\nSo P (Z &gt; 1.833) = 0.5 &#8211; 0.4664 = 0.0336<br \/>\nThe number of bulbs likely to bum for more than 2150 hours is 2000 \u00d7 0.0336 = 67.2 ~ 67<\/p>\n<p>(ii) We want P (X &lt; 1950)<br \/>\n= P(\\(\\frac{\\mathrm{X}-\\mu}{\\sigma}&lt;\\frac{1950-2040}{60}\\))<br \/>\n= P(Z &lt; -1.5)<br \/>\nThe area between Z = -1.5 and Z = 0 is same as area between Z = 0 and Z = 1.5.<br \/>\nFrom the tables, area between Z = 0 and Z = 1.5 is 0.4332<br \/>\nP(Z &lt; -1.5) = 0.5 &#8211; 0.4332 = 0.0668<br \/>\nHence the number of bulbs likely to bum for less than 1950 hours is 2000 \u00d7 0.0668 = 133.6 ~ 134<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41300\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.1.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q5.1\" width=\"378\" height=\"146\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.1.png 378w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.1-300x116.png 300w\" sizes=\"(max-width: 378px) 100vw, 378px\" \/><\/p>\n<p>(iii) We want P(1920 &lt; X &lt; 2100)<br \/>\nWhen X = 1920,<br \/>\n\\(Z=\\frac{X-\\mu}{\\sigma}=\\frac{1920-2040}{60}=-2\\)<br \/>\nWhen X = 2100,<br \/>\n\\(\\mathrm{Z}=\\frac{2100-2040}{60}=\\frac{60}{60}=1\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41301\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.2.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q5.2\" width=\"379\" height=\"151\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.2.png 379w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q5.2-300x120.png 300w\" sizes=\"(max-width: 379px) 100vw, 379px\" \/><br \/>\nSo P(1920 &lt; X &lt; 2100) = P(-2 &lt; Z &lt; 1)<br \/>\n= P(-2 &lt; Z &lt; 0) + P(0 &lt; Z &lt; 1)<br \/>\n= P (0 &lt; Z &lt; 2) + P (0 &lt; Z &lt; 1)<br \/>\n= 0.4772 + 0.3413<br \/>\n= 0.8185<br \/>\nHence the number of bulbs likely to bum for more than 1920 hours but less than 2100 hours is 2000 \u00d7 0.8185 = 1637.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheerkalvi.Guru_.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3\" width=\"195\" height=\"28\" \/><\/p>\n<p>Question 6.<br \/>\nIn a distribution, 30% of the items are under 50 and 10% are over 86. Find the mean and standard deviation of the distribution.<br \/>\nSolution:<br \/>\nLet X be the normal random variable denoting the number of items in the distribution.<br \/>\nIt is given that 30% of items are under 50<br \/>\n\u21d2 P (X &lt; 50) = 30% = 0.3.<br \/>\nAlso, 10% are over 86<br \/>\n\u21d2 P (X &gt; 86) = 10% = 0.1.<br \/>\nWe have to find \u00b5 and \u03c3.<br \/>\nRepresenting the given data diagrammatically,<br \/>\nWhere Z<sub>1<\/sub> = \\(\\frac{50-\\mu}{\\sigma}\\) and Z<sub>2<\/sub> = \\(\\frac{86-\\mu}{\\sigma}\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41303\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q6.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q6\" width=\"382\" height=\"169\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q6.png 382w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q6-300x133.png 300w\" sizes=\"(max-width: 382px) 100vw, 382px\" \/><br \/>\nFrom the diagram,<br \/>\nP (-Z<sub>1<\/sub> &lt; Z &lt; 0) = 0.2<br \/>\nBy symmetry P (0 &lt; Z &lt; Z<sub>1<\/sub>) = 0.2<br \/>\nZ<sub>1<\/sub> = 0.525 (from the normal table)<br \/>\nHence -0.525 = \\(\\frac{50-\\mu}{\\sigma}\\)<br \/>\n(i.e.) 50 &#8211; \u00b5 = -0.525\u03c3 &#8230;&#8230;.. (1)<br \/>\nAgain P(0 &lt; Z &lt; Z<sub>2<\/sub>) = 0.4<br \/>\nZ<sub>2<\/sub> = 1.28<br \/>\nHence \\(\\frac{86-\\mu}{\\sigma}\\) = 1.28<br \/>\n(or) 86 &#8211; \u00b5 = 1.28\u03c3 &#8230;&#8230;. (2)<br \/>\nSolving (1) and (2)<br \/>\n50 &#8211; \u00b5 = -0.525\u03c3<br \/>\n86 &#8211; \u00b5 = 1.28\u03c3<br \/>\nSubtracting,<br \/>\n36 = 1.28\u03c3 + 0.525\u03c3<br \/>\n36 = 1.805\u03c3<br \/>\n\u03c3 = 19.94<br \/>\nUsing this in (2),<br \/>\n86 &#8211; \u00b5 = 1.28 (19.94)<br \/>\n86 &#8211; \u00b5 = 25.52<br \/>\n\u00b5 = 60.48<br \/>\nHence the mean of the distribution is 60.48 and standard deviation is 19.94.