The sum of 100 observations and the sum of their squares are 400 and 2475, respectively.  Later on, three observations, 3, 4 and 5, were found to be incorrect.  If the incorrect observations are omitted, then the variance of the remaining observations is : Option 1) 8.25 Option 2) 8.50 Option 3) 8.00 Option 4) 9.00

As we learnt in

ARITHMETIC Mean -

For the values x1, x2, ....xn of the variant x the arithmetic mean is given by

$\dpi{100} \bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}$

in case of discrete data.

-

Short cut Method for Variance -

In case of discrete frequency distribution

$\sigma ^{2}= \frac{1}{N}\sum_{i=1}^{n}f_{i}d_{i}^{2}-\left ( \frac{1}{N}\sum_{i=1}^{n} f_{i}d_{i}\right )^{2}$

-

$\sum x_{i}=400$

$\frac{\sum x_{i}}{100}=4$

Omitting 3, 4, 5 we get $\frac{\sum x_{i}}{97}=\frac{400-12}{97}=4$

Sum of squares = 2475-50=2425

$\frac{\sum x_{i^{2}}}{100}=25$

Variance = $\frac{\sum x_{i^{2}}}{100}-\left( \frac{\sum x_{i}}{100}\right )^{2}$

=25-$4^{2}=9$

Option 1)

8.25

This option is incorrect

Option 2)

8.50

This option is incorrect

Option 3)

8.00

This option is incorrect

Option 4)

9.00

This option is correct

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-