{"id":4570,"date":"2023-11-02T11:29:37","date_gmt":"2023-11-02T05:59:37","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4570"},"modified":"2023-11-06T11:47:12","modified_gmt":"2023-11-06T06:17:12","slug":"kseeb-solutions-class-8-maths-chapter-13-ex-13-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-13-ex-13-3\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 13 Statistics Ex 13.3"},"content":{"rendered":"

Students can Download Maths Chapter 13 Statistics Ex 13.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 13 Statistics Ex 13.3<\/h2>\n

Question 1.
\nRuns scored by 10 batsmen in a one-day cricket match are given. Find the average run scored. 23, 54, 08, 94, 60, 18, 29, 44, 05, 86
\nAnswer:
\n\u03a3 x = 23 + 54 + 08 + 94 + 60 + 18 + 29 + 44 + 05 + 86 = 421
\nN = 10
\nAverage = Mean =\u00a0\"\\overline{X}=\\frac{\\Sigma<\/p>\n

Question 2.
\nFind the mean weight form the following table:<\/p>\n\n\n\n\n
Weight (kg)<\/td>\n29<\/td>\n30<\/td>\n31<\/td>\n32<\/td>\n33<\/td>\n<\/tr>\n
No. of children<\/td>\n02<\/td>\n01<\/td>\n04<\/td>\n03<\/td>\n05<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:<\/p>\n\n\n\n\n\n\n\n\n\n
Weigh t(kg)<\/td>\nNo. of children(x)<\/td>\n‘ f<\/td>\n<\/tr>\n
29<\/td>\n02<\/td>\n58<\/td>\n<\/tr>\n
30<\/td>\n01<\/td>\n30<\/td>\n<\/tr>\n
31<\/td>\n04.<\/td>\n124<\/td>\n<\/tr>\n
32<\/td>\n03<\/td>\n96<\/td>\n<\/tr>\n
33<\/td>\n05<\/td>\n164<\/td>\n<\/tr>\n
<\/td>\nN= 15<\/td>\nZf = 473<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Mean = \\(\\frac{\\Sigma f_{X}}{N}=\\frac{473}{15}=31.53\\)<\/p>\n

Question 3.
\nCalculate the mean for the following frequency distribution.<\/p>\n\n\n\n\n
Mark<\/td>\n10-20<\/td>\n20-30<\/td>\n30-40<\/td>\n40-50<\/td>\n50-60<\/td>\n60-70<\/td>\n70-80<\/td>\n<\/tr>\n
Frequency<\/td>\n3<\/td>\n7<\/td>\n10<\/td>\n6<\/td>\n8<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Marks<\/td>\nFrequency<\/td>\nMidpoint<\/td>\nfx<\/td>\n<\/tr>\n
10-20<\/td>\n3<\/td>\n15<\/td>\n45<\/td>\n<\/tr>\n
20-30<\/td>\n7<\/td>\n25<\/td>\n175<\/td>\n<\/tr>\n
30-40<\/td>\n10<\/td>\n35<\/td>\n350<\/td>\n<\/tr>\n
40-50<\/td>\n6<\/td>\n45<\/td>\n270<\/td>\n<\/tr>\n
50-60<\/td>\n8<\/td>\n55<\/td>\n440<\/td>\n<\/tr>\n
60-70<\/td>\n2<\/td>\n65<\/td>\n130<\/td>\n<\/tr>\n
70-80<\/td>\n4<\/td>\n75<\/td>\n300<\/td>\n<\/tr>\n
<\/td>\nN = 40<\/td>\n<\/td>\nEfx = 1710<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Mean = \\(\\frac{\\Sigma f_{x}}{N}=\\frac{1710}{40}=42.75\\)<\/p>\n

Question 4.
\nCalculate the mean for the following frequency distribution.<\/p>\n\n\n\n\n
Mark<\/td>\n15-19<\/td>\n20-24<\/td>\n25-29<\/td>\n30-34<\/td>\n35-39<\/td>\n40-44<\/td>\n<\/tr>\n
Frequency<\/td>\n6<\/td>\n5<\/td>\n9<\/td>\n12<\/td>\n6<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:<\/p>\n\n\n\n\n\n\n\n\n\n\n
Marks.<\/td>\nFrequency<\/td>\nMidpoint<\/td>\nf(X)<\/td>\n<\/tr>\n
15-19<\/td>\n6<\/td>\n17<\/td>\n102<\/td>\n<\/tr>\n
20-24<\/td>\n5<\/td>\n22<\/td>\n*110<\/td>\n<\/tr>\n
25-29<\/td>\n9<\/td>\n27<\/td>\n243<\/td>\n<\/tr>\n
30-34<\/td>\n12<\/td>\n32<\/td>\n384<\/td>\n<\/tr>\n
35-39<\/td>\n6<\/td>\n37<\/td>\n222<\/td>\n<\/tr>\n
40-44<\/td>\n2<\/td>\n42<\/td>\n84<\/td>\n<\/tr>\n
<\/td>\nN = 40<\/td>\n<\/td>\nEfx = 1145<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Mean = \\(\\overline{X}=\\frac{\\Sigma f_{X}}{N}=\\frac{1145}{40}=28.625\\)<\/p>\n

