Mean Deviation - Statistics Questions and Answers

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Statistics – Mean Deviation”.

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
View Answer

Answer: b
Explanation: Mean deviation = \(\frac{1}{n}\)[Σ_{i = n} |x_I – 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

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a) 0
b) 3
c) 1
d) ½
View Answer

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

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

Mean deviation = \(\frac{1}{n}\)[Σ_{i = 5} |x_I – 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
View Answer

Answer: c
Explanation: If x₁, x₂, …, x_n are the observations, then the new observations are (1.035) x₁, (1.035) x₂, ……, (1.035) x_n.
Therefore, the new mean is (1.035) x̄
Now, Mean deviation = \(\frac{1}{n}\)[Σ_{i = n} |x_I – mean|]
⇒ New mean deviation = \(\frac{1}{n}\)[Σ_i = n|(1.035)x_I – (1.035) x̄|] = (1.035) × \(\frac{1}{n}\)[Σ_{i = n} |x_I – mean|] = 1.035 x 150 = 155.25.

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

x_i	3	5	20	25	27
f_i	5	12	20	8	15

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

Answer: a
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 f_i = 60	Σ f_ix_i = 1080		Σ f_i\|x_i – 15\| = 462

Now, Mean = \(\frac{1}{n}\) Σ f_ix_i = 1080/60 = 18
⇒ Mean deviation = \(\frac{1}{n}\) Σ f_i|x_i – 18| = 462/60 = 7.7.

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

Answer: b
Explanation: Geometric Mean = (5 x 5² x …… x 5ⁿ)^1/n = [5^(1+2+…+n)]^1/n = [5^n(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
View Answer

Answer: c
Explanation: Combined mean = (Σ x_in_i)/(Σ n_i) = (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)d^2/3
b) n(n + 1)d²/2
c) a + n(n + 1)d²/2
d) n(n + 1)d/(2n + 1)
View Answer

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

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