Question

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

Answer 1

To Find: Find the correct mean and standard deviation. 

Explanation: Here, n=100, \(\bar X\) = 40, σ = 5.1

Now, 

Corrected   \(\Sigma x_i\) = 4000 - 50 + 40

3990

So, The mean of correct score = \(\frac{corrected\,sum}{n}\) 

\(\frac{3990}{100}\) 

Mean = 39.9

Now, Standard variance σ = 5.1 

Since, Variance = σ²

Variance = 26.01

Corrected Variance = \(\Big(\frac{1}{n}\times\,corrected\,\Sigma x_i\Big)-(corrected\,mean)^2\) \(\Big(\frac{1}{100}\times162591\Big)-(39.9)^2\)

1625.91-1592.01

Corrected variance =34 (Approx) 

Now, Corrected Standard Deviation = \(\sqrt{Corrected\,variance}\)

σ = \(\sqrt{34}\)

Correct Deviation is 5.83 

Hence, The correct Mean is 39.9 and Correct SD is 5.83