Question 5.
\nFind the median of the data 15,22, 9,20, 6,18,11,25,14.
\nAnswer:
\n6, 9, 11, 14,(15), 18, 20, 22,25 (Ascending order)
\nN = 9,
\n\"\\frac{N+1}{2}+\\frac{9+1}{2}=\\frac{10}{2}=5^{\\text
\nMedian = 15.<\/p>\n

Question 6.
\nFind the median of the data 22,28,34, 49, 44, 57,18,10,33, 41, 66, 59.
\nAnswer:
\n10, 18, 22, 28, 33, 34, 41, 44, 49, 57, 59, 66 (Ascending order)
\nN = 10
\n\u2234 Median =\u00a0\"\\frac{34+41}{2}=\\frac{75}{2}=37.5\"<\/p>\n

Question 7.
\nFind the median for the following frequency distribution table.<\/p>\n\n\n\n\n
Class interval<\/td>\n110-119<\/td>\n120-129<\/td>\n130-139<\/td>\n140-149<\/td>\n150-159<\/td>\n160-169<\/td>\n<\/tr>\n
Frequency<\/td>\n6<\/td>\n8<\/td>\n15<\/td>\n10<\/td>\n6<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:<\/p>\n\n\n\n\n\n\n\n\n\n
Class interval<\/td>\nFrequency (f)<\/td>\nCumulative frequency (fc)<\/td>\n<\/tr>\n
110-119<\/td>\n6<\/td>\n6<\/td>\n<\/tr>\n
120-129<\/td>\n8<\/td>\n14<\/td>\n<\/tr>\n
130-139<\/td>\n15<\/td>\n29<\/td>\n<\/tr>\n
140-149<\/td>\n10<\/td>\n45<\/td>\n<\/tr>\n
160-169<\/td>\n5<\/td>\n50<\/td>\n<\/tr>\n
<\/td>\n\u00a0N = 50<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

N = 50
\n\\(\\frac{\\mathrm{N}}{2}=\\frac{50^{25}}{\\not 2}=25\\)
\n\u2234 Median class is 130 – 139
\nLRL = 129.5
\nFc = 14
\nFm = 15
\ni = 10
\n\"KSEEB<\/p>\n

Question 8.
\nFind the median for the following frequency distribution table.<\/p>\n\n\n\n\n
Class interval<\/td>\n0-5<\/td>\n5-10<\/td>\n10-15<\/td>\n15-20<\/td>\n20-25<\/td>\n25-30<\/td>\n<\/tr>\n
Frequency<\/td>\n5<\/td>\n3<\/td>\n9<\/td>\n10<\/td>\n8<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:<\/p>\n\n\n\n\n\n\n\n\n\n\n
Class interval<\/td>\nFrequency (t)<\/td>\nCumulative frequency (fc)<\/td>\n<\/tr>\n
0-5<\/td>\n5<\/td>\n5<\/td>\n<\/tr>\n
5-10<\/td>\n3<\/td>\n8<\/td>\n<\/tr>\n
10-15<\/td>\n9 ,<\/td>\n17<\/td>\n<\/tr>\n
15-20<\/td>\n10<\/td>\n27<\/td>\n<\/tr>\n
20-25<\/td>\n8<\/td>\n35<\/td>\n<\/tr>\n
25-30<\/td>\n5<\/td>\n40<\/td>\n<\/tr>\n
<\/td>\nN = 40<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

N = 40
\n\\(\\frac{N}{2}=\\frac{40}{2}=20\\)
\n\u2234 Median class is 15 – 20
\nLRL = 15
\nFc = 17
\nFm = 9
\ni = 5
\n\"KSEEB<\/p>\n

Question 9.
\nFind the mode for the following data.
\n(i) 4,3,1,5,3, 7, 9,6 Answer: Mode = 3
\n(ii) 22,36,18,22,20,34,22, 42, 46,42
\nAnswer:
\nMode = 22<\/p>\n

Question 10.
\nFind the mode for the following data<\/p>\n\n\n\n\n
X<\/td>\n5<\/td>\n10<\/td>\n12<\/td>\n15<\/td>\n20<\/td>\n30<\/td>\n40<\/td>\n<\/tr>\n
f<\/td>\n4<\/td>\n8<\/td>\n11<\/td>\n13<\/td>\n16<\/td>\n12<\/td>\n9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\nMode = 20 (It has the highest frequency)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 13 Statistics Ex 13.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 13 Statistics Ex 13.3 Question 1. Runs scored by 10 batsmen in …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4570"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=4570"}],"version-history":[{"count":1,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4570\/revisions"}],"predecessor-version":[{"id":90909,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4570\/revisions\/90909"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4570"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4570"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4570"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}