{"id":24931,"date":"2020-03-03T13:00:17","date_gmt":"2020-03-03T07:30:17","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=24931"},"modified":"2021-07-02T15:21:22","modified_gmt":"2021-07-02T09:51:22","slug":"kseeb-solutions-for-class-10-maths-chapter-13-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-13-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 13 Statistics Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 13 Statistics Additional Questions, Notes Pdf,\u00a0KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 13 Statistics Additional Questions<\/h2>\n

I. Multiple Choice Questions:<\/span><\/p>\n

Question 1.
\nThe most frequently used measure of central tendency is
\na. Mean
\nb. Mode
\nc. Median
\nd. None of these
\nAnswer:
\na. Mean<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nThe item value which occurs most frequently is called
\na. Mean
\nb. Median
\nc. Mode
\nd. None of these
\nAnswer:
\nc. Mode<\/p>\n

Question 3.
\nThe mean of first ten odd natural number is
\na. 5
\nb. 10
\nc. 20
\nd. 19
\nAnswer:
\nb. 10<\/p>\n

Question 4.
\nThe mean of first n natural number is
\na. \\(\\frac{n+1}{2}\\)
\nb. \\(\\frac{n(n+1)}{2}\\)
\nc. \\(\\frac{n(n-1)}{2}\\)
\nd. n2<\/sup>
\nAnswer:
\na. \\(\\frac{n+1}{2}\\)<\/p>\n

Question 5.
\nThe class mark of the class interval 20 – 29 is
\na. 20
\nb. 29
\nc. 25
\nd. 24.5
\nAnswer:
\nd. 24.5<\/p>\n

Question 6.
\nIn the distribution, the frequency of the class 0 – 5 is
\na. 1
\nb.2
\nc. 3
\nd. 4
\nAnswer:
\nd. 4<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nThe median of the following series is 520, 20, 340, 190, 35, 800, 12, 10, 50, 80
\na. 1210
\nb. 520
\nc. 190
\nd. 80
\nAnswer:
\nd. 80<\/p>\n

Question 8.
\nIf the mean of 5, 7, 9, x is 9 then the value of x is
\na. 11
\nb. 15
\nc. 18
\nd. 16
\nAnswer:
\nb. 15<\/p>\n

Question 9.
\nThe mode of the distribution 3, 5, 7, 4, 2, 1, 4, 3, 4 is
\na. 7
\nb. 4
\nc. 3
\nd. 1
\nAnswer:
\nb. 4<\/p>\n

Question 10.
\nIf the median of the following data arranged in ascending order is 18, then the value of x in 8, 11, 12, 16, 16 + x, 20, 25, 30 is
\na. 1
\nb. 2
\nc. 3
\nd. 4
\nAnswer:
\nd. 4<\/p>\n

Question 11.
\nThe median of first 10 odd numbers is
\na. 10
\nb. 8
\nc. 9
\nd. 11
\nAnswer:
\na. 10<\/p>\n

Question 12.
\nThe mean of five numbers is 18. If one 19 number is removed, then the mean becomes 16. The removed number is
\na. 22
\nb. 24
\nc. 25
\nd. 26
\nAnswer:
\na. 22<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nThe mean of 13 numbers is 24. If 3 is added to each of the numbers. Then, the new mean is
\na. 21
\nb. 24
\nc. 27
\nd. 30
\nAnswer:
\nc. 27<\/p>\n

Question 14.
\nThe mean of 1, 2, 3, 4, 5x is
\na. x + 1
\nb. 5
\nc. 5x + 1
\nd. x + 2
\nAnswer:
\nd. x + 2<\/p>\n

Question 15.
\nIf the mean of 10 observations is 15, then their algebraic sum is
\na. 150
\nb. 15
\nc. 1.5
\nd. 1500
\nAnswer:
\na. 150<\/p>\n

Question 16.
\nThe mean of 10 numbers is 16. If one number 36 of these is changed to 26, then new mean is
\na. 15
\nb. 16
\nc. 10
\nd. 26
\nAnswer:
\na. 15<\/p>\n

Question 17.
\nThe mode of a frequency distribution can be determined graphically from.
\na. histogram
\nb. frequency polygon
\nc. frequency curve
\nd. ogive
\nAnswer:
\na. histogram<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\nIf the mode of the following distribution is 2.8, then find the value of x. 2.5, 2.5, 2.1, 2.7, 2.8, 2.5, x, 2.8, 2.8, 2.7
\na. 2.7
\nb. 2.5
\nc. 2.8
\nd. 2.1
\nAnswer:
\nc. 2.8<\/p>\n

Question 19.
\nWhich of the following is not a measure of central tendency.
\na. Mean
\nb. Median
\nc. Mode
\nd. Range
\nAnswer:
\nd. Range<\/p>\n

Question 20.
\nRelationship among mean, median and mode is
\na. 3 Median = 2 Mode + Mean
\nb. 3 Median = 2 Mean + Mode
\nc. Mode = 2 Mean – Median
\nd. Mode = 3 Mean – 2 Median
\nAnswer:
\nb. 3 Median = 2 Mean + Mode<\/p>\n

Question 21.
\nThe mean and median of a data are respectively 20 and 22. The value of mode is
\na. 20
\nb. 26
\nc. 22
\nd. 21
\nAnswer:
\nb. 26<\/p>\n

Question 22.
\nThe class, whose cumulative frequency is greater than (and nearest to) \\(\\frac{N}{2}\\), is called
\na. average class
\nb. Modal class
\nc. Mean class
\nd. Median class
\nAnswer:
\nd. Median class<\/p>\n

