While calculating mean and variance of 10 readings, a student wrongly used reading 52 for correct re

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13.Statistics

hard

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Option A

Option B

Option C

Option D

Solution

Given $n=10, \bar{x}=45$ and $\sigma^{2}=16$

$\begin{array}{c}\bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\\Rightarrow \quad \frac{\Sigma x_{i}}{10}=45 \Rightarrow \quad \Sigma x_{i}=450 \\\text { Corrected } \Sigma x_{i}=450-52+25=423\end{array}$

$\therefore \quad$ Corrected mean, $\bar{x}=\frac{423}{10}=42.3$

$\Rightarrow \quad \sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}$

$\begin{array}{ll}\Rightarrow & 16=\frac{\Sigma x_{i}^{2}}{10}-(45)^{2} \\ \Rightarrow & \Sigma x_{i}^{2}=20410\end{array}$

$\therefore \quad$ Corrected $\Sigma x_{i}^{2}=20410-(53)^{2}+(25)^{2}=18331$

And Corrected $\sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81$

Standard 11

Mathematics

The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$

Height Weight

Mean $162.6\,cm$ $52.36\,kg$

Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

Can we say that the weights show greater variation than the heights?

medium

If each observation of a raw data whose variance is ${\sigma ^2}$, is multiplied by $\lambda$, then the variance of the new set is

normal

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to

hard

(JEE MAIN-2023)

If the variance of the first $n$ natural numbers is $10$ and the variance of the first m even natural numbers is $16$, then $m + n$ is equal to

hard

(JEE MAIN-2020)

If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to

medium

(JEE MAIN-2022)

	Height	Weight
Mean	$162.6\,cm$	$52.36\,kg$
Variance	$127.69\,c{m^2}$	$23.1361\,k{g^2}$

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Solution

Similar Questions

The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$ Height Weight Mean $162.6\,cm$ $52.36\,kg$ Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$ Can we say that the weights show greater variation than the heights?

If each observation of a raw data whose variance is ${\sigma ^2}$, is multiplied by $\lambda$, then the variance of the new set is

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to

If the variance of the first $n$ natural numbers is $10$ and the variance of the first m even natural numbers is $16$, then $m + n$ is equal to

If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to

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The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$

Height Weight

Mean $162.6\,cm$ $52.36\,kg$

Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

Can we say that the weights show greater variation than the heights?