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Question

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

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Solution

Given thet number of observations (n) = 100

Incorrect mean (¯¯¯x) = 40,

Incorrect standard deviation (σ) = 5.1

We know that,

¯¯¯x=1nni=1xi

i.e., 40=1100100i=1xi or 100i=1xi=4000

i.e., Incorrect sum of observations = 4000

Thus, the correct sum of observations = Incorrect sum - 50+40

= 4000-50+40 = 3990

Hence, correct mean = correct sum100=3990100=39.9

Also, standard deviation σ

=1nni=1x2i1n2(ni=1)2

=1nni=1x2i(¯¯¯x)2

5.1=1100×Incorrectni=1x2i(40)2

or 26.01 = 1100×Incorrectni=1x2i1600

Therefore, Incorrect ni=1x2i=100(26.01+1600)=162601

Now, Correct ni=1x2i=Incorrectni=1x2i(50)2+(40)2

= 162601-2500+1600= 161701

Therefore , correct standard deviation

=correctx2in(Correctmean)2

=161701100(39.9)2

=1617.011592.01=25=5


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