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.
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.
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 $\sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81$
Similar Questions
The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......
The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......
- [JEE MAIN 2022]
Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
- [AIEEE 2012]
Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
- [KVPY 2013]
The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
If it is replaced by $12$
The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
If it is replaced by $12$
If the mean and variance of the data $65,68,58,44$, $48,45,60, \alpha, \beta, 60$ where $\alpha>\beta$ are $56$ and $66.2$ respectively, then $\alpha^2+\beta^2$ is equal to
If the mean and variance of the data $65,68,58,44$, $48,45,60, \alpha, \beta, 60$ where $\alpha>\beta$ are $56$ and $66.2$ respectively, then $\alpha^2+\beta^2$ is equal to
- [JEE MAIN 2024]