{"id":3003,"date":"2023-05-11T15:18:14","date_gmt":"2023-05-11T09:48:14","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=3003"},"modified":"2023-05-12T17:29:15","modified_gmt":"2023-05-12T11:59:15","slug":"ap-ssc-10th-class-maths-solutions-chapter-14-ex-14-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-ssc-10th-class-maths-solutions-chapter-14-ex-14-3\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 14 Statistics Ex 14.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.3<\/h2>\n

10th Class Maths 14th Lesson Statistics Ex 14.3 Textbook Questions and Answers<\/h3>\n

Question 1.
\nThe following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

\"AP
\nSum of the frequencies = 68
\n\u2234 \\(\\frac{n}{2}\\) = \\(\\frac{68}{2}\\) = 34
\nHence median class = 125 – 145
\nLower boundary of the median class, l = 125
\ncf – cumulative frequency of the class preceding the median class = 22
\nf – frequency of the median class = 20
\nh = class size = 20
\nMedian = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\n= 125 + \\(\\frac{[34-22]}{20}\\) \u00d7 20
\n\u2234 Median = 125 + 12 = 137
\nMaximum number of consumers lie in the class 125 – 145
\nModal class is 125 -145
\nl – lower limit of the modal class =125
\nf1<\/sub> – frequency of the modal class = 20
\nf0<\/sub> – frequency of the class preceding the modal class =13
\nf2<\/sub> – frequency of the class succeeding the modal class =14
\nh – size of the class = 20
\nMode (Z) = \\(l+\\frac{f_{1}-f_{0}}{\\left(f_{1}-f_{0}\\right)+\\left(f_{1}-f_{2}\\right)} \\times h\\)
\nMode (Z) = 125 + \\(\\frac{20-13}{(20-13)+(20-14)} \\times 20\\)
\n= 125 + \\(\\frac{7}{7+6}\\) \u00d7 20
\n= 125 + \\(\\frac{140}{13}\\)
\n= 125 + 10.76923
\n\u2234 Mode = 135.769
\nMean \\(\\overline{\\mathrm{x}}=\\mathrm{a}+\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}} \\mathrm{u}_{\\mathrm{i}}}{\\Sigma \\mathrm{f}_{\\mathrm{i}}} \\times \\mathrm{h}\\)
\na = assumed mean = 135
\n\u2234 \\(\\overline{\\mathbf{x}}\\) = 135 + \\(\\frac{7}{68}\\)
\n= 135 + 0.102941
\n\u2243 135.1
\nMean, Median and Mode are approximately same in this case.<\/p>\n

\"AP<\/p>\n

Question 2.
\nIf the median of 60 observations, given below is 28.5, find the values of x and y.
\n\"AP
\nAnswer:
\n\"AP
\nMedian = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\nIt is given that \u2211f = n = 60
\nSo, 45 + x + y = 60
\nx + y = 60 – 45 = 15
\nx + y = 15 ….. (1)
\nThe median is 28.5 which lies be-tween 20 and 30.
\nMedian class = 20 – 30
\nLower boundary of the median class ‘l’ = 20
\n\\(\\frac{N}{2}\\) = \\(\\frac{60}{2}\\) = 30
\ncf – cumulative frequency = 5 + x
\nh = 10
\nMedian = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\n\u21d2 28.5 = 20 + \\(\\frac{30-5-x}{20}\\) \u00d7 10
\n\u21d2 28.5 = 20 + \\(\\frac{25-x}{2}\\)
\n\\(\\frac{25-x}{2}\\) = 28.5 – 20 = 8.5
\n25 – x = 2 \u00d7 8.5
\nx = 25- 17 = 8
\nalso from (1); x + y = 15
\n8 + y = 15
\ny = 7
\n\u2234 x = 8; y = 7.<\/p>\n

\"AP<\/p>\n

Question 3.
\nA life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.
\n\"AP
\nAnswer:
\n\"AP
\nThe given distribution being of the less than type, 25, 30, 35, give the upper limits of corresponding class intervals. So the classes should be 20 – 25, 25 – 30, 30 – 35, ………. 55 – 60.
\nObserve that from the given distribution 2 persons with age less than 20.
\ni.e., frequency of the class below 20 is 2.
\nNow there are 6 persons with age less than 25 and 2 persons with age less than 20.
\n\u2234 The number of persons with age in the interval 20 – 25 is 6 – 2 = 4.
\nSimilarly, the frequencies can be calculated as shown in table.
\nNumber of observations = 100
\nn = 100
\n\\(\\frac{n}{2}\\) = \\(\\frac{100}{2}\\) = 50, which lies in the class 35-40
\n\u2234 35 – 40 is the median class and lower boundary l = 35
\ncf = 45;
\nh = 5;
\nf = 33
\nMedian = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\n= 35 + \\(\\frac{50-45}{33}\\) \u00d7 5
\n= 35 + \\(\\frac{5}{33}\\) \u00d7 5
\n= 35 + 0.7575
\n= 35.7575
\n\u2234 Median \u2243 35.76<\/p>\n

