Question:
The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted
Answer:
Given mean and SD of 100 observations are 20 and 3 respectively.
Now
∑ xi/100 = 20 (1<= i < =100)
=> ∑ xi = 100*20
=> ∑ xi = 2000
3 observations 21, 21 and 18 are recorded incoorectly.
So ∑ xi = 2000 - 21 - 21 - 18
=> ∑ xi = 2000 - 60
=> ∑ xi = 1940
Now new mean is
∑ xi/100 = 1940/97 = 20
So new mean is 20.SD
Again given SD = 3
So varience = (SD)2 = 9
=> ∑ xi2 /100 - (∑ xi/100)2 = 9
=> ∑ xi2 /100 - (20)2 = 9
=> ∑ xi2 /100 - 400 = 9
=> ∑ xi2 /100 = 400 + 9
=> ∑ xi2 /100 = 409
=> ∑ xi2 = 40900
3 observations 21, 21 and 18 are recorded incoorectly.
Now
∑ xi2 = 40900 - 212 - 212 - 182
=> ∑ xi2 = 40900 - 441 - 441 - 324
=> ∑ xi2 = 40900 - 1206
=> ∑ xi2 = 39694
Now varience of remaing observations are
∑ xi2 /97 - (∑ xi/97)2
= 39694/97 - 202
= 409.216 - 400
= 9.216
So new SD = √9.216 = 3.035
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