CBSE Class 10 Maths MCQ Questions of Statistics with answers accessible here. Multiple Choice Questions for Class 10 Statistics with answers Set up according to the most recent syllabus recommended by CBSE, NCERT. MCQ Questions for a significant part of the examination for class 10 Statistics and whenever practiced appropriately can assist you with getting better grades. Refer to more Chapter-wise MCQs for Class 10 Statistics and furthermore download CBSE Class 10 Sample papers for all subjects.

CBSE Class 10 Maths MCQ Questions with answers will help students to score good marks in board exams. Check out below lists of MCQ Questions which is helpful in exams:-

Practice Class 10 Maths MCQ Questions of Statistics

1. The method used to find the mean of a given data is(are):

(a) direct method
(b) assumed mean method
(c) step deviation method
(d) all the above

1. The ________ of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.

(a) Class mark
(b) Class height
(c) Average frequency
(d) Cumulative frequency

1. Mode is the

(a) middle most frequent value
(b) least frequent value
(c) maximum frequent value
(d) none of these

1. One of the methods for determining mode is

(a) Mode = 2 Median -3 Mean
(b) Mode = 3 Median – 2 Mean
(c) Mode = 2 Mean – 3 Median
(d) Mode = 3 Mean – 2 Median

1. The algebraic sum of the deviations of a frequency distribution from its mean is always,

(a) greater than zero
(b) less than zero
(c) zero
(d) a non-zero number

1. While computing mean of the grouped data, we assume that the frequencies are:

(a) evenly distributed over all the classes
(b) centered at the class marks of the classes
(c) centered at the upper limits of the classes
(d) centered at the lower limits of the classes

1. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency of a grouped data gives its:

(a) Mean
(b) Median
(c) Mode
(d) All of these

1. If 35 is removed from the data, 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by:

(a) 2
(b) 1.5
(c) 1
(d) 0.5

1. Which of the following can not be determined graphically?

(a) Mean
(b) Median
(c) Mode
(d) None of these

1. The empirical relationship between the three measures of central tendency is

(a) 3 Median = Mode + 2 Mean
(b) 2 Median = Mode + 2 Mean
(c)  3 Median = Mode + Mean
(d)  3 Median = Mode – 2 Mean

1. Construction of a cumulative frequency table is useful in determining the

(a) mean
(b) median
(c) mode
(d) all the above three measures

1. If the sum of frequencies is 24, then the value of x in the observation: x, 5,6,1,2, will be;

(a) 4
(b) 6
(c) 8
(d) 10

1. The class interval of a given observation is 10 to 15, then the class mark for this interval will be:

(a) 11.5
(b) 12.5
(c) 12
(d) 14

1. If AM of a, a+3, a+6, a+9 and a+12 is 10, then a is equal to;

(a) 1
(b) 2
(c) 3
(d) 4

1. If the mean of first n natural numbers is 3n/5, then the value of n is:

(a) 3
(b) 4
(c) 5
(d) 6

1. The median of the data 13, 15, 16, 17, 19, 20 is:

(a) 30/2
(b) 31/2
(c) 33/2
(d) 35/2

1. Mode is the value of the variable which has:

(a) maximum frequency
(b) minimum frequency
(c) mean frequency
(d) middle most frequency

1. Mode and mean of a data are 12k and 15A. Median of the data is

(a) 12k
(b) 14k
(c) 15k
(d) 16k

1. If mean = (3 median – mode) . k, then the value of k is

(a) 1
(b) 2
(C) 12
(d) 32

1. The mean of the data: 4, 10, 5, 9, 12 is;

(a) 8
(b) 10
(c) 9
(d) 15

1. Answer: (d) all the above

Explanation: The mean for a given data can be calculated using either of the following methods.

Direct method

Assumed mean method

Step deviation method

Explanation: The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.

1. Answer: (c) maximum frequent value
2. Answer: (b) Mode = 3 Median – 2 Mean
4. Answer: (b) centered at the class marks of the classes

Explanation: In computing the mean of grouped data, the frequencies are centred at the class marks of the classes.

Explanation: Since the intersection point of less than type ogive and more than ogive gives the median on the abscissa.

Explanation: We have

30, 34, 35, 36, 37, 38, 39, 40

The data has 8 observations, so there are two middle terms, 4th and 5th term i.e. 36 and 37.

The median is the mean of both these terms.

Median = (36 + 37)/2

Median = 36.5

When 35 is removed from given data as 30, 35, 36, 37, 38, 39, 40 then the number of observations becomes 7.

Now the median is the middle most i.e 4th term which is equal to 37.

Therefore median is increased by 37 – 36.5 = 0.5

2. Answer: (a) 3 Median = Mode + 2 Mean

Explanation: The empirical relationship between the three measures of central tendency is 3 Median = Mode + 2 Mean.

Explanation: Construction of a cumulative frequency table is useful in determining the median.

Explanation: Given,

∑fi = 24

∑fi = x + 5 + 6 + 1 + 2 = 14 + x

24 = 14 + x

x = 24 – 14 = 10

Explanation: Class mark = (Upper limit + Lower limit)/2

= (15 + 10)/2

= 25/2

= 12.5

Explanation: Mean of AM = 10

(a + a + 3 + a + 6 + a + 9 + a + 12)/5 = 10

5a + 30 = 50

5a = 20

a = 4

Explanation: Sum of natural numbers = n(n + 1)/2

Given, mean = 3n/5

Mean = sum of natural numbers/n

3n/5 = n(n + 1)/2n

3n/5 = (n + 1)/2

6n = 5n + 5

n = 5

Explanation: For the given data, there are two middle terms, 16 and 17.

Hence, median = (16 + 17)/2 = 33/2

Explaination: ∵ Mode = 3 median – 2 mean
⇒ 12k = 3 median – 2 × 15k
⇒ 42k = 3 median
⇒ Median = 14k.

Explaination: ∵ Mode = 3 median – 2 mean
⇒ 2 mean = 3 median – mode
⇒ mean = 12 (3 median – mode)
⇒ k= 12