Sampling Distributions and Central Limit

Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 14 days.

In what range would you expect to find the middle 50% of most pregnancies?

If you were to draw samples of size 34 from this population, in what range would you expect to find the middle 50% of most averages for the lengths of pregnancies in the sample?

Answer 1

In any distribution, the "middle 50%" refers to the range of values between the first and third quartiles.

For a Normal distribution, the First Quartile has a z-score of approximately -0.67. If you aren't sure where this comes from, it's the cutoff score for the bottom 25%; you can find that using a table of values or your technology's Inverse Normal function. Similarly, the Third Quartile has a z-score of approximately +0.67.

To convert a z-score back to an actual value, multiply by the Standard Deviation and then add the Mean. For the distribution of individual pregnancies:

* First Quartile = (-0.67)*(14) + 267 = 258 days

* Third Quartile = (0.67)*(14) + 267 = 276 days

For samples of size 34, the averages have the same Mean as individuals (267 days). The Standard Deviation for the averages is the individual Standard Deviation (14) divided by the square root of the sample size (34); here this is 14/sqrt(34) = 2.401.

Thus for the distribution of averages (for sample size 34):

* First Quartile = (-0.67)*(2.401) + 267 = 233 days

* Third Quartile = (0.67)*(2.401) + 267 = 301 days

Note that it's probably easier to use the Inverse Normal function provided by your technology, if that's an option.