SOLUTION: Who's online? Tutor, please help. The mean and standard deviation of 15 observations are found to be 10 and 5 respectively. On rechecking, it was found that one of the observation

Question 1204909: Who's online? Tutor, please help.
The mean and standard deviation of 15 observations are found to be 10 and 5 respectively. On rechecking, it was found that one of the observations with value 7 was incorrect. Calculate the correct mean and standard deviation if the correct observation is 22
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

mean = (sum of values)/(number of values)
mean = S/n
mean = S/15
S/15 = 10
S = 15*10
S = 150

The value 7 is removed and replaced with 22.
The net change for the sum is -7+22 = +15
The sum 150 increases to S = 150+15 = 165

Correct mean = (sum of values)/(number of values)
Correct mean = S/n
Correct mean = 165/15
Correct mean = 11

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Your teacher has not specified if you are using sample standard deviation or population standard deviation.
I'll assume the 2nd option.

sigma = = population standard deviation

One formula to calculate sigma is

I'll replace the portion with A and replace the portion with S

The formula updates to

Notice the S/n part is really just the mean.

Let's determine the value of A based on the incorrect sigma = 5 value mentioned and the sample size n = 15
The value of S/n is the old incorrect mean of 10 (do NOT use 11 just yet).
This is because we want to peel back the layers to see what incorrect A value was found. From there we can make the proper adjustment.

This indicates the incorrect sum of squares is based on the incorrect value of 7.

Remove the 7 and replace it with 22.
We lose 7^2 from that sum of squares value, and gain 22^2.
The net change is -7^2+22^2 = +435
Meaning the 1875 should be 1875+435 = 2310 which is the correct sum of squares.

With the proper observation (22) in place, the sum of squares should be

Let's now recalculate sigma based on that corrected value of A and corrected value of S (2310 and 165 respectively).
In other words, the S/n part is 11 now.
Refer to the previous section.
The value n = 15 stays the same.

So,

approximately
Round this however your teacher instructs.

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Answers:

mean = 11
standard deviation = 5.744562646 approximately

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