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13.Statistics
hard
For a statistical data $x _1, x _2, \ldots, x _{10}$ of $10$ values, a student obtained the mean as $5.5$ and $\sum_{i=1}^{10} x _{ i }^2=371$. He later found that he had noted two values in the data incorrectly as $4$ and $5$ , instead of the correct values $6$ and $8$ , respectively. The variance of the corrected data is
A$7$
B$4$
C$9$
D$5$
(JEE MAIN-2025)
Solution
$\text { Mean } \overline{x}=5.5$
$=\sum_{i=1}^{10} x_{i}=5.5 \times 10=55$
$=\sum_{i=1}^{10} x_{i}^2=371$
$\left(\sum x_{i}\right)_{\text {new }}=55-(4+5)+(6+8)=60$
$\left(\sum x_{i}\right)_{\text {new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430$
$\text { Variance } \sigma^2=\frac{\sum x_{i}^2}{10}-\left(\frac{\sum x_{i}}{10}\right)^2$
$\sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2$
$\sigma^2=43-36$
$\sigma^2=7$
$=\sum_{i=1}^{10} x_{i}=5.5 \times 10=55$
$=\sum_{i=1}^{10} x_{i}^2=371$
$\left(\sum x_{i}\right)_{\text {new }}=55-(4+5)+(6+8)=60$
$\left(\sum x_{i}\right)_{\text {new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430$
$\text { Variance } \sigma^2=\frac{\sum x_{i}^2}{10}-\left(\frac{\sum x_{i}}{10}\right)^2$
$\sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2$
$\sigma^2=43-36$
$\sigma^2=7$
Standard 11
Mathematics
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 :
hard
