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 $(\sigma)=\sqrt{2.86}=1.69$
Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
Let $\mathrm{X}$ be a random variable with distribution.
$\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
$\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals
For two data sets, each of size $5$, the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$, respectively. The variance of the combined data set is
The mean of the numbers $a, b, 8,5,10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean, then $25\; M$ is equal to