<\/p>\n<p>Question 7.<br \/>\nX is normally distributed with mean 12 and SD 4. Find P(X \u2264 20) and P(0 \u2264 X \u2264 12).<br \/>\nSolution:<br \/>\nGiven X ~ N (12, 42), (i.e) mean (\u00b5) = 12 and s.d (\u03c3) = 4.<br \/>\nP(X \u2264 20)<br \/>\n= P(\\(\\frac{X-\\mu}{\\sigma} \\leq \\frac{20-12}{4}\\))<br \/>\n= P(X \u2264 2)<br \/>\nNow P(Z \u2264 2) = P(Z \u2264 0) + P(0 \u2264 Z \u2264 2)<br \/>\n= 0.5 + 0.4772<br \/>\n= 0.9772<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41305\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q7.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q7\" width=\"380\" height=\"173\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q7.png 380w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q7-300x137.png 300w\" sizes=\"(max-width: 380px) 100vw, 380px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheerkalvi.Guru_.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3\" width=\"195\" height=\"28\" \/><\/p>\n<p>P(0 \u2264 X \u2264 12)<br \/>\n= P(\\(\\frac{0-12}{4} \\leq \\frac{X-\\mu}{\\sigma} \\leq \\frac{12-12}{4}\\))<br \/>\n= P (-3 \u2264 Z \u2264 0)<br \/>\n= P(0 \u2264 Z \u2264 3)<br \/>\nP(0 \u2264 Z \u2264 3) = 0.49865 (from normal tables)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41304\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q7.1.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q7.1\" width=\"382\" height=\"183\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q7.1.png 382w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q7.1-300x144.png 300w\" sizes=\"(max-width: 382px) 100vw, 382px\" \/><\/p>\n<p>Question 8.<br \/>\nIf 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<br \/>\n(a) greater than 72 inches<br \/>\n(b) less than or equal to 64 inches<br \/>\n(c) between 65 and 71 inches.<br \/>\nSolution:<br \/>\nGiven X is the normal random variables denoting the height of the students with mean \u00b5 = 68 and s.d (\u03c3) = 3.<br \/>\n(a) To find P(X &gt; 72)<br \/>\n= P(Z &gt; \\(\\frac{72-68}{3}\\))<br \/>\n= P(Z &gt; \\(\\frac{4}{3}\\))<br \/>\n= P(Z &gt; 1.33)<br \/>\nNow P(Z &gt; 1.33) = 0.5 &#8211; P(0 &lt; Z &lt; 1.33)<br \/>\n= 0.5 &#8211; 0.4082<br \/>\n= 0.0918<br \/>\nHence number of students whose height is greater than 72 inches is 500 \u00d7 0.0918 = 45.9 ~ 46<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41308\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q8\" width=\"383\" height=\"175\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.png 383w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8-300x137.png 300w\" sizes=\"(max-width: 383px) 100vw, 383px\" \/><\/p>\n<p>(b) To find P(X \u2264 64)<br \/>\n= P(Z \u2264 \\(\\frac{64-68}{3}\\))<br \/>\n= P(Z \u2264 -1.33)<br \/>\nBy Symmetry,<br \/>\nP(Z \u2264 -1.33) = P(Z \u2265 1.33) = 0.0918 (see before section)<br \/>\nHence number of students whose heights are less than or equal to 64 inches is 0.0918 \u00d7 500 = 46<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41306\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.1.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q8.1\" width=\"381\" height=\"175\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.1.png 381w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.1-300x138.png 300w\" sizes=\"(max-width: 381px) 100vw, 381px\" \/><\/p>\n<p>(c) To find P (65 &lt; X &lt; 71)<br \/>\n= P(\\(\\frac{65-68}{3}\\) &lt; Z &lt; \\(\\frac{71-68}{3}\\))<br \/>\n= P(-1 &lt; Z &lt; 1)<br \/>\n= P(-1 &lt; Z &lt; 0) + P(0 &lt; Z &lt; 1)<br \/>\n= 2P(0 &lt; Z &lt; 1) {By symmetry}<br \/>\n= 2 (0.3413)<br \/>\n= 0.6826<br \/>\nHence number of students with height between 65 and 71 inches is (500) (0.6826) = 341.3 ~ 342.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41307\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.2.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q8.2\" width=\"383\" height=\"173\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.2.png 383w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q8.2-300x136.png 300w\" sizes=\"(max-width: 383px) 100vw, 383px\" \/><\/p>\n<p>Question 9.