The data is obtained in tabular form as follows.
${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |
The data is obtained in tabular form as follows.
${x_i}$ | ${f_i}$ | ${f_i} = \frac{{{x_i} - 64}}{1}$ | ${y_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
$60$ | $2$ | $-4$ | $16$ | $-8$ | $32$ |
$61$ | $1$ | $-3$ | $9$ | $-3$ | $9$ |
$62$ | $12$ | $-2$ | $4$ | $-24$ | $48$ |
$63$ | $29$ | $-1$ | $1$ | $-29$ | $29$ |
$64$ | $25$ | $0$ | $0$ | $0$ | $0$ |
$65$ | $12$ | $1$ | $1$ | $12$ | $12$ |
$66$ | $10$ | $2$ | $4$ | $20$ | $40$ |
$67$ | $4$ | $3$ | $9$ | $12$ | $36$ |
$68$ | $5$ | $4$ | $16$ | $20$ | $80$ |
$100$ | $220$ | $0$ | $286$ |
Mean, $\bar x = A\frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 64 + \frac{0}{{100}} \times 1 = 64 + 0 = 64$
Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}^2} } \right)} } \right]$
$=\frac{1}{100^{2}}[100 \times 286-0]$
$=2.86$
$\therefore$ Standard deviation $(\sigma)=\sqrt{2.86}=1.69$
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
Suppose values taken by a variable $x$ are such that $a \le {x_i} \le b$, where ${x_i}$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, ...n.$ Then..
The mean and variance of $7$ observations are $8$ and $16$ respectively. If two observations are $6$ and $8 ,$ then the variance of the remaining $5$ observations is:
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.
If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is