{"id":68124,"date":"2021-08-30T10:24:00","date_gmt":"2021-08-30T04:54:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68124"},"modified":"2021-08-31T12:48:23","modified_gmt":"2021-08-31T07:18:23","slug":"rd-sharma-solutions-class-11-maths-chapter-32-exercise-32-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-exercise-32-1\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121393\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-32-Exercise-32.1.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-32-Exercise-32.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-32-Exercise-32.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1: <\/strong>We&#8217;ll discuss about dispersion, range, and mean deviation in this exercise. <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\"><strong>RD Sharma Class 11 Maths Solutions<\/strong><\/a> can help students gain a better understanding of the concepts and develop a good understanding of the subject. Subject experts have prepared the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-statistics\/\"><strong>RD Sharma Solutions Class 11 Maths Chapter 32<\/strong><\/a> Exercise 32.1 in straightforward, easy-to-understand language to fulfil the needs of students and assist them in achieving good grades on their board exams. The answers to this exercise are available in pdf format, which can be simply downloaded from the links provided below.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69da539a22221\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" 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Chapter 32 Statistics Exercise 32.1 Free PDF\">Download RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-exercise-32-1\/#access-answers-to-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1- Important Question with Answers\">Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-exercise-32-1\/#faqs-on-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321\" title=\"FAQs on RD 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Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 for free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-exercise-32-1\/#where-can-i-download-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 free PDF?\">Where can I download RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-11-solutions-chapter-32-statistics-exercise-321-free-pdf\"><\/span>Download RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-32-Ex-32.1-1.pdf\">RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-32-Ex-32.1-1.pdf\", \"#example1\");<\/script><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Calculate the mean deviation about the median of the following observation :<br>(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000<\/strong><\/p>\n<p><strong>(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44<\/strong><\/p>\n<p><strong>(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51<\/strong><\/p>\n<p><strong>(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42<\/strong><\/p>\n<p><strong>(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)&nbsp;<\/strong>3011, 2780, 3020, 2354, 3541, 4150, 5000<\/p>\n<p>To calculate the Median (M), let us arrange the numbers in ascending order.<\/p>\n<p>Median is the middle number of all the observation.<\/p>\n<p>2354, 2780, 3011, 3020, 3541, 4150, 5000<\/p>\n<p>So, Median = 3020 and n = 7<\/p>\n<p>By using the formula to calculate Mean Deviation,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 3020|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>3011<\/p>\n<\/td>\n<td>\n<p>9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>2780<\/p>\n<\/td>\n<td>\n<p>240<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>3020<\/p>\n<\/td>\n<td>\n<p>0<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>2354<\/p>\n<\/td>\n<td>\n<p>666<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>3541<\/p>\n<\/td>\n<td>\n<p>521<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4150<\/p>\n<\/td>\n<td>\n<p>1130<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>5000<\/p>\n<\/td>\n<td>\n<p>1980<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>4546<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/7 \u00d7 4546<\/p>\n<p>= 649.42<\/p>\n<p>\u2234 The Mean Deviation is 649.42.<\/p>\n<p><strong>(ii)&nbsp;<\/strong>38, 70, 48, 34, 42, 55, 63, 46, 54, 44<\/p>\n<p>To calculate the Median (M), let us arrange the numbers in ascending order.<\/p>\n<p>Median is the middle number of all the observation.