JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Statistics

Let the mean and variance of 12 observations be

\frac{9}{2}

and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is

\frac{m}{n}

, where m and n are co-prime, then

m + n

is equal to

(A)316
(B)314
(C)317✓ Correct
(D)315

Answer: (C)

Solution

Given mean

\overset{x}{ˉ} = \frac{9}{2}

and variance

σ^{2} = 4

.

First, find the corrected mean:

\overset{x}{ˉ}_{new} = \frac{12 \times \frac{9}{2} + 7 + 14 - 9 - 10}{12} = \frac{56}{12} = \frac{14}{3}

From the original variance:

σ^{2} = \frac{\sum x _{i}^{2}}{12} - (\frac{9}{2})^{2} = 4

\frac{\sum x _{i}^{2}}{12} = 4 + \frac{81}{4} = \frac{97}{4}

\sum x_{i}^{2} = 291

Corrected sum of squares:

\sum (x_{i}^{2})_{new} = 291 - 9^{2} - 1 0^{2} + 7^{2} + 1 4^{2} = 291 - 81 - 100 + 49 + 196 = 355

Corrected variance:

σ_{new}^{2} = \frac{355}{12} - (\frac{14}{3})^{2} = \frac{355}{12} - \frac{196}{9} = \frac{1065 - 784}{36} = \frac{281}{36}

Since

g cd (281, 36) = 1

, we have

m = 281

and

n = 36

.

Therefore,

m + n = 281 + 36 = 317

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Statistics →

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Statistics

Solution

More JEE Main questions on Statistics