CBSE Class 10 Mathematics Statistics MCQs Set 06

Question. The median class for the following data is
Class interval: 20-40, 40-60, 60-80, 80-100
Frequency:        10,      12,       20,       22
(a) 20–40
(b) 40–60
(c) 60–80
(d) 80–100
Answer: C

Question. For a frequency distribution, mean, median and mode are connected by the relation
(a) Mode = 3 Mean – 2 Median
(b) Mode = 2 Median – 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean
Answer: C

Question. The mean and mode of a frequency distribution are 28 and 16 respectively. The median is
(a) 22
(b) 23.5
(c) 24
(d) 24.5
Answer: C

Question. If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 10
Answer: B

Question. The mean of 1, 2, 3, 4, ........, \(n\) is given by
(a) \(\frac{n(n + 1)}{2}\)
(b) \(\frac{(n + 1)}{4}\)
(c) \(\frac{n}{2}\)
(d) \(\frac{(n + 1)}{2}\)
Answer: D

Question. The mean of 15 numbers is 25. If each number is multiplied by 4, mean of the new numbers is
(a) 60
(b) 100
(c) 10
(d) none of the options
Answer: B

Question. The correct formula for finding the mode of a grouped frequency distribution is
(a) \(Mode = h + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times l\)
(b) \(Mode = f_1 + \left( \frac{f_1 - f_0}{2h - f_1 - f_2} \right) \times l\)
(c) \(Mode = l - \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
(d) \(Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
Answer: D

Question. Consider the following frequency distribution.
Class interval: 1-7, 8-14, 15-21, 22-28, 29-35
Frequency:       3,    10,       5,         8,       12
The upper limit of the median class is
(a) 14.5
(b) 14.5
(c) 28
(d) 28.5
Answer: D

Question. Extreme value of a given data
(a) affect the median
(b) do not affect the median
(c) nothing can be said
(d) none of the options
Answer: B

Question. One of the properties of mode is
(a) Not easy to calculate
(b) It is not affected by greatest and least values
(c) Algebraic
(d) Difference of greatest and least values
Answer: B

Question. The mean of \(n\) observations is \(\bar{X}\). If the first item is increased by 1, second by 2 and so on, then the new mean is
(a) \(\bar{X} + n\)
(b) \(\bar{X} + \frac{n}{2}\)
(c) \(\bar{X} + \frac{n + 1}{2}\)
(d) none of the options
Answer: C

Question. Look at the frequency distribution table given below.
Class interval: 35-45, 45-55, 55-65, 65-75
Frequency:         8,       12,       20,      10
The median of the above distribution is
(a) 56.5
(b) 57.5
(c) 58.5
(d) 59
Answer: B

Question. The mean, mode and median of the observations, 7, 7, 5, 7 and \(x\) are the same. Then the observation \(x\) is
(a) 10
(b) 9
(c) 8
(d) 7
Answer: B

Question. Mean of 20 observations is 15. If each observation is multiplied by \(\frac{2}{3}\), then the mean of new observations is
(a) 10
(b) 30
(c) 45
(d) 15
Answer: A

Question. The mean of six numbers : \(x – 5, x – 1, x, x + 2, x + 4\) and \(x + 12\) is 15. Find the mean of first four numbers.
(a) 11
(b) 12
(c) 13
(d) 14
Answer: B

Question. The numbers are arranged in the descending order : 108, 94, 88, 82, \(x + 7, x – 7\), 60, 58, 42, 39. If the median is 73, the value of \(x\) is
(a) 72
(b) 73
(c) 76
(d) 75
Answer: B

Question. The mean of \(x_1, x_2,.......,x_n\) is \(M\). If \(x_i\), \(i = 1,2,......, n\) is replaced by \(5x_i\), the mean becomes \(M_1\), then \(M_1\) is equal to
(a) \(5M\)
(b) \(M + 5\)
(c) \(M + 100\)
(d) \(10 M\)
Answer: A

Question. If mean of ten consecutive odd numbers is 120, then the mean of first five odd numbers among them is
(a) 113
(b) 115
(c) 114
(d) 116
Answer: B

Question. The numbers 3, 5, 7 and 9 have their respectively frequencies \(x – 2, x + 2\), and \(x – 3, x + 3\). If the mean is 6.5, then the value of \(x\) is
(a) 3
(b) 4
(c) 5
(d) 6
Answer: C

Assertion & Reasoning Based MCQs

Directions : In these questions, a statement of Assertion is followed by a statement of Reason is given.
Choose the correct answer out of the following choices :
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.

Question. Assertion : If the arithmetic mean of 5, 7, \(x\), 10, 15 is \(x\), then \(x = 9.25\).
Reason : If \(x_1, x_2, x_3, ..., x_n\) are \(n\) values of a variable \(X\), then the arithmetic mean of these values is given by \(\frac{x_1 + x_2 + x_3 + ... + x_n}{2n}\).
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: C