{"id":1386,"date":"2020-09-15T10:38:22","date_gmt":"2020-09-15T05:08:22","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1386"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-13-ex-13-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-13-ex-13-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3<\/strong>\u00a0are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3<\/h2>\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\"KSEEB
\nSolution:
\n\"KSEEB
\n(i) n = 68, \u2234 \\(\\frac{n}{2}\\) = 34
\nThe median is in class interval (125 – 145)
\nl = 125, n = 68, f = 20, cf = 22, h = 20
\n\"KSEEB
\n= 125 + 12
\n\u2234 Median = 137 units.<\/p>\n

(ii) To find out Mode:
\n\"KSEEB
\nHere, Mode is in class interval (125 – 145)
\nMaximum frequency, l = 125, f<sub>1<\/sub> = 20, f<sub>0<\/sub> = 13, f<sub>2\/sub> = 14, h = 20
\n\"KSEEB
\n= 125 + 10.76
\n= 135.76
\n\u2234 Mode = 135.76 units.<\/p>\n

Question 2.
\nIf the median of the distribution given below is 28.5, find the values of ‘x’ and ‘y’
\n\"KSEEB
\nSolution:
\nMedian is 28.5
\nClass interval which has median is = (20 – 30)
\nl = 20, n = 60, f = 20, cf = 5 + a \u00d7 h = 10
\n\"KSEEB
\n17 = 25 – x
\n\u2234 x = 8
\n5 + x + 20 + 15 + y + 5 = 60
\nx + y – 45 = 60
\nx + y = 15
\ny = 15 – x
\ny = 15 – 8
\n\u2234 y = 7
\n\u2234 x = 8, y = 7<\/p>\n

Question 3.
\nA life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
\n\"KSEEB
\nSolution:
\n\"KSEEB
\nn = 100 \u2234 \\(\\frac{n}{2}\\) = 50
\nMedian in which C.I. is (35 – 40)
\nl = 35, n = 100, f = 33, cf = 45, h = 5
\n\"KSEEB
\n= 35 + 0.76
\n\u2234 Median = 35.76 years<\/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\"KSEEB
\nFind the median length of the leaves.
\n(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)
\nSolution:
\n\"KSEEB
\nn = 40 \u2234 \\(\\frac{n}{2}\\) = 20
\nC.I. which has median is (144.5 – 153.5)
\nl = 144.5, n = 40, f= 12, cf = 17, h = 9
\n\"KSEEB
\n= 144.5 + 2.25
\n\u2234 Median = 146.75 mm.<\/p>\n

Question 5.
\nThe following table gives the distribution of the lifetime of 400 neon lamps:
\n\"KSEEB
\nFind the median life time of a lamp.
\nSolution:
\n\"KSEEB
\nn = 400 \u2234 \\(\\frac{n}{2}=\\frac{400}{2}=200\\)
\nClass interval having median is = (3000 – 3500)
\nl = 3000, n = 400, f = 86, cf = 130, h = 500
\n\"KSEEB
\n= 3000 + 406.98
\n\u2234 Median = 3406.98 hours<\/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 alphabets in the surnames was obtained as follows :
\n\"KSEEB
\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.
\nSolution:
\n\"KSEEB
\n(i) n = 100, \u2234 \\(\\frac{n}{2}\\) = 50
\nClass interval having median is (7 – 10)
\nl = 7, n = 100, f = 40, cf= 36, h = 3
\n\"KSEEB
\n= 7 + 1.05
\n\u2234 Median = 8.05 Letters<\/p>\n

(ii) Clall interval which has mode is (7 – 10)
\nMaximum frequency, l = 7, f1<\/sub> = 40, f0<\/sub> = 30, f2<\/sub> = 16, h = 3
\n\"KSEEB
\n= 7 + 0.88
\n\u2234 Mode = 7.88<\/p>\n

(iii) Mean(\\(\\overline{X}\\)) : Step Deviation Method:
\n\"KSEEB
\n\"KSEEB
\n= 8.5 – 0.18
\n= 8.32
\n\u2234 Mean = 8.32
\n\u2234 (i) Median = 8.05 letters
\n(ii) Mode = 7.88
\n(iii) Mean = 8.32.<\/p>\n

Question 7.
\nThe distribution below gives the weights of 30 students of a class. Find the median weight of the students.
\n\"KSEEB
\nSolution:
\n\"KSEEB
\n(i) n = 30, \u2234 \\(\\frac{\\mathbf{n}}{2}\\) = 15
\nClass interval having median is (55 – 60)
\nl = 55, n = 30, f = 6, cf = 13, h = 5
\n\"KSEEB
\n= 55 + 1.67
\n\u2234 Median = 56.67 kg.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3\u00a0are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3. Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3 Question 1. The following frequency distribution gives the monthly consumption …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1386"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1386"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1386\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1386"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1386"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1386"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}