<\/p>\n<p>34, 38, 42, 44, 46, 48, 54, 55, 63, 70<\/p>\n<p>Here the Number of observations are Even then Median = (46+48)\/2 = 47<\/p>\n<p>Median = 47 and n = 10<\/p>\n<p>By using the formula to calculate Mean Deviation,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 47|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>38<\/p>\n<\/td>\n<td>\n<p>9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>70<\/p>\n<\/td>\n<td>\n<p>23<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>48<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>34<\/p>\n<\/td>\n<td>\n<p>13<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>55<\/p>\n<\/td>\n<td>\n<p>8<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>63<\/p>\n<\/td>\n<td>\n<p>16<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>46<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>54<\/p>\n<\/td>\n<td>\n<p>7<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>44<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>86<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 86<\/p>\n<p>= 8.6<\/p>\n<p>\u2234 The Mean Deviation is 8.6.<\/p>\n<p><strong>(iii)&nbsp;<\/strong>34, 66, 30, 38, 44, 50, 40, 60, 42, 51<\/p>\n<p>To calculate the Median (M), let us arrange the numbers in ascending order.<\/p>\n<p>Median is the middle number of all the observation.<\/p>\n<p>30, 34, 38, 40, 42, 44, 50, 51, 60, 66<\/p>\n<p>Here the Number of observations are Even then Median = (42+44)\/2 = 43<\/p>\n<p>Median = 43 and n = 10<\/p>\n<p>By using the formula to calculate Mean Deviation,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 43|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>30<\/p>\n<\/td>\n<td>\n<p>13<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>34<\/p>\n<\/td>\n<td>\n<p>9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>38<\/p>\n<\/td>\n<td>\n<p>5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>40<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>44<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>50<\/p>\n<\/td>\n<td>\n<p>7<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>51<\/p>\n<\/td>\n<td>\n<p>8<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>60<\/p>\n<\/td>\n<td>\n<p>17<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>66<\/p>\n<\/td>\n<td>\n<p>23<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>87<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 87<\/p>\n<p>= 8.7<\/p>\n<p>\u2234 The Mean Deviation is 8.7.<\/p>\n<p><strong>(iv)&nbsp;<\/strong>22, 24, 30, 27, 29, 31, 25, 28, 41, 42<\/p>\n<p>To calculate the Median (M), let us arrange the numbers in ascending order.<\/p>\n<p>Median is the middle number of all the observation.<\/p>\n<p>22, 24, 25, 27, 28, 29, 30, 31, 41, 42<\/p>\n<p>Here the Number of observations are Even then Median = (28+29)\/2 = 28.5<\/p>\n<p>Median = 28.5 and n = 10<\/p>\n<p>By using the formula to calculate Mean Deviation,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 28.5|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>22<\/p>\n<\/td>\n<td>\n<p>6.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>24<\/p>\n<\/td>\n<td>\n<p>4.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>30<\/p>\n<\/td>\n<td>\n<p>1.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>27<\/p>\n<\/td>\n<td>\n<p>1.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>29<\/p>\n<\/td>\n<td>\n<p>0.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>31<\/p>\n<\/td>\n<td>\n<p>2.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>25<\/p>\n<\/td>\n<td>\n<p>3.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>28<\/p>\n<\/td>\n<td>\n<p>0.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>41<\/p>\n<\/td>\n<td>\n<p>12.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>13.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>47<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 47<\/p>\n<p>= 4.7<\/p>\n<p>\u2234 The Mean Deviation is 4.7.<\/p>\n<p><strong>(v)&nbsp;<\/strong>38, 70, 48, 34, 63, 42, 55, 44, 53, 47<\/p>\n<p>To calculate the Median (M), let us arrange the numbers in ascending order.<\/p>\n<p>Median is the middle number of all the observation.<\/p>\n<p>34, 38, 43, 44, 47, 48, 53, 55, 63, 70<\/p>\n<p>Here the Number of observations are Even then Median = (47+48)\/2 = 47.5<\/p>\n<p>Median = 47.5 and n = 10<\/p>\n<p>By using the formula to calculate Mean Deviation,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 47.5|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>38<\/p>\n<\/td>\n<td>\n<p>9.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>70<\/p>\n<\/td>\n<td>\n<p>22.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>48<\/p>\n<\/td>\n<td>\n<p>0.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>34<\/p>\n<\/td>\n<td>\n<p>13.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>63<\/p>\n<\/td>\n<td>\n<p>15.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>5.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>55<\/p>\n<\/td>\n<td>\n<p>7.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>44<\/p>\n<\/td>\n<td>\n<p>3.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>53<\/p>\n<\/td>\n<td>\n<p>5.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>47<\/p>\n<\/td>\n<td>\n<p>0.