Standard deviation measures how spread out values are from the mean. The values 20 and 80 are the minimum and maximum—the farthest from the mean. What happens to the spread when you remove the most extreme values?

Removing values that are symmetric around the mean keeps the mean the same. Fr‌оm аni‍ko.ai Removing the values farthest from the mean reduces the spread.

Enter a list of 25 values with min 20, max 80, and mean 50:

$A = [20, 30, 35, 40, 45, 48, 49, 50, 50, 50, 50, 50, 50, 50, 50, 50, 50, 50, 51, 52, 55, 60, 65, 70, 80]$

Verify: $\operatorname{mean}(A) = 50$

Remove the first and last values:

$B = [30, 35, 40, 45, 48, 49, 50, 50, 50, 50, 50, 50, 50, 50, 50, 50, 50, 51, 52, 55, 60, 65, 70]$

Calculate:

$\operatorname{mean}(B)$

$\operatorname{stdev}(A)$

$\operatorname{stdev}(B)$

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Confirm that the means are equal and that $\operatorname{stdev}(B) < \operatorname{stdev}(A)$.

The mean of data set $A$ is 50.

We're removing 20 and 80. Notice that:

• 20 is 30 below the mean
• 80 is 30 above the mean

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These two values are symmetric around the mean, so their average is exactly 50.

When you remove values whose average equals the mean, the mean stays the same.

Therefore, the mean of $B$ is also 50.

Standard deviation measures how spread out the data is from the mean.

• Values close to the mean contribute little to the spread
• Values far from the mean contribute a lot to the spread

The values being removed (20 and 80) are the minimum and maximum—they're the farthest possible values from the mean of 50.

Removing the most extreme values always reduces the spread of the data.

• Mean: Stays the same (removed values average to 50)
• Standard deviation: Decreases (removed the most extreme values)

The answer is: The mean of data set $B$ is the same as the mean of data set $A$, and the standard deviation of data set $B$ is less than the standard deviation of data set $A$.