Class 10 Maths Chapter 14 Statistics MCQ

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

ANSWER= D. Cumulative frequency

 

2. 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

ANSWER= D. all the above

 

3. If x1, x2, x3,….., xn are the observations of a given data. Then the mean of the observations will be:

A. Sum of observations/Total number of observations
B. Total number of observations/Sum of observations
C. Sum of observations +Total number of observations
D. None of the above

ANSWER= A. Sum of observations/Total number of observations Explanation: The mean or average of observations will be equal to the ratio of sum of observations and total number of observations.

 

4. Cumulative frequency curve is also called

A. histogram
B. ogive
C. bar graph
D. median

ANSWER= B. ogive

 

5. The relationship between mean, median and mode for a moderately skewed distribution is

A. mode = median – 2 mean
B. mode = 3 median – 2 mean
C. mode = 2 median – 3 mean
D. mode = median – mean

ANSWER= B. mode = 3 median – 2 mean

 

6. The median of set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set

A. is increased by 2
B. is decreased by 2
C. is two times of the original number
D. Remains the same as that of the original set.

ANSWER= D. Remains the same as that of the original set.

 

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

A. 12k
B. 14k
C. 15k
D. 16k

ANSWER= B. 14k

 

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

A. mean
B. median
C. mode
D. all the three above

ANSWER= B. median

 

9. While computing mean of grouped data, we assume that the frequencies are [NCERT Exemplar Problems]

A. evenly distributed over all the classes
B. centred at the classmarks of the classes
C. centred at the upper limits of the classes
D. centred at the lower limits of the classes

ANSWER= B. centred at the classmarks of the classes

 

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

A. centred at the upper limits of the classes
B. centred at the lower limits of the classes
C. centred at the classmarks of the classes
D. evenly distributed over all the classes

ANSWER= C. centred at the classmarks of the classes

 

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

A. Mean
B. Median
C. Mode
D. None of these

ANSWER= A. Mean

 

12. Mode is the

A. middle most frequent value
B. least frequent value
C. maximum frequent value
D. none of these

ANSWER= C. maximum frequent value

 

13. If the mean of frequency distribution is 7.5 and ∑fi xi = 120 + 3k, ∑fi = 30, then k is equal to:

A. 40
B. 35
C. 50
D. 45

ANSWER= B. 35 Explanation: As per the given question, Xmean = ∑fi xi /∑fi 7.5 = (120+3k)/30 225 = 120+3k 3k = 225-120 3k= 105 k=35

 

14. The mode and mean is given by 7 and 8, respectively. Then the median is:

A. 1/13
B. 13/3
C. 23/3
D. 33

ANSWER= C. 23/3

 

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

A. 8
B. 10
C. 9
D. 15

ANSWER= A. 8 Explanation: mean = (4 + 10 + 5 + 9 + 12)/5 = 40/5 = 8 15. The median of the data 13, 15, 16, 17, 19, 20 is: A. 30/2 B. 31/2 C. 33/2 D. 35/2

 

16. 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

ANSWER= C. 5 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

 

17. 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

ANSWER= D. 4 Explanation: Mean of AM = 10 (a + a + 3 + a + 6 + a + 9 + a + 12)/5 = 10 5a + 30 = 50 5a = 20 a = 4

 

18. 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

ANSWER= B. 12.5 Explanation: Class mark = (Upper limit + Lower limit)/2 = (15 + 10)/2 = 25/2 = 12.5

 

19. 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

ANSWER= D. 10 Explanation: Given, ∑fi = 24 ∑fi = x + 5 + 6 + 1 + 2 = 14 + x 24 = 14 + x x = 24 – 14 = 10

 

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

A. mean
B. median
C. mode
D. all the above three measures

ANSWER= B. median

 

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

A. centred at the class marks of the classes
B. evenly distributed over all the classes
C. centred at the upper limits of the classes
D. centred at the lower limits of the classes

ANSWER= A. centred at the class marks of the classes

 

22. 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

ANSWER= A. 3 Median = Mode + 2 Mean

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