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>84<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 84<\/p>\n<p>= 8.4<\/p>\n<p>\u2234 The Mean Deviation is 8.4.<\/p>\n<p><strong>2. Calculate the mean deviation from the mean for the following data :<br>(i) 4, 7, 8, 9, 10, 12, 13, 17<\/strong><\/p>\n<p><strong>(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17<\/strong><\/p>\n<p><strong>(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44<\/strong><\/p>\n<p><strong>(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49<br>(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)&nbsp;<\/strong>4, 7, 8, 9, 10, 12, 13, 17<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]\/8<\/p>\n<p>= 80\/8<\/p>\n<p>= 10<\/p>\n<p>Number of observations, \u2018n\u2019 = 8<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 10|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4<\/p>\n<\/td>\n<td>\n<p>6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>7<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>8<\/p>\n<\/td>\n<td>\n<p>2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>9<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>10<\/p>\n<\/td>\n<td>\n<p>0<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>12<\/p>\n<\/td>\n<td>\n<p>2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>13<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>17<\/p>\n<\/td>\n<td>\n<p>7<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>24<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/8 \u00d7 24<\/p>\n<p>= 3<\/p>\n<p>\u2234 The Mean Deviation is 3.<\/p>\n<p><strong>(ii)&nbsp;<\/strong>13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17]\/12<\/p>\n<p>= 168\/12<\/p>\n<p>= 14<\/p>\n<p>Number of observations, \u2018n\u2019 = 12<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 14|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>13<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>17<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>16<\/p>\n<\/td>\n<td>\n<p>2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>14<\/p>\n<\/td>\n<td>\n<p>0<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>11<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>13<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>10<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>16<\/p>\n<\/td>\n<td>\n<p>2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>11<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>18<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>12<\/p>\n<\/td>\n<td>\n<p>2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>17<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>28<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/12 \u00d7 28<\/p>\n<p>= 2.33<\/p>\n<p>\u2234 The Mean Deviation is 2.33.<\/p>\n<p><strong>(iii)&nbsp;<\/strong>38, 70, 48, 40, 42, 55, 63, 46, 54, 44<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44]\/10<\/p>\n<p>= 500\/10<\/p>\n<p>= 50<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 50|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>38<\/p>\n<\/td>\n<td>\n<p>12<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>70<\/p>\n<\/td>\n<td>\n<p>20<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>48<\/p>\n<\/td>\n<td>\n<p>2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>40<\/p>\n<\/td>\n<td>\n<p>10<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>8<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>55<\/p>\n<\/td>\n<td>\n<p>5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>63<\/p>\n<\/td>\n<td>\n<p>13<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>46<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>54<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>44<\/p>\n<\/td>\n<td>\n<p>6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>84<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 84<\/p>\n<p>= 8.4<\/p>\n<p>\u2234 The Mean Deviation is 8.4.<\/p>\n<p><strong>(iv)&nbsp;<\/strong>36, 72, 46, 42, 60, 45, 53, 46, 51, 49<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [36 + 72 + 46 + 42 + 60 + 45 + 53 + 46 + 51 + 49]\/10<\/p>\n<p>= 500\/10<\/p>\n<p>= 50<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 50|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>36<\/p>\n<\/td>\n<td>\n<p>14<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>72<\/p>\n<\/td>\n<td>\n<p>22<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>46<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>8<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>60<\/p>\n<\/td>\n<td>\n<p>10<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>45<\/p>\n<\/td>\n<td>\n<p>5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>53<\/p>\n<\/td>\n<td>\n<p>3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>46<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>51<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>49<\/p>\n<\/td>\n<td>\n<p>1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>72<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 72<\/p>\n<p>= 7.2<\/p>\n<p>\u2234 The Mean Deviation is 7.2.