{"id":17500,"date":"2022-05-24T19:30:27","date_gmt":"2022-05-24T14:00:27","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17500"},"modified":"2022-05-16T15:30:15","modified_gmt":"2022-05-16T10:00:15","slug":"mcq-questions-for-class-11-maths-chapter-15","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-15\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Statistics Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 15\u00a0Statistics Objective Questions.<\/p>\n

Statistics Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Statistics Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Statistics Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Statistics Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nIf the varience of the data is 121 then the standard deviation of the data is
\n(a) 121
\n(b) 11
\n(c) 12
\n(d) 21<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 11
\nGiven, varience of the data = 121
\nNow, the standard deviation of the data = \u221a(121)
\n= 11<\/p>\n<\/details>\n

\n

Question 2.
\nThe mean deviation from the mean for the following data: 4, 7, 8, 9, 10, 12, 13 and 17 is
\n(a) 2
\n(b) 3
\n(c) 4
\n(d) 5<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 3
\nMean = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)\/10 = 80\/10 = 8
\n|xi<\/sub> – mean|= |4 – 10| + |7 – 10| + |8 – 10| + |9 – 10| + |10 – 10| + |12 – 10| + |13 – 10| + |17 – 10|
\n= 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24
\nNow, mean deviation form mean = 24\/8 = 3<\/p>\n<\/details>\n

\n

Question 3.
\nThe mean of 1, 3, 4, 5, 7, 4 is m the numbers 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then, p + q =
\n(a) 4
\n(b) 5
\n(c) 6
\n(d) 7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 7
\nThe mean of 1, 3, 4, 5, 7, 4 is m
\n\u21d2 (1 + 3 + 4 + 5 + 7 + 4)\/6 = m
\n\u21d2 m = 24\/6
\n\u21d2 m = 4
\nThe numbers 3, 2, 2, 4, 3, 3, p have mean m – 1
\n\u21d2 (3 + 2 + 2 + 4 + 3 + 3 + p)\/7 = m – 1
\n\u21d2 (17 + p)\/7 = 4 – 1
\n\u21d2 (17 + p)\/7 = 3
\n\u21d2 17 + p = 7 \u00d7 3
\n\u21d2 17 + p = 21
\n\u21d2 p = 21 – 17
\n\u21d2 p = 4
\nThe numbers 3, 2, 2, 4, 3, 3, p have median q.
\n\u21d2 The numbers 2, 2, 3, 3, 3, 4, 4 have median q
\n\u21d2 (7 + 1)\/2 th term = q
\n\u21d2 4th term = q
\n\u21d2 q = 3
\nNow p + q = 4 + 3 = 7<\/p>\n<\/details>\n

\n

Question 4.
\nIf the difference of mode and median of a data is 24, then the difference of median and mean is
\n(a) 12
\n(b) 24
\n(c) 8
\n(d) 36<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 12
\nGiven the difference of mode and median of a data is 24
\n\u21d2 Mode – Median = 24
\n\u21d2 Mode = Median + 24
\nNow, Mode = 3 \u00d7 Median – 2 \u00d7 Mean
\n\u21d2 Median + 24 = 3 \u00d7 Median – 2 \u00d7 Mean
\n\u21d2 24 = 3 \u00d7 Median – 2 \u00d7 Mean – Median
\n\u21d2 24 = 2 \u00d7 Median – 2 \u00d7 Mean
\n\u21d2 Median – Mean = 24\/2
\n\u21d2 Median – Mean = 12<\/p>\n<\/details>\n

\n

Question 5.
\nThe coefficient of variation is computed by
\n(a) S.D\/.Mean \u00d7 100
\n(b) S.D.\/Mean
\n(c) Mean.\/S.D \u00d7 100
\n(d) Mean\/S.D.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) S.D.\/Mean
\nThe coefficient of variation = S.D.\/Mean<\/p>\n<\/details>\n

\n

Question 6.
\nThe geometric mean of series having mean = 25 and harmonic mean = 16 is
\n(a) 16
\n(b) 20
\n(c) 25
\n(d) 30<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 20
\nThe relationship between Arithmetic mean (AM), Geometric mean (GM) And Harmonic mean (HM) is
\nGM\u00b2 = AM \u00d7 HM
\nGiven AM = 25
\nHM = 16
\nSo GM\u00b2 = 25 \u00d7 16
\n\u21d2GM = \u221a(25 \u00d7 16)
\n= 5 \u00d7 4
\n= 20
\nSo, Geometric mean = 20<\/p>\n<\/details>\n

