Question:
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
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
Show Hint
The shortcut formula for variance is
\[
\sigma^2
=
\frac{\sum x_i^2}{n}
-
\left(\frac{\sum x_i}{n}\right)^2
\]
which avoids calculating each deviation individually.
Updated On: Jun 25, 2026
- \(1500\)
- \(1600\)
- \(1700\)
- \(1800\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Write the formula for variance.
For \(n\) observations, \[ \text{Variance} = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \] Given, \[ n=20 \] \[ \sum x_i=1000 \] and \[ \sum x_i^2=84000 \]
Step 2: Calculate the mean.
Mean is \[ \bar{x}=\frac{\sum x_i}{n} \] Thus, \[ \bar{x}=\frac{1000}{20}=50 \]
Step 3: Calculate variance.
Now, \[ \frac{\sum x_i^2}{n} = \frac{84000}{20} = 4200 \] Also, \[ \left(\frac{\sum x_i}{n}\right)^2 = 50^2 = 2500 \] Therefore, \[ \text{Variance}=4200-2500 \] Hence, \[ \text{Variance}=1700 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{1700} \]
For \(n\) observations, \[ \text{Variance} = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \] Given, \[ n=20 \] \[ \sum x_i=1000 \] and \[ \sum x_i^2=84000 \]
Step 2: Calculate the mean.
Mean is \[ \bar{x}=\frac{\sum x_i}{n} \] Thus, \[ \bar{x}=\frac{1000}{20}=50 \]
Step 3: Calculate variance.
Now, \[ \frac{\sum x_i^2}{n} = \frac{84000}{20} = 4200 \] Also, \[ \left(\frac{\sum x_i}{n}\right)^2 = 50^2 = 2500 \] Therefore, \[ \text{Variance}=4200-2500 \] Hence, \[ \text{Variance}=1700 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{1700} \]
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