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Statistics — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSStatistics4 Marks
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.
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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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