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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:
$\frac{16}{3}$
$4$
$\frac{17}{3}$
$5$
Solution
$\text { Given } 32+8 \alpha+9 \beta=(8+\alpha+\beta) \times 6$
$\Rightarrow 2 \alpha+3 \beta=16 \quad \ldots \text { (i) }$
$\text { Also, } 4 \times 16+4 \times \alpha+9 \beta=(8+\alpha+\beta) \times 6.8$
$\Rightarrow 640+40 \alpha+90 \beta=544+68 \alpha+68 \beta$
$\Rightarrow 28 \alpha-22 \beta=96$
$\Rightarrow 14 \alpha-11 \beta=48 \quad \ldots (ii)$
from $(i)\, \, (ii)$
$\alpha=5 \, \,\beta=2$
so, new mean $=\frac{32+35+18}{15}=\frac{85}{15}=\frac{17}{3}$
Similar Questions
Find the mean, variance and standard deviation using short-cut method
| Height in cms | $70-75$ | $75-80$ | $80-85$ | $85-90$ | $90-95$ | $95-100$ | $100-105$ | $105-110$ | $110-115$ |
| No. of children | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
Find the mean and variance for the data
| ${x_i}$ | $92$ | $93$ | $97$ | $98$ | $102$ | $104$ | $109$ |
| ${f_i}$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
