There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Marks } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & x-2 & x & x^{2} & (x+1)^{2} & 2 x & x+1 \\ \hline \end{array}$
where $x$ is a positive integer. Determine the mean and standard deviation of the marks.
Sum of frequencies,
$x-2+x+x^{2}+(x+1)^{2}+2 x+x+1=60$
$2 x^{2}+7 x-60=0$
$(2 x+15)(x-4)=0$
$x=4$
$\begin{array}{|c|c|c|c|c|} \hline x _{ i } & f_{i} & d_{i}=x_{i}-3 & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 0 & 2 & -3 & -6 & 18 \\ \hline 1 & 4 & -2 & -8 & 16 \\ \hline 2 & 16 & -1 & -16 & 16 \\ \hline A=3 & 25 & 0 & 0 & 0 \\ \hline 4 & 8 & 1 & 8 & 8 \\ \hline 5 & 5 & 2 & 10 & 20 \\ \hline \text { Total } & \Sigma f_{i}=60 & & \Sigma f_{i}=-12 & \Sigma f_{i} d_{i}^{2}=78 \\ \hline \end{array}$
Mean $=A+\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}=3+\left(\frac{-12}{60}\right)=2.8$
Standard Deviation,
$\sigma$=$\sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\right)^{2}}=\sqrt{\frac{78}{60}-\left(\frac{-12}{60}\right)^{2}}=\sqrt{1.3-0.04}=\sqrt{1.26}=1.12$
The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is
Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}<\mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the $5$ observations $9$,$25$, $a$, $b$, $c$ be $18$,$4$ and $\frac{136}{5}$, respectively. Then $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to ..............
The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is
Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to
Let $n \geq 3$. A list of numbers $x_1, x, \ldots, x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers $y_1, y_2, \ldots, y_n$ is made as follows $y_1=\frac{x_1+x_2}{2}, y_2=\frac{x_1+x_2}{2}$ and $y_j=x_j$ for $j=3,4, \ldots, n$.
The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Then, which of the following is necessarily true?