JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Statistics

WhatsApp Email

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :

(A)10
(B)36
(C)43✓ Correct
(D)60

Answer: (C)

Solution

Let the incorrect observation be

x_{1}

and correct be

x_{1}^{'}

. Given

x_{1} + x_{1}^{'} = 70

.

From the incorrect mean:

\sum x_{i} = 50 \times 15 = 750

.
From the correct mean:

\sum^{'} x_{i} = 50 \times 16 = 800

.

So

x_{1}^{'} - x_{1} = 50

. Combined with

x_{1} + x_{1}^{'} = 70

, we get

x_{1}^{'} = 60, x_{1} = 10

.

From

σ^{2} = 4

\frac{\sum x _{i}^{2}}{50} - 225 = 4

, so

\frac{\sum x _{i}^{2}}{50} = 229

.

For correct variance:

σ^{'2} = \frac{\sum ^{'} x _{i}^{2}}{50} - 256 = \frac{\sum x _{i}^{2} - 100 + 3600}{50} - 256 = 229 + 70 - 256 = 43

.

Answer: (C)

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Statistics →

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Statistics

Solution

More JEE Main questions on Statistics