This set of Class 11 Maths Chapter 15 Multiple Choice Questions & Answers (MCQs) focuses on “Statistics – Variance and Standard Deviation”.
1. Find the variance of the observation values taken in the lab.
| 4.2 | 4.3 | 4 | 4.1 |
a) 0.27
b) 0.28
c) 0.3
d) 0.31
View Answer
Explanation: X = (4.2 + 4.3 + 4 + 4.1)/4 = 4.15
| xi | xi – X | (xi – X)2 |
|---|---|---|
| 4.2 | 0.05 | 0.0025 |
| 4.3 | 1.04 | 1.0816 |
| 4 | -0.15 | 0.0225 |
| 4.1 | -0.05 | 0.0025 |
| Σ (xi – X)2 = 1.11 |
Variance = \(\frac{1}{n}\) Σ (xi – X)2 = 1.11/4 = 0.28
2. If the standard deviation of a data is 0.012. Find the variance.
a) 0.144
b) 0.00144
c) 0.000144
d) 0.0000144
View Answer
Explanation: S.D. = √variance
⇒ (0.012)2 = Variance
⇒ Variance = 0.000144
3. The mean of two samples of the sizes 250 and 320 were found to be 20,12 respectively. Their standard deviations were 2 & 5, respectively. Find the variance of combined sample of size 650.
a) 10.23
b) 32.5
c) 65
d) 26.17
View Answer
Explanation: Combined Mean = X = (n1x1 + n2x2)/(n1 + n2) = (250 × 20 + 320 × 12)/650 = 13.6
Let d1 = x1 – X = 20-13.6 = 6.4, d2 = x2 – X = 12-13.6 = -1.6
Variance = [250(6.4 + 40.96) + 320(13.6 + 2.56)]/650 = 26.17.
4. Find the variance of the first 10 natural numbers.
a) 7.25
b) 7
c) 8.25
d) 8
View Answer
Explanation: Variance = \(\frac{1}{10}\) [12 + 22 +…+ 102] – \(\frac{1}{20}\)[1 + 2 +…. 10]2 = 38.5 – 30.25 = 8.25.
5. If the standard deviation of the numbers 2, 4, 5 & 6 is a constant a, then find the standard deviation of the numbers 4, 6, 7 & 8.
a) a + 2
b) 2a
c) 4a
d) a
View Answer
Explanation: The standard deviation is independent of the change of origin. So, the standard deviation for the new set will remain the same.
6. Assuming the variance of four numbers w, x, y, and z as 9. Find the variance of 5w, 5x, 5y and 5z.
a) 225
b) 5/9
c) 9/5
d) 54
View Answer
Explanation: (σx)2 = h2(σu)2, if u = (x – h)/a
Now, h = (1/5). ⇒ Variance, (σu)2 = 9 × 25 = 225
7. A fisherman is weighing each of 50 fishes. Their mean weight worked out is 50 gm and a standard deviation of 2.5 gm. Later it was found that the measuring scale was misaligned and always under reported every fish weight by 2.5 gm. Find the mean and standard deviation of fishes.
a) 52.5,2.5
b) 30,5
c) 50,5
d) 48.5,2.5
View Answer
Explanation: Since mean(X + b) = meanX + b and Var(X + b) = VarX, so we get correct mean as 50 + 2.5 = 52.5 gm and S.D. is 2.5 gm.
8. What is the median and standard deviation of a distribution are 50 and 5 respectively, if each item is increased by 4.
a) Median will increase and S.D. will increase
b) Both will remain same
c) median will go up by 2 but S.D. will remain same
d) median will increase and S.D. will decrease
View Answer
Explanation: Median will change if the observations are changes but standard deviation is unaffected by the origin. So, here the median will go up by 4 and S.D. will remain same.
9. If the S.D. is a set of observations is 4 and if each observation is divided by 4, find the S.D. of the new observations.
a) 4
b) 3
c) 2
d) 1
View Answer
Explanation: σ1 = 4 and d = (X – 0)/4
And σ1 = |4|σd
⇒ σd = 4/4 = 1.
10. The change in which of following terms does not affect the standard deviation?
a) Origin
b) Scale
c) Origin and scale
d) Neither origin nor scale
View Answer
Explanation: Change in origin does not affect the standard deviation, whereas standard deviation is affected by scale.
11. The mean and Standard deviation of a sample were found to be 9.5 and 2.5, respectively. Later, an additional observation 15 was added to the original data. Find the S.D. of the 11 observation.
a) 2.6
b) 2.8
c) 2.86
d) 3.24
View Answer
Explanation: \(\frac{1}{n}\) (Σ xi) = 9.5,
\(\frac{1}{n}\) (Σ [(xi)2 – (X)2] = 6.25, Σ xi = 95 and \(\frac{1}{10}\) Σ[(X)2] = 96.5
Corrected mean = (95 + 15)/11 = 10
Corrected Variance = [(1/11) x (965 + 225)] – 100 = 90/11
⇒ Standard deviation = √Variance = √(90/11) = 2.86.
12. The mean of 5 observations is 3 and variance is 2. If three of the five observations are 1, 3, 5, find the other two.
a) 2, 6
b) 3, 3
c) 1, 5
d) 2, 4
View Answer
Explanation: \(\frac{1}{5}\) (Σ xi) = 3 = 1 + 3 + 5 + x4 + x5
⇒ x4 + x5 = 15 ———— (i)
Now, \(\frac{1}{5}\) Σ (xi)2 – \([\frac{1}{5}\) Σ xi]2 = 4
⇒ (x4)2 + (x5)2 = 30 —————-(ii)
Solving equation i and ii, x4 = 2 and x5 = 4.
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