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Mathematics Question on Variance and Standard Deviation

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On respectively, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

A

3.86\sqrt{3.86}

B

1.8

C

3.96\sqrt{3.96}

D

1.94

Answer

3.96\sqrt{3.96}

Explanation

Solution

The given mean is:

xˉ=10  ⟹  Σxi20=10.\bar{x} = 10 \implies \frac{\Sigma x_i}{20} = 10.

Thus:

Σxi=10×20=200.\Sigma x_i = 10 \times 20 = 200.

When the incorrect observation (8) is replaced with the correct value (12):

Σxi=200−8+12=204.\Sigma x_i = 200 - 8 + 12 = 204.

The corrected mean is:

xˉ=Σxi20=20420=10.2.\bar{x} = \frac{\Sigma x_i}{20} = \frac{204}{20} = 10.2.

The standard deviation (S.D.) is given as:

S.D.2=Variance=22=4.\text{S.D.}^2 = \text{Variance} = 2^2 = 4.

From the variance formula:

Σxi220−(Σxi20)2=4.\frac{\Sigma x_i^2}{20} - \left(\frac{\Sigma x_i}{20}\right)^2 = 4.

Substitute:

Σxi220−102=4.\frac{\Sigma x_i^2}{20} - 10^2 = 4. Σxi220=104  ⟹  Σxi2=2080.\frac{\Sigma x_i^2}{20} = 104 \implies \Sigma x_i^2 = 2080.

After replacing 8 with 12:

Σxi2=2080−82+122=2080−64+144=2160.\Sigma x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160.

The corrected variance is:

Σxi220−(Σxi20)2.\frac{\Sigma x_i^2}{20} - \left(\frac{\Sigma x_i}{20}\right)^2. 216020−(10.2)2.\frac{2160}{20} - (10.2)^2. Σxi220=108,(10.2)2=104.04.\frac{\Sigma x_i^2}{20} = 108, \quad (10.2)^2 = 104.04. Variance=108−104.04=3.96.\text{Variance} = 108 - 104.04 = 3.96.

The corrected standard deviation is:

S.D.=3.96.\text{S.D.} = \sqrt{3.96}.