\"AP<\/p>\n

Question 4.
\nThe lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
\n\"AP
\nFind the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5,…, 171.5 – 180.5.)
\nAnswer:
\nSince the formula, Median = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\) assumes continuous classes assumes continuous class, the data needs to be converted to continuous classes.
\nThe classes then changes to 117.5 – 126.5; 126.5 – 133.5, …… 171.5 – 180.5.
\n\"AP
\n\u2211fi<\/sub> = n = 40
\n\\(\\frac{n}{2}\\) = \\(\\frac{40}{2}\\) = 20
\n\\(\\frac{n}{2}\\)th observation lie in the class 144.5- 153.5
\n\u2234 Median class = 144.5 – 153.5
\nLower boundary, l = 144.5
\nFrequency of the median class, f = 12
\nc.f. = 17
\nh = 9
\n\u2234 Median = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\n= 144.5 + \\(\\frac{20-17}{12}\\) \u00d7 9
\n= 144.5 + \\(\\frac{3}{12}\\) \u00d7 9
\n= 144.5 + \\(\\frac{9}{4}\\)
\n= 144.5 + 2.25
\n\u2234 Median length = 146.75 mm.<\/p>\n

\"AP<\/p>\n

Question 5.
\nThe following table gives the distribution of the life-time of 400 neon lamps.
\n\"AP
\nFind the median life-time of a lamp.
\nAnswer:
\n\"AP
\nTotal observations are n = 400
\n\\(\\frac{n}{2}\\)th observation i.e \\(\\frac{400}{2}\\) = 200
\n200 lies in the class 3000 – 3500
\n\u2234 Median class = 3000 – 3500
\nLower boundary l = 3000
\nfrequency of the median class f = 86
\nc.f = 130
\nClass size, h = 500
\nMedian = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\n= 3000 + \\(\\frac{200-130}{86}\\) \u00d7 500
\n= 3000 + \\(\\frac{70}{86}\\) \u00d7 500
\n= 3000 + 406.977
\n= 3406.98
\n\u2234 Median life \u2243 3406.98 hours<\/p>\n

\"AP<\/p>\n

Question 6.
\n100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows.
\n\"AP
\nDetermine the median number of letters in the surnames. Find the mean number of letters in the surnames ? Also, find the modal size of the surnames.
\nAnswer:
\nNumber of letters in the surnames.
\nAlso find the modal size of the surnames.
\n\"AP
\nTotal observations are n = 100
\n\\(\\frac{n}{2}\\) = \\(\\frac{100}{2}\\) = 50
\n50 lies in the class 7 – 10
\n\u2234 Median class = 7 – 10
\nl – lower boundary = 7
\nf – frequency of the median class = 40
\ncf = 36
\nClass size h = 3
\nMedian:
\nMedian = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\n= 7 + \\(\\frac{50-36}{40}\\) \u00d7 3
\n= 7 + \\(\\frac{14}{40}\\) \u00d7 3
\n= 7 + \\(\\frac{42}{40}\\)
\n= 7 + 1.05
\n= 8.05
\n\u2234 Median = 8.05.<\/p>\n

Mean:
\nAssumed mean, a = 8.5
\nMean \\(\\overline{\\mathrm{x}}=\\mathrm{a}+\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}} \\mathrm{d}_{\\mathrm{i}}}{\\Sigma \\mathrm{f}_{\\mathrm{i}}}\\)
\n= 8.5 + \\(\\frac{(-18)}{100}\\)
\n= 8.5 – 0.18
\n= 8.32
\n\u2234 Mean = 8.32.<\/p>\n

Mode:
\nMaximum number of surnames = 40
\n\u2234 Modal class = 7-10
\nl – lower boundary of the modal class = 7
\nMode (Z) = \\(l+\\frac{f_{1}-f_{0}}{\\left(f_{1}-f_{0}\\right)+\\left(f_{1}-f_{2}\\right)} \\times h\\)
\nl = 7; f1<\/sub> = 40, f0<\/sub> = 30, f2<\/sub> = 16, h = 3
\nMode (Z) = 7 + \\(\\frac{40-30}{(40-30)+(40-16)}\\) \u00d7 3
\n= 7 + \\(\\frac{10}{10+24}\\) \u00d7 3
\n= 7 + \\(\\frac{30}{34}\\)
\n= 7 + 0.882
\n= 7.882<\/p>\n

Median = 8.0.5; Mean = 8.32; Modal size = 7.88.<\/p>\n

\"AP<\/p>\n

Question 7.
\nThe distribution below gives the weights of 30 students of a class. Find the median weight of the students.
\n\"AP
\nAnswer:
\n\"AP
\nNumber of observations (n) = \u2211fi<\/sub>
\n\\(\\frac{n}{2}\\) = \\(\\frac{30}{2}\\) = 15
\n15 lies in the class 50 – 55
\n\u2234 Median class = 50-55
\nl – lower boundary of the median class = 55
\nf – frequency of the median class = 8
\nc.f = 5
\nClass size h = 6
\nMedian = \\(l+\\frac{\\left[\\frac{\\mathrm{n}}{2}-\\mathrm{cf}\\right]}{\\mathrm{f}} \\times \\mathrm{h}\\)
\n= 50 + \\(\\frac{15-5}{8}\\) \u00d7 6
\n= 50 + 7.5
\n= 57.5
\n= 50 + 7.5 = 57.5
\n\u2234 Median weight = 57.5 kg.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.3 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.3 10th Class Maths 14th Lesson Statistics Ex 14.3 Textbook Questions and Answers Question 1. The following frequency distribution gives the monthly consumption of … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/3003"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=3003"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/3003\/revisions"}],"predecessor-version":[{"id":40491,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/3003\/revisions\/40491"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=3003"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=3003"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=3003"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}