- Home
- Standard 11
- Mathematics
13.Statistics
hard
The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is $............$.
A
$265$
B
$269$
C
$264$
D
$289$
(JEE MAIN-2023)
Solution
Sol. $\bar{x}=50$
$\sum x_i=500$
$\sum x_{i \text { correct }}=500+20+25-45-50=450$
$\sigma^2=144$
$\frac{\sum x_i^2}{10}-(50)^2=144$
$\sum x_{i c o r r e c t}^2=\left(144+(50)^2\right) \times 10-(45)^2-(50)^2+(20)^2+(25)^2$
$22940$
Correct variance $=\frac{\sum\left(x_{\text {icorrect }}\right)^2}{10}-\left(\frac{\sum x_{\text {icorrect }}}{10}\right)^2$
$=2294-(45)^2$
$=2294-2025=269$
Standard 11
Mathematics
