{"id":36970,"date":"2024-12-29T04:32:39","date_gmt":"2024-12-28T23:02:39","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=36970"},"modified":"2024-12-30T10:09:31","modified_gmt":"2024-12-30T04:39:31","slug":"samacheer-kalvi-12th-business-maths-guide-chapter-7-ex-7-3","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-12th-business-maths-guide-chapter-7-ex-7-3\/","title":{"rendered":"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3"},"content":{"rendered":"<p>Tamilnadu State Board New Syllabus\u00a0<a href=\"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-12th-business-maths-guide\/\">Samacheer Kalvi 12th Business Maths Guide<\/a> Pdf Chapter 7 Probability Distributions Ex 7.3 Text Book Back Questions and Answers, Notes.<\/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 mean \u00b5 and varaince \u03c3\u00b2, if its probability density function is given by<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36972\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-1.png?resize=392%2C82&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 1\" width=\"392\" height=\"82\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-1.png?w=392&amp;ssl=1 392w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-1.png?resize=300%2C63&amp;ssl=1 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" data-recalc-dims=\"1\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2020\/01\/SamacheerKalvi.Guide_.png?resize=207%2C20&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3\" width=\"207\" height=\"20\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 2.<br \/>\nDefine standard normal variate.<br \/>\nSolution:<br \/>\nA random variable Z = (X &#8211; \u00b5)\/\u03c3 follows the standard normal distribution. Z is called the standard normal variate with mean 0 and standard deviation 1 i.e Z &#8211; N (0, 1). Its Probability density function is given by:<br \/>\n\u03c6(z) = \\(\\frac { 1 }{\\sqrt {2\u03c0}}\\) e<sup>-x\u00b2<\/sup>\/2 -\u221e &lt; z &lt; \u221e<\/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. m<br \/>\nWrite down any five characteristics of the normal probability curve.<br \/>\nSolution:<br \/>\nChief Characteristics or Properties of Normal Probability distribution and Normal probability Curve.<br \/>\nThe normal probability curve with mean \u00b5 and standard deviation \u03c3 has the following properties:<br \/>\n(i) the curve is bell-shaped and symmetrical about the line x = u.<br \/>\n(ii) Mean, median, and mode of the distribution coincide.<br \/>\n(iii) x-axis is an asymptote to the curve, (tails of the curve never touches the horizontal (x) axis)<br \/>\n(iv) No portion of the curve lies below the x-axis as f(x) being the probability function can never be negative.<br \/>\n(v) The points of inflexion of the curve are x = \u00b5 \u00b1 \u03c3<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2020\/01\/SamacheerKalvi.Guide_.png?resize=207%2C20&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3\" width=\"207\" height=\"20\" data-recalc-dims=\"1\" \/><\/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 denote the burning of the bulb follows normal distribution with mean 2,040 and standard deviation 60 hours.