JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Statistics

Let in a series of

2 n

observations, half of them are equal to a and remaining half are equal to -a. Also by adding a constant

b

in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of

a^{2} + b^{2}

is equal to :

Answer: (C)

Solution

The

2 n

observations are

n

copies of

a

and

n

copies of

- a

, so their mean is

0

and their variance is

\frac{n ( a - 0 ) ^{2} + n ( - a - 0 ) ^{2}}{2 n} = a^{2}

giving standard deviation

∣ a ∣

. Adding the constant

b

to every observation shifts the mean by

b

and leaves every deviation from the mean unchanged, so the new mean is

b = 5

and the new standard deviation is

∣ a ∣ = 20

. Hence

a^{2} = 400

and

a^{2} + b^{2} = 400 + 25 = 425

The value of

a^{2} + b^{2}

425

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option