JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Statistics

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Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is

(A)32
(B)38✓ Correct
(C)40
(D)36

Answer: (B)

Solution

For set

A

\sum a_{i} = 5 \cdot 5 = 25

and, from the variance,

\sum a_{i}^{2} = 5 (12 + 5^{2}) = 185

. For set

B

\sum b_{i} = 5 \cdot 8 = 40

and

\sum b_{i}^{2} = 5 (20 + 8^{2}) = 420

. Subtracting

3

from each element of

A

leaves the variance

12

with new mean

2

, so

\sum (a_{i} - 3)^{2} = 5 (12 + 2^{2}) = 80

; adding

2

to each element of

B

leaves the variance

20

with new mean

10

, so

\sum (b_{i} + 2)^{2} = 5 (20 + 1 0^{2}) = 600

. For the combined set

C

10

elements, the mean is

\frac{5 \cdot 2 + 5 \cdot 10}{10} = 6

, and the variance is

\frac{80 + 600}{10} - 6^{2} = 68 - 36 = 32

. Hence the sum of the mean and the variance of

C

6 + 32 = 38

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Statistics

Solution

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