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

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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) About Me  (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 = sigma = population standard deviation

One formula to calculate sigma is
sigma+=+sqrt%28+sum%28x%5E2%29%2Fn+-+%28sum%28x%29%2Fn%29%5E2+%29

I'll replace the sum%28x%5E2%29 portion with A and replace the sum%28x%29 portion with S

The formula updates to
sigma+=+sqrt%28+A%2Fn+-+%28S%2Fn%29%5E2+%29
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.

sigma+=+sqrt%28+A%2Fn+-+%28S%2Fn%29%5E2+%29

5+=+sqrt%28+A%2F15+-+%28150%2F15%29%5E2+%29

5+=+sqrt%28+A%2F15+-+%2810%29%5E2+%29

5%5E2+=+A%2F15+-+100

25+=+A%2F15+-+100

25%2B100+=+A%2F15

A%2F15+=+125

A+=+15%2A125

A+=+1875

This indicates the incorrect sum of squares is sum%28x%5E2%29+=+1875 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 sum%28x%5E2%29+=+2310

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,
sigma+=+sqrt%28+A%2Fn+-+%28S%2Fn%29%5E2+%29

sigma+=+sqrt%28+2310%2F15+-+%28165%2F15%29%5E2+%29

sigma+=+sqrt%28+154+-+%2811%29%5E2+%29

sigma+=+sqrt%28+154+-+121+%29

sigma+=+sqrt%28+33+%29

sigma+=+5.744562646 approximately
Round this however your teacher instructs.

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Answers:

mean = 11
standard deviation = 5.744562646 approximately