Mean and standard deviation of 10 observations are 20 and 84 respectively. Later on, it was observed

English

Gujarati

Hindi

13.Statistics

hard

The mean and standard deviation of $10$ observations are $20$ and $84$ respectively. Later on, it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is:

A

$14$

B

$13$

C

$12$

D

$11$

(JEE MAIN-2023)

Solution

$\mu=20, \sigma=8$

$\mu_{\text {Corrected }}=\frac{200-50+40}{10}=19$

$\sigma^2=\frac{1}{10} \sum x_i^2-20^2$

$(64+400) 10=\sum x_i^2$

$\sigma_{\text {Corrected }}^2=\frac{1}{10}[(64+400) 10-2500+1600]-19^2$

$=374-361$

$=13$

Standard 11

Mathematics

Share

Similar Questions

Find the variance of the following data: $6,8,10,12,14,16,18,20,22,24$

medium

Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}<\mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the $5$ observations $9$,$25$, $a$, $b$, $c$ be $18$,$4$ and $\frac{136}{5}$, respectively. Then $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to …………..

hard

(JEE MAIN-2024)

For a given distribution of marks mean is $35.16$ and its standard deviation is $19.76$. The co-efficient of variation is..

easy

The varience of data $1001, 1003, 1006, 1007, 1009, 1010$ is –

medium

If the variance of observations ${x_1},\,{x_2},\,……{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}…….,\,a{x_n}$, $\alpha \ne 0$ is

easy