\"KSEEB<\/p>\n

Question 23.
\nThe curve less than ogive in a cumulative frequency distribution is
\na. decreasing
\nb. constant
\nc. Increasing
\nd. straight line
\nAnswer:
\nc. Increasing<\/p>\n

Question 24.
\nThe curve more than ogive in a cumulative frequency distribution is
\na. decreasing
\nb. constant
\nc. increasing
\nd. straight line
\nAnswer:
\na. decreasing<\/p>\n

Question 25.
\nWhich measure of central tendency is given by the x – coordinate of the point of intersection of the ‘more than ogive’ and ‘less than ogive’?
\na. Mean
\nb. Median
\nc. Mode
\nd. Range
\nAnswer:
\nb. Median<\/p>\n

II. Short Answer Questions:<\/span><\/p>\n

Question 1.
\nWrite the formula to find mean for grouped data direct method.
\nAnswer:
\nMean =
\n\"KSEEB<\/p>\n

Question 2.
\nWrite the formula to find mean for grouped data using step deviation method.
\nAnswer:
\nMean =
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nWrite the formula to find mode for grouped data
\nAnswer:
\nMode =
\n\"KSEEB<\/p>\n

Question 4.
\nWrite the formula to find median for grouped data.
\nAnswer:
\nMedian =
\n\"KSEEB<\/p>\n

Question 5.
\nFind the class mark of the class interval 30 – 40.
\nAnswer:
\nClass mark
\n\"KSEEB<\/p>\n

Question 6.
\nFind the mean of the first 7 natural numbers.
\nAnswer:
\nMean
\n\"KSEEB<\/p>\n

Question 7.
\nAdata has 19 observations arranged in descending order which observation represents the median data?
\nAnswer:
\nTotal number of observations = 19 (odd).
\n\u2234 Median will be
\n\"KSEEB
\n10th term will be the median.<\/p>\n

Question 8.
\nWrite the modal class for the given frequency distribution
\n\"KSEEB
\nAnswer:
\nModal class is 15 – 20. It has highest frequency.<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nWrite the median class for the given frequency distribution.
\n\"KSEEB
\nAnswer:
\nCumulative frequency > \\(\\frac{N}{2}=\\frac{40}{2}\\) = 20 is 25.
\nMedian class is 25 – 30<\/p>\n

Question 10.
\nWrite the empirical relationship between the three measures of central tendency?
\nAnswer:
\n3 median = mode + 2 mean<\/p>\n

III. Long Answer Questions:<\/span><\/p>\n

Question 1.
\nIf the mean of the following distribution is 6, find the value of P.
\n\"KSEEB
\nAnswer:
\n\\(\\overline{\\mathbf{X}}\\) = 6
\n\"KSEEB
\n66 = 2P + 52
\n2P = 66 – 52
\n2P = 14
\nP = \\(\\frac{14}{2}\\) = 7
\nP = 7<\/p>\n

Question 2.
\nIf xi<\/sub>‘s are the mid points of the class intervals of the class intervals of a grouped data, fi<\/sub>‘s are the corresponding
\nfrequencies and \\(\\overline{\\mathbf{X}}\\) is the mean, then find \u03a3fi<\/sub>(xi<\/sub> – \\(\\overline{\\mathbf{X}}\\)).
\nAnswer:
\n\"KSEEB
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nThe following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.<\/p>\n

\"KSEEB
\nAnswer:
\n\"KSEEB
\nAssumed mean = a = 70
\nClass size = h = 10
\n\"KSEEB
\n\u2234 Mean literacy rate is 69.43%<\/p>\n

Question 4.
\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 years.
\n\"KSEEB
\nAnswer:
\nGiven cumulative frequency distribution table convert into frequency distribution table
\n\"KSEEB
\nCumulative greater than \\(\\frac{\\mathrm{n}}{2}=\\frac{100}{2}\\) = 50 is 78.
\n\u2234 Median class is 35 – 40
\nl = 35, h = 5, f = 33 and cf=45
\n\"KSEEB
\n\"KSEEB
\n= 35 + 0.76
\n= 35.76
\n\u2234 Median age is 35.76 years.<\/p>\n

Question 5.
\nThe mean of the following frequency table is 50. But the frequencies f1<\/sub> and f2<\/sub> in class 20 – 40 and 60 – 80 respectively are missing. Find the missing frequencies.
\n\"KSEEB
\nAnswer:
\nAssumed mean = a = 50 and h = 20
\n\"KSEEB
\n\u03a3fi<\/sub> = 68 + f1<\/sub> + f2<\/sub> = 120
\nf1<\/sub> + f2<\/sub> = 120 – 68
\nf1<\/sub> + f2<\/sub> = 52 \u2192 (1)
\n\"KSEEB
\n0 = 4 – f1<\/sub> + f2<\/sub>.
\nf1<\/sub> – f2<\/sub> = 4 \u2192 (2)
\nAdding equation (1) and (2)
\nf1<\/sub> + f2<\/sub> = 52
\n\"KSEEB
\nPut f1<\/sub> = 28 in equation (1)
\n28 + f2<\/sub> = 52
\nf2<\/sub> = 52 – 28
\nf2<\/sub> = 24
\n\u2234 missing frequencies f1<\/sub> is 28 and f2<\/sub> is 24.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 13 Statistics Additional Questions, Notes Pdf,\u00a0KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 13 Statistics Additional Questions I. Multiple Choice Questions: Question 1. The most frequently used …<\/p>\n","protected":false},"author":5,"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\/24931"}],"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=24931"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/24931\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=24931"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=24931"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=24931"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}