<br \/>\nHere m = 2040; \u03c3 = 60 and N = 2000<br \/>\nThe standard normal variate<br \/>\nz = \\(\\frac { x-\u00b5 }{\u03c3}\\) = \\(\\frac { x-2040 }{60}\\)<br \/>\n(i) p(morethan 2,150 hours)<br \/>\np(x &gt; 2150)<br \/>\nwhen x = 2150<br \/>\nz = \\(\\frac { 2150-2040 }{60}\\) = \\(\\frac { 110 }{60}\\)= 1.833<br \/>\np(x &gt; 2150) = p(z &gt; 1.833)<br \/>\n= p(0 &lt; z &lt; \u221e) &#8211; p(0 &lt; z &lt; 1.833)<br \/>\n= 0.5 &#8211; 0.4664<br \/>\n= 0.0336<br \/>\n\u2234 Number of bulbs whose burning time is more than 2150 hours=0.0336 \u00d7 2000<br \/>\n= 67.2 = 67(approximately)<\/p>\n<p>(ii) p(less than 1950 hours)<br \/>\np(x &lt; 1950)<br \/>\nwhen x = 1950<br \/>\nz = \\(\\frac { 1950-3040 }{60}\\) = \\(\\frac { -90 }{60}\\)= -1.5<br \/>\np(x &lt; 1950) = p(z &lt; -1.5) = p(z &gt; 1.5)<br \/>\n= 0.5 &#8211; 0.4332<br \/>\n= 0.068<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36973\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-2.png?resize=374%2C120&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 2\" width=\"374\" height=\"120\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-2.png?w=374&amp;ssl=1 374w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-2.png?resize=300%2C96&amp;ssl=1 300w\" sizes=\"(max-width: 374px) 100vw, 374px\" data-recalc-dims=\"1\" \/><br \/>\nNumbers of bulbs whose burning time is less than<br \/>\n1950 = 0.0668 \u00d7 2000 = 133.6<br \/>\n= 134 (approximately)<\/p>\n<p>(iii) p(more 1,920 hours but less than 2,100 hours)<br \/>\n= p(1920 &lt; x &lt; 2100)<br \/>\nwhen x = 1950<br \/>\nz = \\(\\frac { 1920-2040 }{60}\\) = \\(\\frac { -120 }{60}\\)= -2<br \/>\nwhen x = 2100<br \/>\nz = \\(\\frac { 2100-2040 }{60}\\) = \\(\\frac { -60 }{60}\\)= -2<br \/>\n\u2234 p(1920 &lt; x &lt; 2040) = p(-2 &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 \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36974\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-3.png?resize=371%2C120&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 3\" width=\"371\" height=\"120\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-3.png?w=371&amp;ssl=1 371w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-3.png?resize=300%2C97&amp;ssl=1 300w\" sizes=\"(max-width: 371px) 100vw, 371px\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 Number of bulbs whose burning time more than 1920 hours but less than 2100 hours) = 0.8185 \u00d7 2000<br \/>\n= 1637<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2020\/01\/SamacheerKalvi.Guide_.png?resize=207%2C20&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3\" width=\"207\" height=\"20\" data-recalc-dims=\"1\" \/><\/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 \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36975\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-4.png?resize=367%2C147&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 4\" width=\"367\" height=\"147\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-4.png?w=367&amp;ssl=1 367w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-4.png?resize=300%2C120&amp;ssl=1 300w\" sizes=\"(max-width: 367px) 100vw, 367px\" data-recalc-dims=\"1\" \/><br \/>\nz = \\(\\frac { x-\u00b5 }{\u03c3}\\)<br \/>\nGiven that<br \/>\np(x &lt; 50) = 0.3 p(x &gt; 86) = 0.1<br \/>\np(z &lt; -c) = 0.3<br \/>\np(-c &lt; z &lt; 0) = 0.5 &#8211; 0.3<br \/>\np(-c &lt; z &lt; 0) = 0.2 {from the table}<br \/>\np(0 &lt; z &lt; c) = 0.2<br \/>\nc = 0.53<br \/>\nthen -c = -0.53<br \/>\n\u2234 \\(\\frac { 50-\u00b5 }{\u03c3}\\) = -0.53<br \/>\n50 &#8211; \u00b5 = -0.53\u03c3<br \/>\n\u00b5 &#8211; 0.53\u03c3 = 50 \u2192 1<br \/>\np(x &lt; 50) = 0.1<br \/>\np(0 &lt; z &lt; \u221e) = -p(0 &lt; z &lt; c<sub>1<\/sub>) = 0.1<br \/>\np(0 &lt; z &lt; \u221e) = p(0 &lt; z &lt; c<sub>1<\/sub>) + 0.1<br \/>\n0.5 = p(0 &lt; z &lt; c<sub>1<\/sub>) + 0.1<br \/>\np(0 &lt; z &lt; c<sub>1<\/sub>) = 0.5 &#8211; 0.1<br \/>\np = (0 &lt; z &lt; c<sub>1<\/sub>) = 0.4<br \/>\nc<sub>1<\/sub> = 1.29<br \/>\n\u2234 \\(\\frac { 86-\u00b5 }{\u03c3}\\) = 1.29<br \/>\n86 &#8211; \u00b5 = 1.29 \u03c3<br \/>\n\u00b5 + 1.29\u03c3 = 86 \u2192 2<br \/>\nsolving eqn 1 &amp; 2<br \/>\neqn 2 \u21d2 m + 1.29\u03c3 = 86<br \/>\neqn 1 \u21d2 m + 0.53\u03c3 = 50<br \/>\n&#8211; + &#8211; &#8230;&#8230;&#8230;&#8230;.<br \/>\n&#8230;&#8230;&#8230;&#8230; 1.82 &#8230;&#8230;&#8230;&#8230; \u03c3 = 36 &#8230;&#8230;&#8230;&#8230;.<br \/>\n\u03c3 = \\(\\frac { 36 }{1.82}\\) \u2234 = 19.78<br \/>\nSubstitute \u03c3 = 19.78 in eqn 1<br \/>\n\u00b5 &#8211; 0.53(19.78) = 50<br \/>\n\u00b5 -10.48 = 50<br \/>\n\u00b5 = 50 + 10.48<br \/>\n\u00b5 = 60.48<br \/>\nMean = 60.48 and standard deviation = 19.78<\/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 \/>\nx is normally distribution with mean 12 and sd 4<br \/>\n\u2234 \u00b5 = 12 and \u03c3 = 4<br \/>\nStandard normal variable<br \/>\nz = \\(\\frac { x-\u00b5 }{\u03c3}\\) = \\(\\frac { x-12 }{4}\\)<br \/>\n(i) p(x \u2264 20)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36976\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-5.png?resize=377%2C163&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 5\" width=\"377\" height=\"163\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-5.png?w=377&amp;ssl=1 377w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-5.png?resize=300%2C130&amp;ssl=1 300w\" sizes=\"(max-width: 377px) 100vw, 377px\" data-recalc-dims=\"1\" \/><br \/>\nwhen x = 20<br \/>\nz = \\(\\frac { 20-12 }{4}\\) = \\(\\frac { 8 }{4}\\) = 2<br \/>\nv(x \u2264 20) = \\(\\frac { 8 }{4}\\) = 2<br \/>\np(x \u2264 20) = p(z \u2264 2)<br \/>\n= 0.5 + p(0 &lt; z &lt; 2)<br \/>\n= 0.5 + 0.4772<br \/>\n= 0.9772<\/p>\n<p>(ii) p(0 &lt; x &lt; 12 )<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36977\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-6.png?resize=371%2C171&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 6\" width=\"371\" height=\"171\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-6.png?w=371&amp;ssl=1 371w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-6.png?resize=300%2C138&amp;ssl=1 300w\" sizes=\"(max-width: 371px) 100vw, 371px\" data-recalc-dims=\"1\" \/><br \/>\nwhen x = 0<br \/>\nz = \\(\\frac { 0-12 }{4}\\) = \\(\\frac { -12 }{4}\\) = -3<br \/>\nwhen x = 12<br \/>\nz = \\(\\frac { 12-12 }{4}\\) = \\(\\frac { 0 }{4}\\) = 0<br \/>\np(0 \u2264 x \u2264 12) = p(-3 \u2264 z \u2264 0)<br \/>\n= p(0 \u2264 z \u2264 3)<br \/>\n= 0.4987<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2020\/01\/SamacheerKalvi.Guide_.png?resize=207%2C20&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3\" width=\"207\" height=\"20\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 8.<br \/>\nIf the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height<br \/>\n(a) greater than 2 inches<br \/>\n(b) less than or equal to 64 inches<br \/>\n(c) between 65 and 71 inches.<br \/>\nSolution:<br \/>\nlet x denote the height of a student N = 500; m = 68.0 inches and \u03c3 = 3.0 inches the standard normal variate<br \/>\nz = \\(\\frac { x-\u00b5 }{\u03c3}\\) = \\(\\frac { x-68 }{3}\\)<br \/>\na(greater than 72 inches)<br \/>\np = p(x &gt; 72)<br \/>\nwhen x = 72<br \/>\nz = \\(\\frac { 72-68 }{3}\\) = \\(\\frac { 4 }{3}\\) = 1.33<br \/>\np(x &gt; 72) = p(z &gt; 1.33)<br \/>\n= 0.5 &#8211; 0.4082<br \/>\n= 0.0918<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36978\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-7.png?resize=374%2C164&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 