\n

Question 7.
\nWhen tested the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623. The mean of the lives of 5 bulbs is
\n(a) 1445
\n(b) 1446
\n(c) 1447
\n(d) 1448<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1446
\nGiven, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
\nNow, mean = (1357 + 1090 + 1666 + 1494 + 1623)\/5
\n= 7230\/5
\n= 1446<\/p>\n<\/details>\n

\n

Question 8.
\nMean of the first n terms of the A.P. a + (a + d) + (a + 2d) + \u2026\u2026\u2026 is
\n(a) a + nd\/2
\n(b) a + (n – 1)d
\n(c) a + (n \u2212 1)d\/2
\n(d) a + nd<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) a + (n \u2212 1)d\/2
\nMean of the first n terms of the A.P. {a + (a + d) + (a + 2d) + \u2026\u2026\u2026 a + (n-1)d}\/n
\n= (n\/2){2a + (n – 1)d}\/n
\n= (1\/2){2a + (n – 1)d}
\n= a + (n – 1)d\/2<\/p>\n<\/details>\n

\n

Question 9.
\nThe mean of a group of 100 observations was found to be 20. Later on, it was found that three observations were incorrect, which was recorded as 21, 21 and 18. Then the mean if the incorrect observations are omitted is
\n(a) 18
\n(b) 20
\n(c) 22
\n(d) 24<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 20
\nGiven mean of 100 observations is 20
\nNow
\n\u2211 xi\/100 = 20 (1 <= i <= 100)
\n\u21d2 \u2211 xi = 100 \u00d7 20
\n\u21d2 \u2211 xi = 2000
\n3 observations 21, 21 and 18 are recorded incorrectly.
\nSo \u2211 xi = 2000 – 21 – 21 – 18
\n\u21d2 \u2211 xi = 2000 – 60
\n\u21d2 \u2211 xi = 1940
\nNow new mean is
\n\u2211 xi\/100 = 1940\/97 = 20
\nSo, the new mean is 20<\/p>\n<\/details>\n

\n

Question 10.
\nIf covariance between two variables is 0, then the correlation coefficient between them is
\n(a) nothing can be said
\n(b) 0
\n(c) positive
\n(d) negative<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 0
\nThe relationship between the correlation coefficient and covariance for two variables as shown below:
\nr(x, y)<\/sub> = COV (x, y)\/{sx<\/sub> \u00d7 sy<\/sub>}
\nr(x, y)<\/sub> = correlation of the variables x and y
\nCOV (x, y) = covariance of the variables x and y
\nsx<\/sub> = sample standard deviation of the random variable x
\nsy<\/sub> = sample standard deviation of the random variable y
\nNow given COV (x, y) = 0
\nThen r(x, y)<\/sub> = 0<\/p>\n<\/details>\n

\n

Question 11.
\nThe mean of 1, 3, 4, 5, 7, 4 is m the numbers 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then, p + q =
\n(a) 4
\n(b) 5
\n(c) 6
\n(d) 7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 7
\nThe mean of 1, 3, 4, 5, 7, 4 is m
\n\u21d2 (1 + 3 + 4 + 5 + 7 + 4)\/6 = m
\n\u21d2 m = 24\/6
\n\u21d2 m = 4
\nThe numbers 3, 2, 2, 4, 3, 3, p have mean m – 1
\n\u21d2 (3 + 2 + 2 + 4 + 3 + 3 + p)\/7 = m – 1
\n\u21d2 (17 + p)\/7 = 4 – 1
\n\u21d2 (17 + p)\/7 = 3
\n\u21d2 17 + p = 7 \u00d7 3
\n\u21d2 17 + p = 21
\n\u21d2 p = 21 – 17
\n\u21d2 p = 4
\nThe numbers 3, 2, 2, 4, 3, 3, p have median q.
\n\u21d2 The numbers 2, 2, 3, 3, 3, 4, 4 have median q
\n\u21d2 (7 + 1)\/2th term = q
\n\u21d2 4th term = q
\n\u21d2 q = 3
\nNow p + q = 4 + 3 = 7<\/p>\n<\/details>\n

