Question

For a group of 200 candidates, the mean and the standard deviations of scores were found to be 40 and 15 respectively. Later on, it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.

Answer 1

To Find: Find the correct mean and standard deviation. 

Explanation: Here, n=200, \(\bar X\) = 40, σ = 15

Since, the score was incorrect, 

Now, The sum is incorrect 

Corrected \(\Sigma x_i\) = 8000 - 34 - 53 + 43 + 35

8000 - 7 

The correct score is 7993 

So, The mean of correct score = \(\frac{\Sigma x}{n}\) 

\(\frac{7993}{200}\) 

Mean = 39.95 

Now, Standard variance = 15 

Since, Variance = \(\sigma^2\) 

Variance = 255

Now, the correct \(\Sigma (x_i)^2 \) = 365000 - 34²- 53² + 43² + 35² 

365000 - 1156 - 2809+1849 + 1225

\(\Sigma (x_i)^2 = 364109\) 

Corrected Variance = 
\(\Big(\frac{1}{n}\times corrected \ \Sigma x_i\Big)\) - (Corrected mean)²

\(\Big(\frac{1}{200} \times 364109\Big) - (39.95)^2\) 

1820.54 - 1596.40

Corrected variance =224.14 

Now, Corrected Standard Deviation 

= \(\sqrt{Corrected\ variance}\) 

^{\(\sigma = \sqrt{224.14}\)} 

Correct Deviation is 14.97