A student scores the following marks in five | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the sixth test score be $x$. Given the mean of six tests is $48$,we have: $\frac{54+45+41+43+57+x}{6} = 48$ $240 + x = 288$ $x = 48$. The mark

Vedclass · Accepted Answer

$\frac{10}{\sqrt{3}}$

Vedclass · Answer

(B) Let the sixth test score be $x$. Given the mean of six tests is $48$,we have: $\frac{54+45+41+43+57+x}{6} = 48$ $240 + x = 288$ $x = 48$. The marks are $54, 45, 41, 43, 57, 48$. The mean $\bar{x} = 48$. The standard deviation $\sigma = \sqrt{\frac{1}{n} \sum (x_i - \bar{x})^2}$. $\sigma = \sqrt{\frac{(54-48)^2 + (45-48)^2 + (41-48)^2 + (43-48)^2 + (57-48)^2 + (48-48)^2}{6}}$ $\sigma = \sqrt{\frac{6^2 + (-3)^2 + (-7)^2 + (-5)^2 + 9^2 + 0^2}{6}}$ $\sigma = \sqrt{\frac{36 + 9 + 49 + 25 + 81 + 0}{6}}$ $\sigma = \sqrt{\frac{200}{6}} = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}$.

$A$ student scores the following marks in five tests: $54, 45, 41, 43, 57$. His score is not known for the sixth test. If the mean score is $48$ in six tests,then the standard deviation of marks in six tests is

Similar Questions

Find the variance of the sequence $a, a + d, a + 2d, \dots, a + 2nd$.

In a series of $2n$ observations,half of them are equal to $a$ and the remaining half are equal to $-a$. If the standard deviation of the observations is $2$,then $|a|$ equals:

If the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$,then the variance of the data is:

The mean and variance of six observations are $8$ and $16$ respectively. If each observation is multiplied by $3$,then the new variance of the resulting observations is:

If the mean and variance of the following data: $6, 10, 7, 13, a, 12, b, 12$ are $9$ and $\frac{37}{4}$ respectively,then $(a-b)^{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module