While calculating the mean and variance of 10 readings, a student… - Vidyadip

Question

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

✓

Answer

Mean = 45
Variance = 16
n = 10
$\sum\text{x}_{\text{i}}=450$
Corrected Sum = 450 - 52 + 25 = 423
Corrected Mean = 42.3
Variance = 16
$16=\frac{\sum\text{x}_{\text{i}}^2}{10}-(45)^2$
Incorrected $\sum\text{x}_{\text{i}}^2=20410$
Corrected $\sum\text{x}_{\text{i}}^2=$
Incorrect $\sum\text{x}_{\text{i}}^2$ - (Sum of squares of incorrect value) + (Sum of squares of corrected value)
Corrected $\sum\text{x}_{\text{i}}^2=20410-2704+625=18331$
Corrected $\sigma=\sqrt{\frac{\text{Corrected}\sum\text{x}_\text{i}^2}{\text{n}}-(\text{Corrected Mean})^2}$
Corrected $\sigma=\sqrt{\frac{18331}{10}-(42.3)^2}=6.62$
Corrected Variance = 6.62 × 6.62 = 43.82

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