- Home
- Standard 11
- Mathematics
The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:
$5$
$8$
$4$
$6$
Solution
$n_{1}=100 \quad n=250$
$\therefore n_{2}=250-100 \Rightarrow n_{2}=150$
$\bar{x}=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}$
$15.6=\frac{100(15)+(150)\left(\bar{x}_{2}\right)}{250}$
$\Rightarrow \overline{\mathrm{x}}_{2}=16$
$\overline{\mathrm{X}}_{1}=15 \quad\quad\quad\quad \Rightarrow$
$\sigma_{1}^{2}=V_{1}(x)=9 \quad \sigma^{2}=\operatorname{Var}(x)=13.44$
$\sigma^{2}=\frac{\mathrm{n}_{1} \sigma_{1}^{2}+\mathrm{n}_{2} \sigma_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}$
$\mathrm{n}_{2}=150, \overline{\mathrm{x}}_{2}=16, \mathrm{~V}_{2}(\mathrm{x})=\sigma_{2}$
$13.44=\frac{100 \times 9+150 \times \sigma_{2}^{2}}{250}+\frac{100 \times 150}{(250)^{2}} \times 1$
$\Rightarrow \sigma_{2}^{2}=16 \Rightarrow \sigma_{2}=4$
Similar Questions
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 :
If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is
$X$ | $c$ | $2c$ | $3c$ | $4c$ | $5c$ | $6c$ |
$f$ | $2$ | $1$ | $1$ | $1$ | $1$ | $1$ |
The diameters of circles (in mm) drawn in a design are given below:
Diameters | $33-36$ | $37-40$ | $41-44$ | $45-48$ | $49-52$ |
No. of circles | $15$ | $17$ | $21$ | $22$ | $25$ |
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]