<\/p>\n<p><strong>(v)&nbsp;<\/strong>57, 64, 43, 67, 49, 59, 44, 47, 61, 59<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59]\/10<\/p>\n<p>= 550\/10<\/p>\n<p>= 55<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 55|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>57<\/p>\n<\/td>\n<td>\n<p>2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>64<\/p>\n<\/td>\n<td>\n<p>9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>43<\/p>\n<\/td>\n<td>\n<p>12<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>67<\/p>\n<\/td>\n<td>\n<p>12<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>49<\/p>\n<\/td>\n<td>\n<p>6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>59<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>44<\/p>\n<\/td>\n<td>\n<p>11<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>47<\/p>\n<\/td>\n<td>\n<p>8<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>61<\/p>\n<\/td>\n<td>\n<p>6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>59<\/p>\n<\/td>\n<td>\n<p>4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>74<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 74<\/p>\n<p>= 7.4<\/p>\n<p>\u2234 The Mean Deviation is 7.4.<\/p>\n<p><strong>3.<\/strong>&nbsp;<strong>Calculate the mean deviation of the following income groups of five and seven members from their medians:<\/strong><\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p><strong>I<\/strong><\/p>\n<p><strong>Income in \u20b9<\/strong><\/p>\n<\/td>\n<td>\n<p><strong>II<\/strong><\/p>\n<p><strong>Income in \u20b9<\/strong><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4000<\/p>\n<\/td>\n<td>\n<p>3800<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4200<\/p>\n<\/td>\n<td>\n<p>4000<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4400<\/p>\n<\/td>\n<td>\n<p>4200<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4600<\/p>\n<\/td>\n<td>\n<p>4400<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4800<\/p>\n<\/td>\n<td>\n<p>4600<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4800<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>5800<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us calculate the mean deviation for the first data set.<\/p>\n<p>Since the data is arranged in ascending order,<\/p>\n<p>4000, 4200, 4400, 4600, 4800<\/p>\n<p>Median = 4400<\/p>\n<p>Total observations = 5<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 M|<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 4400|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4000<\/p>\n<\/td>\n<td>\n<p>400<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4200<\/p>\n<\/td>\n<td>\n<p>200<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4400<\/p>\n<\/td>\n<td>\n<p>0<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4600<\/p>\n<\/td>\n<td>\n<p>200<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4800<\/p>\n<\/td>\n<td>\n<p>400<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>1200<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/5 \u00d7 1200<\/p>\n<p>= 240<\/p>\n<p>Let us calculate the mean deviation for the second data set.<\/p>\n<p>Since the data is arranged in ascending order,<\/p>\n<p>3800, 4000, 4200, 4400, 4600, 4800, 5800<\/p>\n<p>Median = 4400<\/p>\n<p>Total observations = 7<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 M|<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 4400|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>3800<\/p>\n<\/td>\n<td>\n<p>600<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4000<\/p>\n<\/td>\n<td>\n<p>400<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4200<\/p>\n<\/td>\n<td>\n<p>200<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4400<\/p>\n<\/td>\n<td>\n<p>0<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4600<\/p>\n<\/td>\n<td>\n<p>200<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>4800<\/p>\n<\/td>\n<td>\n<p>400<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>5800<\/p>\n<\/td>\n<td>\n<p>1400<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>3200<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/7 \u00d7 3200<\/p>\n<p>= 457.14<\/p>\n<p>\u2234 The Mean Deviation of set 1 is 240 and set 2 is 457.14<\/p>\n<p><strong>4. The lengths (in cm) of 10 rods in a shop are given below:<br>40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2<br>(i)&nbsp;Find the mean deviation from the median.<br>(ii)&nbsp;Find the mean deviation from the mean also.