# Class 11 Maths MCQ – Statistics

This set of Class 11 Maths Chapter 15 Multiple Choice Questions & Answers (MCQs) focuses on “Statistics”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. Find the mean deviation about the median of the scores of a batsman given below.

Innings Scores
1 20
2 56
3 0
4 84
5 11
6 120

a) 10
b)10.5
c) 11
d) 9

Explanation: Mean deviation = $$\frac{1}{n}$$[Σi = n |xI – A|], where A is median or AM
From the given data, Median, A = (20 + 56)/2 = 38
⇒ Mean deviation = 1/6 x (63) = 10.5.

2. What is the mean deviation from the mean for the following data?

 117 156 206 198 223

a) 0
b) 3
c) 1
d) ½

Explanation: Mean = (117 + 156 + 206 + 198 + 223)/5 = 180

 Xi Xi – mean 117 156 206 198 223 -63 -24 26 18 43

Mean deviation = $$\frac{1}{n}$$[Σi = 5 |xI – mean|] = 1/5 x [(-63) + (-24) + 26 + 18 + 43] = 1/5 x [0] = 0.

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3. The mean deviation of an ungrouped data is 150. If each observation is increased by 3.5%, then what is the new mean deviation?
a) 153.5
b) 3.5
c) 155.25
d) 150

Explanation: If x1, x2, …, xn are the observations, then the new observations are (1.035) x1, (1.035) x2, ……, (1.035) xn.
Therefore, the new mean is (1.035) x̄
Now, Mean deviation = $$\frac{1}{n}$$[Σi = n |xI – mean|]
⇒ New mean deviation = $$\frac{1}{n}$$[Σi = n|(1.035)xI – (1.035) x̄|] = (1.035) × $$\frac{1}{n}$$[Σi = n |xI – mean|] = 1.035 x 150 = 155.25.

4. Find the mean deviation about mean from the following data:

 xi fi 3 5 20 25 27 5 12 20 8 15

a) 7.7
b) 15
c) 8.7
d) 6.2

Explanation: From the given data,

xi fi fixi |xi-18| fi|xi-18|
3 5 15 15 75
5 12 60 13 156
20 20 400 2 40
25 8 200 7 56
27 15 405 9 135
Σ fi = 60 Σ fixi = 1080 Σ fi|xi – 15| = 462

Now, Mean = $$\frac{1}{n}$$ Σ fixi = 1080/60 = 18
⇒ Mean deviation = $$\frac{1}{n}$$ Σ fi|xi – 18| = 462/60 = 7.7.

5. What is the geometric mean of 5,52, ….,5n?
a) 5n/2
b) 5(n+1)/2
c) 5n(n+1)/2
d) 5n

Explanation: Geometric Mean = (5 x 52 x …… x 5n)1/n = [5(1+2+…+n)]1/n = [5n(n+1)/2]1/n = 5(n+1)/2.

6. In a class there are 20 juniors, 15 seniors and 5 graduate students. If the junior averaged 65 in the midterm exam, the senior averaged 70 and the graduate students averaged 91, then what is the mean of the centre class approximately?
a) 71
b) 74
c) 70
d) 72

Explanation: Combined mean = (Σ xini)/(Σ ni) = (20 × 65 + 15 × 70 + 5 × 91)/(20 + 15 + 5) = 70.

7. Find the mean deviation from mean of the observations: a, a+d, …., (a+2nd).
a) n(n + 1)d2/3
b) n(n + 1)d2/2
c) a + n(n + 1)d2/2
d) n(n + 1)d/(2n + 1)

Explanation: Mean = $$\frac{1}{n}$$ Σ xi = $$\frac{1}{2n+1}$$ [a + (a + d) + … + (a + 2nd)] = a + nd
⇒ Mean Deviation = $$\frac{1}{2n+1}$$ [2 × d × (1 + 2 + … + n)] = [n (n + 1) (d)]/(2n + 1).

More MCQs on Class 11 Maths Chapter 15:

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