Mathematics Questions and Answers – Statistics – Mean Deviation

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

Answer: b
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
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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

Xi 117 156 206 198 223
Xi – mean -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
View Answer

Answer: c
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 3 5 20 25 27
fi 5 12 20 8 15
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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 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.

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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
View Answer

Answer: b
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
View Answer

Answer: c
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)
View Answer

Answer: d
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).

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter