Search any question & find its solution
Question:
Answered & Verified by Expert
If the total number of observations is $20, \sum x_i=1000$ and $\sum x_i^2=84000$, then the variance of the distribution is
Options:
Solution:
2970 Upvotes
Verified Answer
The correct answer is:
1700
Given number of observation is 20 .
$$
\begin{aligned}
& \mathrm{x}=20, \sum \mathrm{x}_{\mathrm{i}}=1000 \text { and } \sum \mathrm{x}_{\mathrm{i}}^2=84000 \\
& \sigma^2=\frac{1}{\mathrm{x}} \sum_{\mathrm{i}=1}^{\mathrm{x}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2 \\
& =\frac{1}{\mathrm{x}} \sum_{\mathrm{i}=1}^{\mathrm{x}}\left(\mathrm{x}_1^2+\overline{\mathrm{x}}^2-2 \mathrm{x}_{\mathrm{i}} \overline{\mathrm{x}}\right) \\
& \sigma^2=\frac{1}{\mathrm{x}} \sum \mathrm{x}_{\mathrm{i}}^2-\overline{\mathrm{x}}^2=\frac{1}{\mathrm{x}} \sum \mathrm{x}_{\mathrm{i}}^2-\left(\frac{1}{\mathrm{x}} \sum \mathrm{x}_{\mathrm{i}}\right)^2 \\
& \sigma^2=\frac{1}{20} \times 84000-\left(\frac{1}{20} \times 1000\right)^2 \\
& \sigma^2=4200-2500=1700
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{x}=20, \sum \mathrm{x}_{\mathrm{i}}=1000 \text { and } \sum \mathrm{x}_{\mathrm{i}}^2=84000 \\
& \sigma^2=\frac{1}{\mathrm{x}} \sum_{\mathrm{i}=1}^{\mathrm{x}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2 \\
& =\frac{1}{\mathrm{x}} \sum_{\mathrm{i}=1}^{\mathrm{x}}\left(\mathrm{x}_1^2+\overline{\mathrm{x}}^2-2 \mathrm{x}_{\mathrm{i}} \overline{\mathrm{x}}\right) \\
& \sigma^2=\frac{1}{\mathrm{x}} \sum \mathrm{x}_{\mathrm{i}}^2-\overline{\mathrm{x}}^2=\frac{1}{\mathrm{x}} \sum \mathrm{x}_{\mathrm{i}}^2-\left(\frac{1}{\mathrm{x}} \sum \mathrm{x}_{\mathrm{i}}\right)^2 \\
& \sigma^2=\frac{1}{20} \times 84000-\left(\frac{1}{20} \times 1000\right)^2 \\
& \sigma^2=4200-2500=1700
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.