7\" width=\"374\" height=\"164\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-7.png?w=374&amp;ssl=1 374w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-7.png?resize=300%2C132&amp;ssl=1 300w\" sizes=\"(max-width: 374px) 100vw, 374px\" data-recalc-dims=\"1\" \/><br \/>\nNumber of students whose height are greater than 72 inches<br \/>\n= 0.0918 \u00d7 500<br \/>\n= 45.9<br \/>\n= 46 (approximately)<\/p>\n<p>(b) p(less than or equal to 64 inches)<br \/>\np(x \u2264 64)<br \/>\nwhen x = 64<br \/>\nz = \\(\\frac { 64-68 }{3}\\) = \\(\\frac { -4 }{3}\\) = -1.33<br \/>\np(x \u2264 64) = p(z \u2264 -1.33)<br \/>\np(z \u2265 -1.33)<br \/>\n= 0.5 &#8211; 0.4082<br \/>\n= 0.0918<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36979\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-8.png?resize=371%2C167&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 8\" width=\"371\" height=\"167\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-8.png?w=371&amp;ssl=1 371w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-8.png?resize=300%2C135&amp;ssl=1 300w\" sizes=\"(max-width: 371px) 100vw, 371px\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 Number of heights whose ate less than or equal to 64 inches = 0.0918 \u00d7 500<br \/>\n= 45.9<br \/>\n= 46 (approximately)<\/p>\n<p>(c) p(between 65 and 71 inches)<br \/>\np(65 \u2264 x \u2264 71)<br \/>\nwhen x = 65<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36980\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-9.png?resize=372%2C164&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 9\" width=\"372\" height=\"164\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-9.png?w=372&amp;ssl=1 372w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-9.png?resize=300%2C132&amp;ssl=1 300w\" sizes=\"(max-width: 372px) 100vw, 372px\" data-recalc-dims=\"1\" \/><br \/>\nz = \\(\\frac { 65-68 }{3}\\) = \\(\\frac { -3 }{3}\\) = -1<br \/>\nwhen x = 71<br \/>\nz = \\(\\frac { 71-68 }{3}\\) = \\(\\frac { 3 }{3}\\) = 1<br \/>\np(65 \u2264 x \u2264 71) = p(-1 &lt; z &lt; 1)<br \/>\n= p(-1 &lt; z &lt; 0) + p(0 &lt; z &lt; 1)<br \/>\n= p(0 &lt; z &lt; 1) + p(0 &lt; z &lt; 1)<br \/>\n= 2 \u00d7 [p(0 &lt; z &lt; 1)]<br \/>\n= 2 \u00d7 0.3413<br \/>\n= 0.6826<br \/>\n\u2234 Number of students whose height between 65 and 7 inches = 0.6826 \u00d7 500<br \/>\n= 341.3<br \/>\n= 342 (approximately)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2020\/01\/SamacheerKalvi.Guide_.png?resize=207%2C20&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3\" width=\"207\" height=\"20\" data-recalc-dims=\"1\" \/><\/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 random variable have long the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36981\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-10.png?resize=374%2C166&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 10\" width=\"374\" height=\"166\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-10.png?w=374&amp;ssl=1 374w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-10.png?resize=300%2C133&amp;ssl=1 300w\" sizes=\"(max-width: 374px) 100vw, 374px\" data-recalc-dims=\"1\" \/><br \/>\n\u00b5 = 16,28 and \u03c3 = 0.12<br \/>\nThe standard normal variate<br \/>\nz = \\(\\frac { x-\u00b5 }{\u03c3}\\) = \\(\\frac { x-16.28 }{0.12}\\) = 1<br \/>\np(less than 16.35 seconds) = p(x &lt; 16.35)<br \/>\nwhen x = 16.35<br \/>\nz = \\(\\frac { 16.35-16.28 }{0.12}\\) = \\(\\frac { 0.07 }{0.12}\\) = 0.583<br \/>\np(x&lt; 16.35) = p(z &lt; 0.583)<br \/>\n= p(\u2014\u221e &lt; z &lt; 0) + p(0 &lt; z &lt; 0.583)<br \/>\n= 0.5 + 0.2190<br \/>\n= 0.7190<\/p>\n<p>Question 10.