\n

Question 12.
\nIn a series, the coefficient of variation is 50 and standard deviation is 20 then the arithmetic mean is
\n(a) 20
\n(b) 40
\n(c) 50
\n(d) 60<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 40
\nGiven, in a series, the coefficient of variation is 50 and standard deviation is 20
\n\u21d2 (standard deviation\/AM) \u00d7 100 = 50
\n\u21d2 20\/AM = 50\/100
\n\u21d2 20\/AM = 1\/2
\n\u21d2 AM = 2 \u00d7 20
\n\u21d2 AM = 40
\nSo, the arithmetic mean is 40<\/p>\n<\/details>\n

\n

Question 13.
\nThe coefficient of correlation between two variables is independent of
\n(a) both origin and the scale
\n(b) scale but not origin
\n(c) origin but not scale
\n(d) neither scale nor origin<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) both origin and the scale
\nThe coefficient of correlation between two variables is independent of both origin and the scale.<\/p>\n<\/details>\n

\n

Question 14.
\nThe geometric mean of series having mean = 25 and harmonic mean = 16 is
\n(a) 16
\n(b) 20
\n(c) 25
\n(d) 30<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 20
\nThe relationship between Arithmetic mean (AM), Geometric mean (GM) And Harmonic mean (HM) is
\nGM\u00b2 = AM \u00d7 HM
\nGiven AM = 25
\nHM = 16
\nSo GM\u00b2 = 25 \u00d7 16
\n\u21d2 GM = \u221a(25 \u00d7 16)
\n= 5 \u00d7 4
\n= 20
\nSo, Geometric mean = 20<\/p>\n<\/details>\n

\n

Question 15.
\nOne of the methods of determining mode is
\n(a) Mode = 2 Median – 3 Mean
\n(b) Mode = 2 Median + 3 Mean
\n(c) Mode = 3 Median – 2 Mean
\n(d) Mode = 3 Median + 2 Mean<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Mode = 3 Median – 2 Mean
\nWe can calculate the mode as
\nMode = 3 Median – 2 Mean<\/p>\n<\/details>\n

\n

Question 16.
\nIf the correlation coefficient between two variables is 1, then the two least square lines of regression are
\n(a) parallel
\n(b) none of these
\n(c) coincident
\n(d) at right angles<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) coincident
\nIf the correlation coefficient between two variables is 1, then the two least square lines of regression are coincident<\/p>\n<\/details>\n

