JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Statistics

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The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If

α

and

β

are the mean and standard deviation respectively for correct data, then

(α, β)

is:

(A) $(11, 25)$
(B) $(11, 26)$
(C) $(10.5, 26)$ ✓ Correct
(D) $(10.5, 25)$

Answer: (C)

Solution

Given:

n = 20

, incorrect mean

\overset{x}{ˉ} = 10

, incorrect SD

= 2.5

.

Incorrect sum

= 20 \times 10 = 200

.
Correct sum

= 200 - 25 + 35 = 210

.
Correct mean

α = \frac{210}{20} = 10.5

.

For variance:

σ^{2} = \frac{\sum x _{i}^{2}}{n} - \overset{x}{ˉ}^{2}

.

Incorrect:

2. 5^{2} = \frac{\sum x _{i}^{2}}{20} - 100

, so

\sum x_{i}^{2} = 20 (106.25) = 2125

.

Correct

\sum x_{i}^{2} = 2125 - 2 5^{2} + 3 5^{2} = 2125 - 625 + 1225 = 2725

.

Correct variance

β = \frac{2725}{20} - (10.5)^{2} = 136.25 - 110.25 = 26

.

Therefore

(α, β) = (10.5, 26)

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Statistics

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