<br \/>\nIf the heights of 500 students are normally distributed with mean of 68.0 inches and standard deviation of 3.0 inches, how many students have a height (a) greater than 2 inches (b) less than or equal to 64 inches (c) between 65 and 71 inches.<br \/>\nSolution:<br \/>\nLet x be a normal variate with mean of 400 labour days and a standard deviation of 100 labour days<br \/>\nm = 400 and \u03c3 = 100<br \/>\nThe construction work should be completed within 450 days.<br \/>\nThe standard normal variate<br \/>\n\\(\\frac { x-\u00b5 }{6}\\) = \\(\\frac { x-400 }{100}\\)<br \/>\npersonality for 1 labour day = Rs 10,000<br \/>\nIf personality amount is = 2,00,000 than No of excess<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36982\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-11.png?resize=372%2C162&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 11\" width=\"372\" height=\"162\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-11.png?w=372&amp;ssl=1 372w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-11.png?resize=300%2C131&amp;ssl=1 300w\" sizes=\"(max-width: 372px) 100vw, 372px\" data-recalc-dims=\"1\" \/><br \/>\ndays = \\(\\frac { 200000 }{10000}\\) = 20<br \/>\n\u2234 x = 450 + 20 = 470<br \/>\nwhen x = 470<br \/>\nz = \\(\\frac { 470-400 }{100}\\) = \\(\\frac { 70 }{100}\\) = 0.7<br \/>\n= p(x \u2265 470) = p(z \u2265 0.7)<br \/>\n= 0.5 &#8211; 0.2580<br \/>\n= 0.2420<\/p>\n<p>(ii) p(at most 500 days) = p(x \u2264 500 )<br \/>\nwhen x = 500<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-36983\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-12.png?resize=382%2C165&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3 12\" width=\"382\" height=\"165\" srcset=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-12.png?w=382&amp;ssl=1 382w, https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2021\/03\/Samacheer-Kalvi-12th-Business-Maths-Guide-Chapter-7-Probability-Distributions-Ex-7.3-12.png?resize=300%2C130&amp;ssl=1 300w\" sizes=\"(max-width: 382px) 100vw, 382px\" data-recalc-dims=\"1\" \/><br \/>\nz = \\(\\frac { 500-400 }{100}\\) = \\(\\frac { 100 }{100}\\) = 1<br \/>\np(x \u2264 500) = p(z \u2264 1)<br \/>\n= p(\u221e &lt; z &lt; 0) -r- p(0 &lt; z &lt; 1)<br \/>\n= 0.5 + 0.3415<br \/>\n= 0.8413<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/samacheerkalvi.guide\/wp-content\/uploads\/2020\/01\/SamacheerKalvi.Guide_.png?resize=207%2C20&#038;ssl=1\" alt=\"Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.3\" width=\"207\" height=\"20\" data-recalc-dims=\"1\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.3 Text Book Back Questions and Answers, Notes. 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 &hellip;<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[5],"tags":[],"class_list":["post-36970","post","type-post","status-publish","format-standard","hentry","category-class-12"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/36970"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=36970"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/36970\/revisions"}],"predecessor-version":[{"id":41749,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/36970\/revisions\/41749"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=36970"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=36970"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=36970"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}