Estimate the population standard deviation of the grouped data.?

Question

Score Frequency
1-3 /1
4-6 /6
7-9 /3

Answer 1

Find the midpoint of each interval.
For 1-3, the midpoint is (3+1)/2 = 4/2 = 2
For 4-6, the midpoint is (6+4)/2 = 10/2 = 5
For 7-9, the midpoint is (9+7)/2 = 16/2 = 8
2(1) + 5(6) + 8(3)
= 2+30+24
= 56
n = number of data points = 1+6+3 = 10
variance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]
= (1/9)((4+150+192) - 3136/10)
= (1/9)(346 - 313.6)
= (1/9)(32.4)
= 3.6
standard deviation = sqrt(variance)
= sqrt(3.6)
= 1.9

Answer 2

There are two ways you can justify this. The first (and possibly easiest) is that you only have (n-1) degrees of freedom (where the degrees are based upon the number of independent data points collected and the amount of information used). So, in this case, you have n independent data points and have used up one degree in calculating your mean. The other (and more correct) way to think about it is that the standard deviation (and more appropriately the variance) as presented in the equation above is an unbiased estimate of the true value. You are using an estimate of the second central moment, which is the expected value of the square of the difference between the independent data points and the true mean (or, for the estimated value, you use the average or some other estimator). If you expand the equation, you find that the expected value of the second central moment (in this case) equals (n-1) *sigma^2, so your estimate needs to be divided by n-1. If you want to see more about it you can check most standard intermediate stats text books (e.g., Mathematical Statistics w/ Applications by Wackerly, Mendenhall, and Scheaffer). Also, in regards to some of the others comments. You NEVER adjust your statistical estimates to make them larger or smaller because you think they should be. That essentially voids all the methodology behind what is being done. The closest you ever come to that is the finite population correction which is used in survey methodology when the sample fraction becomes appreciably large and thus your variance is too high (since it assumes an infinite sample). Most methods of variance estimation could be thought to have an implicit finite population correction but the fraction is so small as to not impact the estimate.

Answer 3

Around 1.9 to 2.0, approximately