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13.Statistics
easy
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:
A
$\frac{16}{3}$
B
$4$
C
$\frac{17}{3}$
D
$5$
(JEE MAIN-2021)
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 $=\frac{32+35+18}{15}=\frac{85}{15}=\frac{17}{3}$
Standard 11
Mathematics
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