\n70<\/td>\n | ?<\/td>\n | 62<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Find the mean score of XB. \nSolution: \nGiven n1<\/sub> = 30; n2<\/sub> = 35; \\(\\bar{x}_1\\) = 70; \\(\\bar{x}\\)12<\/sub> = 62; \\(\\bar{x}_2\\) = ? \nWe know that \\(\\bar{x}\\)12<\/sub> = \\(\\frac{n_1 \\bar{x}_1+n_2 \\bar{x}_2}{n_1+n_2}\\) \n\u21d2 62 = \\(\\frac{30 \\times 70+35 \\times \\bar{x}_2}{30+35}\\) \n\u21d2 4030 – 2100 = 35\\(\\bar{x}_2\\) \n\u21d2 \\(\\bar{x}_2\\) = \\(\\frac { 1930 }{ 35 }\\) \n= 55.14<\/p>\nQuestion 5. \nTwo samples of sizes 50 and 100 are given. The mean of these samples respectively are 56 and 50. Find the mean of size 150 by combining. \nSolution: \nHere n1<\/sub> = 50; n2<\/sub> = 100; \\(\\bar{x}_1\\) = 56; \\(\\bar{x}_2\\) = 50 \nThen \\(\\bar{x}\\)12<\/sub> = \\(\\frac{n_1 \\bar{x}_1+n_2 \\bar{x}_2}{n_1+n_2}\\) \n\u21d2 \\(\\bar{x}\\)12<\/sub> = \\(\\frac{50 \\times 56+100 \\times 50}{50+100}\\) \n\u21d2 \\(\\bar{x}\\)12<\/sub> = \\(\\frac{2800+5000}{150}\\) = \\(\\frac{7800}{150}\\) = 52<\/p>\n<\/p>\n Question 6. \nThe mean and standard deviation of the marks obtained by the groups of students, consisting of 50 each, are given below. Calculate the mean and standard deviation of the marks obtained by all the 100 students.<\/p>\n \n\n\nGroups<\/td>\n | Mean<\/td>\n | Standard deviation<\/td>\n<\/tr>\n | \n1<\/td>\n | 60<\/td>\n | 8<\/td>\n<\/tr>\n | \n2<\/td>\n | 55<\/td>\n | 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Solution: \nHere n1<\/sub> = n2<\/sub> = 50; \\(\\bar{x}_1\\) = 60; \\(\\bar{x}_2\\) = 55; \u03c31<\/sub> = 8; \u03c31<\/sub> = 7 \nThen combined mean \\(\\bar{x}\\)12<\/sub> = \\(\\frac{n_1 \\bar{x}_1+n_2 \\bar{x}_2}{n_1+n_2}\\) = \\(\\frac{50 \\times 30+50 \\times 44}{50+50}\\) = \\(\\frac{3000+2750}{100}\\) = \\(\\frac{5750}{100}\\) = 57.5 \n\u2234 d1<\/sub> = \\(\\bar{x}\\)12<\/sub> – \\(\\bar{x}_1\\) = 57.5 – 60 = – 2.5 and d2<\/sub> = \\(\\bar{x}\\)12<\/sub> – \\(\\bar{x}_2\\) = 57.5 – 55 = 2.5 \nTherefore, combined S.D = \u03c312<\/sub> = \n<\/p>\nQuestion 7. \nThe mean and standard deviation of distribution of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together. \nSolution: \nHere n1<\/sub> = 100; n2<\/sub> = 150; \\(\\bar{x}_1\\) = 50; \\(\\bar{x}_2\\) = 40 \n\u03c31<\/sub> = 5; \u03c32<\/sub> = 6 \n\u2234 Combined Mean \\(\\bar{x}\\)12<\/sub> = \\(\\frac{n_1 \\bar{x}_1+n_2 \\bar{x}_2}{n_1+n_2}\\) = \\(\\frac{100 \\times 50+150 \\times 40}{100+150}\\) = \\(\\frac{11000}{250}\\) = 44 \n\u2234 d1<\/sub> = \\(\\bar{x}\\)12<\/sub> – \\(\\bar{x}_1\\) = 44 – 50 = – 6 and d2<\/sub> = \\(\\bar{x}\\)12<\/sub> – \\(\\bar{x}_2\\) = 44 – 40 = 4 \n\u2234 Combined S.D = \u03c32<\/sub> = \n<\/p>\n","protected":false},"excerpt":{"rendered":"Students can cross-reference their work with ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(a) to ensure accuracy. S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(a) Question 1. One set of 100 observations has the mean 15 and another set of 150 observations has the mean 16. Find …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[115871],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/169142"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=169142"}],"version-history":[{"count":7,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/169142\/revisions"}],"predecessor-version":[{"id":169157,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/169142\/revisions\/169157"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=169142"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=169142"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=169142"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}} | |