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. If the incorrect observation $8$ is replaced by $12$,calculate the correct mean and standard deviation.
- AMean $= 10.2$,Standard Deviation $= 1.99$
- BMean $= 10.2$,Standard Deviation $= 1.9899$
- CMean $= 10.5$,Standard Deviation $= 1.98$
- DMean $= 10.2$,Standard Deviation $= 1.98$
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An online exam is attempted by $50$ candidates,out of which $20$ are boys. The average marks obtained by boys is $12$ with a variance of $2$. The variance of marks obtained by $30$ girls is also $2$. The average marks of all $50$ candidates is $15$. If $\mu$ is the average marks of girls and $\sigma^{2}$ is the variance of marks of $50$ candidates,then $\mu+\sigma^{2}$ is equal to ...... .
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View SolutionBased on the following statements,choose the correct option.
Statement-$I:$ The variance of the first $n$ even natural numbers is $\frac{n^2-1}{4}$.
Statement-$II:$ The difference between the variance of the first $20$ even natural numbers and their arithmetic mean is $112$.
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View SolutionThe mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted and $a$ and $b$ are respectively the mean and variance of the remaining $6$ observations,then $a+3b-5$ is equal to $..........$.
DifficultJEE MAIN 2023
View SolutionThe 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 if the wrong item is omitted.
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