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.
(N/A) Given $n=10, \bar{x}=45$ and $\sigma^{2}=16$.
$\bar{x} = \frac{\Sigma x_{i}}{n} = 45 \Rightarrow \Sigma x_{i} = 450$.
Corrected $\Sigma x_{i} = 450 - 52 + 25 = 423$.
Corrected mean $\bar{x} = \frac{423}{10} = 42.3$.
Using $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$:
$16 = \frac{\Sigma x_{i}^{2}}{10} - (45)^{2} \Rightarrow \Sigma x_{i}^{2} = 10(16 + 2025) = 20410$.
Corrected $\Sigma x_{i}^{2} = 20410 - (52)^{2} + (25)^{2} = 20410 - 2704 + 625 = 18331$.
Corrected $\sigma^{2} = \frac{18331}{10} - (42.3)^{2} = 1833.1 - 1789.29 = 43.81$.
$\bar{x} = \frac{\Sigma x_{i}}{n} = 45 \Rightarrow \Sigma x_{i} = 450$.
Corrected $\Sigma x_{i} = 450 - 52 + 25 = 423$.
Corrected mean $\bar{x} = \frac{423}{10} = 42.3$.
Using $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$:
$16 = \frac{\Sigma x_{i}^{2}}{10} - (45)^{2} \Rightarrow \Sigma x_{i}^{2} = 10(16 + 2025) = 20410$.
Corrected $\Sigma x_{i}^{2} = 20410 - (52)^{2} + (25)^{2} = 20410 - 2704 + 625 = 18331$.
Corrected $\sigma^{2} = \frac{18331}{10} - (42.3)^{2} = 1833.1 - 1789.29 = 43.81$.
Explore More
Similar Questions
Consider the following frequency distribution:
Value $4$ $5$ $8$ $9$ $6$ $12$ $11$ Frequency $5$ $f_1$ $f_2$ $2$ $1$ $1$ $3$
Suppose that the sum of the frequencies is $19$ and the median of this frequency distribution is $6$. For the given frequency distribution,let $\alpha$ denote the mean deviation about the mean,$\beta$ denote the mean deviation about the median,and $\sigma^2$ denote the variance. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.
List-$I$ List-$II$ $(P) \ 7f_1+9f_2$ is equal to $(1) \ 146$ $(Q) \ 19\alpha$ is equal to $(2) \ 47$ $(R) \ 19\beta$ is equal to $(3) \ 48$ $(S) \ 19\sigma^2$ is equal to $(4) \ 145$ $(5) \ 55$
| Value | $4$ | $5$ | $8$ | $9$ | $6$ | $12$ | $11$ |
| Frequency | $5$ | $f_1$ | $f_2$ | $2$ | $1$ | $1$ | $3$ |
Suppose that the sum of the frequencies is $19$ and the median of this frequency distribution is $6$. For the given frequency distribution,let $\alpha$ denote the mean deviation about the mean,$\beta$ denote the mean deviation about the median,and $\sigma^2$ denote the variance. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.
| List-$I$ | List-$II$ |
| $(P) \ 7f_1+9f_2$ is equal to | $(1) \ 146$ |
| $(Q) \ 19\alpha$ is equal to | $(2) \ 47$ |
| $(R) \ 19\beta$ is equal to | $(3) \ 48$ |
| $(S) \ 19\sigma^2$ is equal to | $(4) \ 145$ |
| $(5) \ 55$ |
An analysis of monthly wages paid to workers in two firms $A$ and $B$,belonging to the same industry,gives the following results:
\text{Parameter}
\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}
$A: 586, B: 648$
\text{Mean of monthly wages}
$Rs. 5253$
\text{Variance of wages}
$A: 100, B: 121$
Which firm,$A$ or $B$,pays a larger total amount as monthly wages?
| \text{Parameter} | \text{Firm } $A$ \text{ and Firm } $B$ |
| \text{No. of wage earners} | $A: 586, B: 648$ |
| \text{Mean of monthly wages} | $Rs. 5253$ |
| \text{Variance of wages} | $A: 100, B: 121$ |
Which firm,$A$ or $B$,pays a larger total amount as monthly wages?
Medium
View SolutionThe mean of a data set consisting of $20$ observations is $40$. If one observation $53$ was wrongly recorded as $33$,then the correct mean will be
DifficultJEE MAIN 2013
View SolutionThe mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$,respectively,by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?
Difficult
View SolutionLet $y_1, y_2, y_3, \dots, y_n$ be $n$ observations. Let $w_i = l y_i + k$ for all $i = 1, 2, 3, \dots, n$,where $l$ and $k$ are constants. If the mean of $y_i$ is $48$ and their standard deviation is $12$,and the mean of $w_i$ is $55$ and their standard deviation is $15$,then the values of $l$ and $k$ are:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo