Class 10 Maths Chapter 14 Statistics MCQs

Class 10 Maths MCQs for Chapter 14 (Statistics) are provided here, online. Students can practice these objective type questions, prepared as per CBSE syllabus and NCERT curriculum. It will help students to score good marks in board exams. All the multiple choice questions are presented here, chapter-wise, with their answers and detailed explanations.

Class 10 Maths MCQs for Statistics

1. 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

Explanation: The mean or average of observations will be equal to the ratio of sum of observations and total number of observations.

xmean=x1+x2+x3+…..+xn/n

2. 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

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

3. 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

Explanation: Using Empirical formula,

Mode = 3Median – 2 Mean

3Median = Mode+2Mean

Median = (Mode+2Mean)/3

Median = (7+2.8)/3 = (7+16)/3 = 23/3

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

(a)8

(b)10

(c)9

(d)15

Answer: a

Explanation: mean = (4+10+5+9+12)/5 = 40/5 = 8

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

(a)30/2

(b)31/2

(c)33/2

(d)35/2

Answer: c

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

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

6. 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

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

7. 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

Explanation: Mean of AM = 10

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

5a+ 30 = 50

5a=20

a=4

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

(a)11.5

(b)12.5

(c)12

(d)14

Answer: b

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

= (15+10)/2

= 25/2

= 12.5

9. 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

Explanation:

Given,

∑fi = 24

∑fi = x+5+6+1+2=14+x

24 = 14+x

x=24-14 = 10

10. The mean of following distribution is:

xi 11 14 17 20
fi 3 6 8 7

(a)15.6

(b)17

(c)14.8

(d)16.4

Answer: d

Explanation:

xi fi fixi
11 3 33
14 6 84
17 8 136
20 7 140
∑fi = 24 ∑fi xi=393

xmean = ∑fi xi/∑fi = 393/24 = 16.4

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