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)&nbsp;<\/strong>Find the mean deviation from the median<\/p>\n<p>Let us arrange the data in ascending order,<\/p>\n<p>15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 M|<\/p>\n<p>The number of observations are Even then Median&nbsp;= (40+52.3)\/2 = 46.15<\/p>\n<p>Median = 46.15<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 46.15|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>40.0<\/p>\n<\/td>\n<td>\n<p>6.15<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>52.3<\/p>\n<\/td>\n<td>\n<p>6.15<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>55.2<\/p>\n<\/td>\n<td>\n<p>9.05<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>72.9<\/p>\n<\/td>\n<td>\n<p>26.75<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>52.8<\/p>\n<\/td>\n<td>\n<p>6.65<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>79.0<\/p>\n<\/td>\n<td>\n<p>32.85<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>32.5<\/p>\n<\/td>\n<td>\n<p>13.65<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>15.2<\/p>\n<\/td>\n<td>\n<p>30.95<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>27.9<\/p>\n<\/td>\n<td>\n<p>19.25<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>30.2<\/p>\n<\/td>\n<td>\n<p>15.95<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>167.4<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 167.4<\/p>\n<p>= 16.74<\/p>\n<p>\u2234 The Mean Deviation is 16.74.<\/p>\n<p><strong>(ii)&nbsp;<\/strong>Find the mean deviation from the mean also.<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [40.0 + 52.3 + 55.2 + 72.9 + 52.8 + 79.0 + 32.5 + 15.2 + 27.9 + 30.2]\/10<\/p>\n<p>= 458\/10<\/p>\n<p>= 45.8<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 45.8|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>40.0<\/p>\n<\/td>\n<td>\n<p>5.8<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>52.3<\/p>\n<\/td>\n<td>\n<p>6.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>55.2<\/p>\n<\/td>\n<td>\n<p>9.4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>72.9<\/p>\n<\/td>\n<td>\n<p>27.1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>52.8<\/p>\n<\/td>\n<td>\n<p>7<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>79.0<\/p>\n<\/td>\n<td>\n<p>33.2<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>32.5<\/p>\n<\/td>\n<td>\n<p>13.3<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>15.2<\/p>\n<\/td>\n<td>\n<p>30.6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>27.9<\/p>\n<\/td>\n<td>\n<p>17.9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>30.2<\/p>\n<\/td>\n<td>\n<p>15.6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>166.4<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 166.4<\/p>\n<p>= 16.64<\/p>\n<p>\u2234 TheMean Deviation is 16.64<\/p>\n<p><strong>5. In question 1(iii), (iv), (v) find the number of observations lying between&nbsp;X\u00af\u00af\u00af\u00af\u2013M.D&nbsp;and&nbsp;X\u00af\u00af\u00af\u00af+M.D, where M.D. is the mean deviation from the mean.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(iii)<\/strong>&nbsp;34, 66, 30, 38, 44, 50, 40, 60, 42, 51<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]\/10<\/p>\n<p>= 455\/10<\/p>\n<p>= 45.5<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 45.5|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>34<\/p>\n<\/td>\n<td>\n<p>11.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>66<\/p>\n<\/td>\n<td>\n<p>20.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>30<\/p>\n<\/td>\n<td>\n<p>15.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>38<\/p>\n<\/td>\n<td>\n<p>7.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>44<\/p>\n<\/td>\n<td>\n<p>1.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>50<\/p>\n<\/td>\n<td>\n<p>4.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>40<\/p>\n<\/td>\n<td>\n<p>5.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>60<\/p>\n<\/td>\n<td>\n<p>14.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>3.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>51<\/p>\n<\/td>\n<td>\n<p>5.5<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>90<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 90<\/p>\n<p>= 9<\/p>\n<p>Now<\/p>\n<p>X\u00af\u00af\u00af\u00af\u2013M.D&nbsp;= 45.5 \u2013 9 = 36.5<\/p>\n<p>X\u00af\u00af\u00af\u00af+M.D&nbsp;= 45.5 + 9 = 54.5<\/p>\n<p>So, There are total 6 observation between&nbsp;X\u00af\u00af\u00af\u00af\u2013M.D&nbsp;and<strong>&nbsp;X\u00af\u00af\u00af\u00af+M.D<\/strong><\/p>\n<p><strong>(iv)&nbsp;<\/strong>22, 24, 30, 27, 29, 31, 25, 28, 41, 42<\/p>\n<p>We know that,<\/p>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]\/10<\/p>\n<p>= 299\/10<\/p>\n<p>= 29.9<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 29.9|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>22<\/p>\n<\/td>\n<td>\n<p>7.9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>24<\/p>\n<\/td>\n<td>\n<p>5.9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>30<\/p>\n<\/td>\n<td>\n<p>0.1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>27<\/p>\n<\/td>\n<td>\n<p>2.9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>29<\/p>\n<\/td>\n<td>\n<p>0.9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>31<\/p>\n<\/td>\n<td>\n<p>1.1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>25<\/p>\n<\/td>\n<td>\n<p>4.9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>28<\/p>\n<\/td>\n<td>\n<p>1.9<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>41<\/p>\n<\/td>\n<td>\n<p>11.1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>12.1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>48.8<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 