The mean and standard deviation of the marks of 10 students were found to be 50 and 12 res…

Question

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is \_\_\_\_\_\_\_\_\_

Answer: 269

Accepted Answer

The initial mean is given by:

$\vec{x}=50$

So, the total sum of the marks initially was:

$\sum {x}_{i}=\vec{x}	imes n=50	imes 10=500$

We later realize that two marks were incorrectly read as 45 and 50, when they should have been 20 and 25. Therefore, the corrected sum of the marks is:

$\sum {x}_{i}=\sum {x}_{i}-45-50+20+25=500-45-50+20+25=450$

The initial variance is given as:

${\sigma }^{2}=144$

We know that variance is calculated as the mean of the squares minus the square of the mean. Therefore, rearranging gives:

$\frac{\sum {x}_{2}^{i}}{n}={\sigma }^{2}+{\vec{x}}^{2}=144+{50}^{2}=2644$

Then, the sum of the squares of the initial marks is:

$\sum {x}_{2}^{i}=n	imes \frac{\sum {x}_{2}^{i}}{n}=10	imes 2594=26440$

The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:

$\sum {x}_{2}^{i}=\sum {x}_{2}^{i}-{45}^{2}-{50}^{2}+{20}^{2}+{25}^{2}=26400-{45}^{2}-{50}^{2}+{20}^{2}+{25}^{2}=22940$

Now we can calculate the corrected variance. The variance is the mean of the squares minus the square of the mean. Using the corrected values gives:

${\sigma }_{2}^{}=\frac{\sum {x}_{2}^{i}}{n}-{(\frac{\sum x{i}_{}}{n})}^{2}=\frac{22940}{10}-{(\frac{450}{10})}^{2}=2294-{45}^{2}=2294-2025=269$

Therefore, the correct variance is 269.

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