Let the mean and variance of the frequency di | vedclass

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Question

$	ext { Given } 32+8 \alpha+9 \beta=(8+\alpha+\beta) 	imes 6$

$\Rightarrow 2 \alpha+3 \beta=16 \quad \ldots 	ext { (i) }$

$	ext { Also, } 4

Vedclass · Accepted Answer

$\frac{17}{3}$

Vedclass · Answer

$	ext { Given } 32+8 \alpha+9 \beta=(8+\alpha+\beta) 	imes 6$

$\Rightarrow 2 \alpha+3 \beta=16 \quad \ldots 	ext { (i) }$

$	ext { Also, } 4 	imes 16+4 	imes \alpha+9 \beta=(8+\alpha+\beta) 	imes 6.8$

$\Rightarrow 640+40 \alpha+90 \beta=544+68 \alpha+68 \beta$

$\Rightarrow 28 \alpha-22 \beta=96$

$\Rightarrow 14 \alpha-11 \beta=48 \quad \ldots (ii)$

from $(i)\, \, (ii)$

$\alpha=5 \, \,\beta=2$

so, new mean $=\frac{32+35+18}{15}=\frac{85}{15}=\frac{17}{3}$

$\mathrm{x}$	$\mathrm{x}_{1}=2$	$\mathrm{x}_{2}=6$	$\mathrm{x}_{3}=8$	$\mathrm{x}_{4}=9$
$\mathrm{f}$	$4$	$4$	$\alpha$	$\beta$

Class:	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$

Let the mean and variance of the frequency distribution

$\mathrm{x}$ $\mathrm{x}_{1}=2$ $\mathrm{x}_{2}=6$ $\mathrm{x}_{3}=8$ $\mathrm{x}_{4}=9$

$\mathrm{f}$ $4$ $4$ $\alpha$ $\beta$

be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:

Similar Questions

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:

If the mean of the frequency distribution

Class: $0-10$ $10-20$ $20-30$ $30-40$ $40-50$

Frequency $2$ $3$ $x$ $5$ $4$

is $28$ , then its variance is $........$.

The frequency distribution:

$\begin{array}{|l|l|l|l|l|l|l|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$

Find the standard deviation.

Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$

The variance of $\alpha$, $\beta$ and $\gamma$ is $9$, then variance of $5$$\alpha$, $5$$\beta$ and $5$$\gamma$ is

Let the mean and variance of the frequency distribution $\mathrm{x}$ $\mathrm{x}_{1}=2$ $\mathrm{x}_{2}=6$ $\mathrm{x}_{3}=8$ $\mathrm{x}_{4}=9$ $\mathrm{f}$ $4$ $4$ $\alpha$ $\beta$ be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be: