Question:
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12.
Answer:
Given mean and SD of 20 observations are 10 and 2 respectively.
Case 1. If wrong items are omitted
Given mean = 10
∑ xi/20 = 10 (1<= i < =20)
=> ∑ xi = 10*20
=> ∑ xi = 200
Observations 8 is recorded incoorectly.
So ∑ xi = 200 - 8
=> ∑ xi = 192
Now new mean is
∑ xi/19 = 192/19 = 10.11
So new mean is 10.11
Again given SD = 2
So varience = (SD)2 = 4
=> ∑ xi2 /20 - (∑ xi/20)2 = 4
=> ∑ xi2 /20 - (10)2 = 4
=> ∑ xi2 /20 - 100 = 4
=> ∑ xi2 /20 = 100 + 4
=> ∑ xi2 /20 = 104
=> ∑ xi2 = 104*20 = 2080
Observation 8 is recorded incoorectly.
Now
∑ xi2 = 2080 - 82
=> ∑ xi2 = 2080 - 64
=> ∑ xi2 = 2016
Now variance of remaining observations are
∑ xi2 /19 - (∑ xi/19)2
=> 2016/19 - 10.112
=> 106.11 - 102.21
=> 3.9
So new SD = √3.9 = 1.975
Case 2. If wrong item 8 is replaced by 12
Given mean = 10
∑ xi/20 = 10 (1<= i < =20)
=> ∑ xi = 10*20
=> ∑ xi = 200
Observations 8 is recorded incoorectly.
So ∑ xi = 200 - 8 + 12
=> ∑ xi = 192 + 12 = 204
Now new mean is
∑ xi/19 = 204/20 = 10.2
So new mean is 10.2
Again given SD = 2
So varience = (SD)2 = 4
=> ∑ xi2 /20 - (∑ xi/20)2 = 4
=> ∑ xi2 /20 - (10)2 = 4
=> ∑ xi2 /20 - 100 = 4
=> ∑ xi2 /20 = 100 + 4
=> ∑ xi2 /20 = 104
=> ∑ xi2 = 104*20 = 2080
Observation 8 is recorded incoorectly.
Now
∑ xi2 = 2080 - 82 + 122
=> ∑ xi2 = 2080 - 64 + 144
=> ∑ xi2 = 2160
Now variance of remaining observations are
∑ xi2 /20 - (∑ xi/20)2
=> 2160/20 - 10.22
=> 108 - 104.04
=> 3.96
So new SD = √3.96 = 1.99
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