<br \/>\nIn 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.<br \/>\nSolution:<br \/>\nLet X be the normal random variable denoting the developing time of the prints.<br \/>\nGiven that mean \u00b5 = 16.28 and s.d \u03c3 = 0.12<br \/>\nTo find P(X &lt; 16.35)<br \/>\nWhen X = 16.35, \\(\\mathrm{Z}=\\frac{\\mathrm{X}-\\mu}{\\sigma}=\\frac{16.35-16.28}{0.12}\\)<br \/>\nZ = 0.58<br \/>\nSo P(X &lt; 16.35) = P(Z &lt; 0.58)<br \/>\nNow P(Z &lt; 0.58) = 0.5 + P(0 &lt; Z &lt; 0.58)<br \/>\n= 0.5 + 0.2190<br \/>\n= 0.7190<br \/>\nThus the probability that it will take less than 16.35 seconds to develop prints is 0.719.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41309\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q9.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q9\" width=\"380\" height=\"170\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q9.png 380w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q9-300x134.png 300w\" sizes=\"(max-width: 380px) 100vw, 380px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheerkalvi.Guru_.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3\" width=\"195\" height=\"28\" \/><\/p>\n<p>Question 10.<br \/>\nTime 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 \u20b9 10,000 for each labour day spent in excess of 450. What is the probability that<br \/>\n(i) the company pays a penalty of at least \u20b9 2,00,000?<br \/>\n(ii) the company takes at most 500 days to complete the flyover?<br \/>\nSolution:<br \/>\nLet X be the normal variate denoting the number of labour days.<br \/>\nGiven mean \u00b5 = 400 and s.d \u03c3 = 100<br \/>\n(i) The company pays penalty of \u20b9 2,00,000 at the rate of \u20b9 10,000 per each extra labour day.<br \/>\nNo. of extra days = \\(\\frac{2,00,000}{10,000}\\) = 20<br \/>\nSo the probability that company pays a penalty of atleast \u20b9 2,00,000 is probability that labour days should be atleast 450 + 20 = 470 days, (i.e.) P (X \u2265 470)<br \/>\nNow P(X \u2265 470) = P(Z \u2265 \\(\\frac{470-400}{100}\\)) = P (Z \u2265 0.7)<br \/>\nP(Z \u2265 0.7) = 0.5 &#8211; P(0 \u2264 Z \u2264 0.7) = 0.5 &#8211; 0.258 = 0.242<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41311\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q10.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q10\" width=\"384\" height=\"172\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q10.png 384w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q10-300x134.png 300w\" sizes=\"(max-width: 384px) 100vw, 384px\" \/><\/p>\n<p>(ii) P (X \u2264 500) = P(Z \u2264 \\(\\frac{500-40}{100}\\)) = P (Z \u2264 1)<br \/>\nP(Z \u2264 1) = 0.5 + P (0 \u2264 Z \u2264 1) = 0.5 + 0.3413 = 0.8413<br \/>\nThus the probability that the company takes at most 500 days to complete the flyover is 0.8413.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-41310\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q10.1.png\" alt=\"Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3 Q10.1\" width=\"377\" height=\"174\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q10.1.png 377w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-12th-Business-Maths-Solutions-Chapter-7-Probability-Distributions-Ex-7.3-Q10.1-300x138.png 300w\" sizes=\"(max-width: 377px) 100vw, 377px\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>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 &hellip;<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[8],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/16464"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=16464"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/16464\/revisions"}],"predecessor-version":[{"id":60754,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/16464\/revisions\/60754"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=16464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=16464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=16464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}