Question

For a group of 200 candidates, the mean and 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.

Solution

We have,

n = 200, ¯¯¯¯¯X=40,σ=15

∴¯¯¯¯¯X=1n∑xi=¯¯¯¯¯X=200×40=8000

Corrected~ ∑xi = Incorrect~ ∑xi - (sum of incorrect values) + (sum of correct values)

= 800 -34-53+43+35 = 7991

∴ Corrected mean = corrected∑xin=7991200=39.955

Now, σ=15

⇒152=1200(∑x2i)−(1200∑xi)2

⇒255=1200(∑x2i)−(8000200)2

⇒255=1200(∑x2i)−1600

⇒∑x2i=200×1825=365000

⇒ Incorrect ∑x2i=365000

∑x2i=(incorrect∑x21)

- (sum of square of incorrect values ) + (sum of squares of correct values )

=365000−(34)2−(53)2=(43)2+(35)2

= 364019

891

so, Correct σ=√1n∑x2i−(1n∑xi)2

=√364109200−(7991200)2

=√1820.545−1596.402=14.97

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