One set containing five numbers has mean 8 an | vedclass

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

Question

(b) Here ${n_1} = 5$, ${\bar x_1} = 8$, $\sigma _1^2 = 18$, ${n_2} = 3$ ${\bar x_2} = 8$, $\sigma _2^2 = 24$

$\bar x = $ combined mean $ = \frac{{5

Vedclass · Accepted Answer

$20.25$

Vedclass · Answer

(b) Here ${n_1} = 5$, ${\bar x_1} = 8$, $\sigma _1^2 = 18$, ${n_2} = 3$ ${\bar x_2} = 8$, $\sigma _2^2 = 24$

$\bar x = $ combined mean $ = \frac{{5 	imes 8 + 3 	imes 8}}{{5 + 3}}$ $ = \frac{{64}}{8} = 8$

Combined variance $ = \frac{{{n_1}(\sigma _1^2 + D_1^2) + {n_2}(\sigma _2^2 + D_2^2)}}{{{n_1} + {n_2}}}$,

where ${D_1} = {\bar x_1} - \bar x$, ${D_2} = {\bar x_2} - \bar x$

Now, ${D_1} = 8 - 8;\,\,{D_2} = 8 - 8 = 0$

Combined variance $ = \frac{{5(18) + 3(24)}}{{5 + 3}}$ $ = \frac{{90 + 72}}{8}$ $ = \frac{{162}}{8}$ = $20.25$.

Class	$10-20$	$20-30$	$30-40$
Frequency	$2$	$x$	$2$

One set containing five numbers has mean $8$ and variance $18$ and the second set containing $3$ numbers has mean $8$ and variance $24$. Then the variance of the combined set of numbers is

Similar Questions

If the variance of the following frequency distribution is $50$ then $x$ is equal to:

Class $10-20$ $20-30$ $30-40$

Frequency $2$ $x$ $2$

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to

If the standard deviation of $0, 1, 2, 3, …..,9$ is $K$, then the standard deviation of $10, 11, 12, 13 …..19$ is

Find the mean and variance of the frequency distribution given below:

$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x<3 & 3 \leq x<5 & 5 \leq x<7 & 7 \leq x<10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$

Let $\mathrm{n}$ be an odd natural number such that the variance of $1,2,3,4, \ldots, \mathrm{n}$ is $14 .$ Then $\mathrm{n}$ is equal to ..... .