\n

Question 17.
\nThe mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. then the remaining two observations are
\n(a) 4, 6
\n(b) 6, 8
\n(c) 8, 10
\n(d) 10, 12<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 6, 8
\nGiven mean and variance of 7 observations are 8 and 16.
\nFive observations are 2, 4, 10, 12, 14.
\nLet the other two observations are x and y.
\nSo 7 observations are : 2, 4, 10, 12, 14 ,x ,y
\nNow
\nMean = (2 + 4 + 10 + 12 + 14 + x + y)\/7
\n\u21d2 8 = (2 + 4 + 10 + 12 + 14 + x + y)\/7
\n\u21d2 8 \u00d7 7 = 2 + 4 + 10 + 12 + 14 + x + y
\n\u21d2 56 = 42 + x + y
\n\u21d2 x + y = 56 – 42
\n\u21d2 x + y = 14 ………….. 1
\nAgain Given varience = 16
\n\u21d2 (1\/7) \u00d7 \u2211 (xi – mean)\u00b2 = 16 (7 <= i <= 1)
\n\u21d2 \u2211 (xi – mean)\u00b2 = 16 \u00d7 7
\n\u21d2 \u2211 (xi – mean)\u00b2 = 112
\n\u21d2 {(2 – 8)\u00b2 + (4 – 8)\u00b2 + (10 – 8)\u00b2 + (12 – 8)\u00b2 + (14 – 8)\u00b2 + (x – 8)\u00b2 + (y – 8)\u00b2} = 112
\n\u21d2 {(-6)\u00b2 +(-4)\u00b2 + (2)\u00b2 + (4)\u00b2 + (6)\u00b2 + x\u00b2 + 64 – 16x + y\u00b2 + 64 – 16y } = 112
\n\u21d2 {36 + 16 + 4 + 16 + 36 + x\u00b2 + y\u00b2 + 64 + 64 – 16(x + y) } = 112
\n\u21d2 {108 + x\u00b2 + y\u00b2 + 128 – (16 \u00d7 14)} = 112 (since x + y = 14)
\n\u21d2 {108 + x\u00b2 + y\u00b2 + 128 – 224} = 112
\n\u21d2 x\u00b2 + y\u00b2 + 236 – 224 = 112
\n\u21d2 x\u00b2 + y\u00b2 + 12 = 112
\n\u21d2 x\u00b2 + y\u00b2 = 12 – 12
\n\u21d2 x\u00b2 + y\u00b2 = 100…………….2
\nSquaring equation 1, we get
\n(x + y)\u00b2 = 196
\n\u21d2 x\u00b2 + y\u00b2 + 2xy = 196
\n\u21d2 100 + 2xy = 196
\n\u21d2 2xy = 196 – 100
\n\u21d22xy = 96
\n\u21d2 xy = 96\/2
\n\u21d2 xy = 48 …………. 3
\nNow (x – y)\u00b2 = x\u00b2 + y\u00b2 – 2xy
\n= 100 – 2 \u00d7 48
\n= 100 – 96
\n= 4
\n\u21d2 x – y = \u221a2
\n\u21d2 x – y = 2, -2
\ncase 1: when x – y = 2 and x + y = 14
\nAfter solving it, we get x = 8, y = 6
\ncase 2: when x – y = -2 and x + y = 14
\nAfter solving it, we get x = 6, y = 8
\nSo, the two numbers are 6 and 8<\/p>\n<\/details>\n

\n

Question 18.
\nRange of a data is calculated as
\n(a) Range = Max Value – Min Value
\n(b) Range = Max Value + Min Value
\n(c) Range = (Max Value – Min Value)\/2
\n(d) Range = (Max Value + Min Value)\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Range = Max Value – Min Value
\nRange of a data is calculated as
\nRange = Max Value – Min Value<\/p>\n<\/details>\n

\n

Question 19.
\nMean deviation for n observations x1<\/sub>, x2<\/sub>, ……….., xn<\/sub> from their mean x is given by
\n(a) \u2211(xi<\/sub> – x) where (1 \u2264 i \u2264 n)
\n(b) {\u2211|xi<\/sub> – x|}\/n where (1 \u2264 i \u2264 n)
\n(c) \u2211(xi<\/sub> – x)\u00b2 where (1 \u2264 i \u2264 n)
\n(d) {\u2211(xi<\/sub> – x)\u00b2}\/n where (1 \u2264 i \u2264 n)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) {\u2211|xi<\/sub> – x|}\/n where (1 \u2264 i \u2264 n)
\nMean deviation for n observations x1<\/sub>, x2<\/sub>, ……….., xn<\/sub> from their mean x is calculated as
\n{\u2211|xi<\/sub> – x|}\/n where (1 \u2264 i \u2264 n)<\/p>\n<\/details>\n

\n

Question 20.
\nIf the mean of the following data is 20.6, then the value of p is
\nx: 10 15 p 25 35
\nf: 3 10 25 7 5
\n(a) 30
\n(b) 20
\n(c) 25
\n(d) 10<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 20
\nMean = \u2211 fi<\/sub> \u00d7 xi \/\u2211 fi<\/sub>
\n\u21d2 20.6 = (10 \u00d7 3 + 15 \u00d7 10 + p \u00d7 25 + 25 \u00d7 7 + 35 \u00d7 5)\/(3 + 10 + 25 + 7 + 5)
\n\u21d2 20.6 = (30 + 150 + 25p + 175 + 175)\/50
\n\u21d2 20.6 = (530 + 25p)\/50
\n\u21d2 530 + 25p = 20.6 \u00d7 50
\n\u21d2 530 + 25p = 1030
\n\u21d2 25p = 1030 – 530
\n\u21d2 25p = 500
\n\u21d2 p = 500\/25
\n\u21d2 p = 20
\nSo, the value of p is 20<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the …<\/p>\n

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