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,
(a) If the wrong item was omitted.
(b) If it is replaced by 12.
(a) If the wrong item was omitted:
When the wrong item 8 is omitted, the correct sum of the observations would be:
Correct sum = (Mean) * (Number of observations)
Correct sum = 10 * 20 = 200
Now, since the incorrect item of 8 was omitted, the correct sum of the remaining 19 observations would be:
Correct sum of 19 observations = 200 - 8 = 192
The correct mean is then:
Correct mean = Correct sum / Number of observations
Correct mean = 192 / 19 = 10.1053
(b) If the wrong item is replaced by 12:
When the wrong item 8 is replaced by 12, the correct sum of the 20 observations would be:
Correct sum = 10 * 20 + 12 - 8 = 200 + 4 = 204
The correct mean is then:
Correct mean = Correct sum / Number of observations
Correct mean = 204 / 20 = 10.2
To calculate the correct standard deviation in each case, we need to use the formula for standard deviation:
Standard deviation = sqrt [(Σ(x - mean)^2) / n]
For case (a) when the wrong item was omitted:
Correct standard deviation = sqrt [(Σ(x - mean)^2) / n]
Correct standard deviation = sqrt [(16/19)] = sqrt(0.8421) ≈ 0.9176
For case (b) when the wrong item was replaced by 12:
Correct standard deviation = sqrt [(Σ(x - mean)^2) / n]
Correct standard deviation = sqrt [(16/20)] = sqrt(0.8) ≈ 0.8944
Therefore, the correct mean and standard deviation would be:
(a) Correct mean = 10.1053, Correct standard deviation ≈ 0.9176
(b) Correct mean = 10.2, Correct standard deviation ≈ 0.8944