JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 26)
The variance $$\sigma^2$$ of the data
| $$x_i$$ | 0 | 1 | 5 | 6 | 10 | 12 | 17 |
|---|---|---|---|---|---|---|---|
| $$f_i$$ | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
is _________.
Answer
29
Explanation
| $$\mathrm{x_i}$$ | $$\mathrm{f_i}$$ | $$\mathrm{f_i x_i}$$ | $$\mathrm{f_i x^2_i}$$ |
|---|---|---|---|
| 0 | 3 | 0 | 0 |
| 1 | 2 | 2 | 2 |
| 5 | 3 | 15 | 75 |
| 6 | 2 | 12 | 72 |
| 10 | 6 | 60 | 600 |
| 12 | 3 | 36 | 432 |
| 17 | 3 | 51 | 867 |
| $$\Sigma \mathrm{f}_{\mathrm{i}}=22$$ | $$\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=2048$$ |
$$\begin{aligned} & \therefore \quad \Sigma \mathrm{f}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=176 \\ & \text { So } \overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{176}{22}=8 \\ & \text { for } \sigma^2=\frac{1}{\mathrm{~N}} \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^2-(\overline{\mathrm{x}})^2 \\ & =\frac{1}{22} \times 2048-(8)^2 \\ & =93.090964 \\ & =29.0909 \\ & \end{aligned}$$
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