48.8<\/p>\n<p>= 4.88<\/p>\n<p>Now<\/p>\n<p><img class=\"alignnone size-full wp-image-121410\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1-1.png\" alt=\"1\" width=\"255\" height=\"64\"><\/p>\n<p>So, There are total 5 observation between <img class=\"alignnone size-full wp-image-121411\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/2.png\" alt=\"2\" width=\"83\" height=\"28\"> and <img class=\"alignnone size-full wp-image-121412\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/3.png\" alt=\"3\" width=\"83\" height=\"28\"><\/p>\n<p><strong>(v)&nbsp;<\/strong>38, 70, 48, 34, 63, 42, 55, 44, 53, 47<\/p>\n<p>We know that,<\/p>\n<p><img class=\"alignnone size-full wp-image-121413\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/a.png\" alt=\"a\" width=\"226\" height=\"57\"><\/p>\n<p>Where, |d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 x|<\/p>\n<p>So, let \u2018x\u2019 be the mean of the given observation.<\/p>\n<p>x = [38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47]\/10<\/p>\n<p>= 494\/10<\/p>\n<p>= 49.4<\/p>\n<p>Number of observations, \u2018n\u2019 = 10<\/p>\n<div>\n<table>\n<tbody>\n<tr>\n<td>\n<p>x<sub>i<\/sub><\/p>\n<\/td>\n<td>\n<p>|d<sub>i<\/sub>| = |x<sub>i<\/sub>&nbsp;\u2013 49.4|<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>38<\/p>\n<\/td>\n<td>\n<p>11.4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>70<\/p>\n<\/td>\n<td>\n<p>20.6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>48<\/p>\n<\/td>\n<td>\n<p>1.4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>34<\/p>\n<\/td>\n<td>\n<p>15.4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>63<\/p>\n<\/td>\n<td>\n<p>13.6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>42<\/p>\n<\/td>\n<td>\n<p>7.4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>55<\/p>\n<\/td>\n<td>\n<p>5.6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>44<\/p>\n<\/td>\n<td>\n<p>5.4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>53<\/p>\n<\/td>\n<td>\n<p>3.6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>47<\/p>\n<\/td>\n<td>\n<p>2.4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Total<\/p>\n<\/td>\n<td>\n<p>86.8<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>MD=1n\u2211ni=1|di|<\/p>\n<p>= 1\/10 \u00d7 86.8<\/p>\n<p>= 8.68<\/p>\n<p>Now<\/p>\n<p><img class=\"alignnone size-full wp-image-121414\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/b-1.png\" alt=\"b\" width=\"519\" height=\"97\"><\/p>\n<p><strong>Statistics Ex 32.1 Q6<\/strong><br><img src=\"https:\/\/farm8.staticflickr.com\/7574\/15967084570_64751597e9_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-32-Statistics-Ex-32.1-Q-6\" width=\"265\" height=\"341\"><\/p>\n<p>We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a> Class 11, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630311165903\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321\"><\/span>How many questions are there in RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 6 questions in RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630311376275\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321-for-free\"><\/span>Is RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 for free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, You can get RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630311433587\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-solutions-class-11-maths-chapter-32-exercise-321-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1: We&#8217;ll discuss about dispersion, range, and mean deviation in this exercise. RD Sharma Class 11 Maths Solutions can help students gain a better understanding of the concepts and develop a good understanding of the subject. Subject experts have prepared the RD Sharma Solutions Class 11 &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-exercise-32-1\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":121393,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68124"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=68124"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68124\/revisions"}],"predecessor-version":[{"id":121596,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68124\/revisions\/121596"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121393"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68124"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68124"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68124"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}