# Calculating expected value and standard deviation when I have frequencies and intervals in percentage

A company made some tests for a material humidity with these results:

HUMIDITY ............... FREQUENCY

• 18.75 - 19.75 % ... 3
• 19.75 - 20.75 % ... 22
• 20.75 - 21.75 % ... 123
• 21.75 - 22.75 % ... 300
• 22.75 - 23.75 % ... 201
• 23.75 - 24.75 % ... 61
• 24.75 - 25.75 % ... 6
• 25.75 - 26.75 % ... 1

Using point estimation calculate the expected value and the standard deviation... I have no idea how to calculate it since I don't have values but intervals in percentage...

[RESULT: expect.value: 22.486, stand.dev. : 0.998%]

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You have to make some assumptions about how the values are distributed inside the intervals. One such assumption is:

• The values all appear at the midpoint of the interval.

and another one is:

• The values are equi-spaced inside the intervals.

Fortunately, you have enough observations and they're regular enough that the result won't materially depend on the assumption that you make. I'd go with the simple option of assuming that the values are all at the midpoint of the intervals if I were you.

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This is commonly called group averages. Each interval represents a group. You take the average of the end points (same as what Chris called the mid point) multiply by the number of points in the interval sum up all the intervals and divide by the total sample size. This is a weighted average of midpoints weighted by the number of points in the group. It is equivalent to assuming either that all the points were at the midpoint or that they are spread uniformly through the interval. I think the second assumption sounds a little more reasonable though it is mathematically equivalent to the first assumption. This is the same result as Chris is suggesting using a slightly different description. For the variance take the computed weighted mean subtract it from each of the midpoints, square the difference weigh it by the number of points in the interval and sum the squares up and divide by n. This is a weighted average of the squared deviations of the midpoints from the sample's weighted average. The weight are the same as what was used to get the mean.

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When I'm calculating the standard deviation, is the sum of the weighted squares of the differences not supposed to be divided by (n-1) instead of n? ...or when should I divide by n and when by (n-1) ? – user1111261 May 25 '12 at 17:03
@user1111261 It is the variance that you divide by n or n-1 not the standard deviation for the standard deviation you take the square root of n or n-1. In elementary statistics courses students are taught to divide by n-1 when computing a variance. i wish it was not taught that way. For the maximum likelihood estimate you divide by n. For an unbiased estimate you divide by n-1. For large n there is not much difference between these estimators and it is not always clear whether or not an unbiased estimator is better. – Michael Chernick May 25 '12 at 17:46
Accuracy is mean square error which is the square of the bias plus the variance, An unbiased estimate with a large variance can be less accurate than a biased etimator with a small variance. – Michael Chernick May 25 '12 at 17:47
Hey Michael, I posted one more question here: stats.stackexchange.com/questions/29168/… Thank you so much!!! :o) – user1111261 May 25 '12 at 19:50