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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=1nxi=¯¯¯¯¯X=200×40=8000

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

 = 800 -34-53+43+35 = 7991

Corrected mean = correctedxin=7991200=39.955

Now, σ=15

152=1200(x2i)(1200xi)2

255=1200(x2i)(8000200)2

255=1200(x2i)1600

x2i=200×1825=365000

  Incorrect x2i=365000

x2i=(incorrectx21)

- (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=1nx2i(1nxi)2

=364109200(7991200)2

=1820.5451596.402=14.97

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