Mathematics Multiple Choice Questions 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
Answer: b
Clarification: 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?
a) 0
b) 3
c) 1
d) ½
Answer: a
Clarification: 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.
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
Answer: c
Clarification: 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 |
a) 7.7
b) 15
c) 8.7
d) 6.2
Answer: a
Clarification: 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.