Notes for February 11 and 13

Reverse look-up of z scores: Percentiles to z-scores, z-scores to raw scores

So far, we only use the percentile look-up table if we know a data set is normally distributed. In these cases, we are given the average mux and the standard deviation sigmax and find a z-score z(x) for a raw score x using this formula.

z(x) = (x - mux)/sigmax

We can then use this z-score to look up a percentile on the orange look-up table.  

Example: For men, the average height is 70.5 inches (5' 10.5") and the standard deviation is 2.8 inches. What is the z-score for a man 6'0" tall (72 inches)?

z(72) = (72 -70.5)/2.8 = 1.5/2.8 = 0.535714..., which rounded to the nearest hundredth is 0.54. Using our lookup table in row 0.5 and column 0.04, we get .7054, which means a man 6'0" is as tall or taller than 70.54% of the male population. Subtracting from 100%, we can also say that 29.46% of the male population is 6'0" or taller.

What if instead we wanted to find the cut-off height for the 75th percentile of men. What we have to find is when the table has two consecutive numbers that look like this

... .74xx .75xx ...

What that would mean is that between those two z-score will be the cutoff for the 75th percentile. Looking on the positive side of the chart we find

0.6 ... .7486 .7517 ...

In the columns 0.06 and 0.07. This means the 75th percentile lies in between 0.66 and 0.67. Since .7486 is .0014 below .7500 and .7517 is .0017 above, it's fair to say the 75th percentile is about half way in between, and 0.665 is a good approximation.

Now that we have a z-score, we use this formula to find the raw score x.

x = mux + z(x) × sigmax

In this case we get

x = 70.5" + 0.665 × 2.8 = 70.5" + 1.862" = 72.362" = 6" 0.4"

This says the 75th percentile of men's heights is at just below six feet and one half inch tall.

What about the percentage of men between 5'11" and 6'2". Here were find the percentages for both hieghts, then subtract the smaller percentage from the larger.

6'2" = 74", and z(74) = (74-70.5)/2.8 = 3.5/2.8 = 1.25

z = 1.25 corresponds to 0.8944

5'11" = 71" and z(71) = (71-70.5)/2.8 = 0.5/2.8 = 0.17857... = 0.18

z = 0.18 corresponds to 0.5714

0.8944 - .5714 = 0.323, which says about 32.3% of men fall between exactly 5'11" and exactly 6'2" tall.

The Central Limit Theorem 

 If we have a normally distributed population and take a random sample, z(x-bar) is different from z(x) by changing the standard deviation. The simplest way to compute z(x-bar) is to take the z-score the normal way then multiply by the square root of n, the size of the sample. For example, if 10 men averaged to a height of 6'0", they would have a different z-score and correspond to a different proportion of the population.

 z(72) = (72 -70.5)/2.8 × sqrt(10) = 1.694..., which rounds to 1.69. Looking up in row 1.6 and column 0.09, we get .9545.  This says a sample of 10 men averaging 6'0" tall is taller than 95.45% of all samples of ten men, and only 4.55% of all samples of ten men are shorter than that.

This rule is called the Central Limit Theorem. We will be using it to try to get estimates of the average of a population using the average of a sample that is assumed to be representative.

Practice problems for homework 3

For pregnancies, assume mux = 280.6 days and sigmax = 9.7 days.

a. What is the z-score for 273 days?
b. What is the proportion that corresponds to 273 days?
c. What is the z-score for 288 days?
d. What is the proportion that corresponds to 288 days?
e. What proportion of pregnancies last between 273 to 288 days?





With the Central Limit Theorem, we need to know the average and standard deviation of a population and the average and size of a sample, x-bar and n, respectively. This gives us a z-score which corresponds to a proportion.

f. What is the z-score for 273 days for a sample where n = 8?
g. What is the proportion that corresponds to 273 days for a sample where n = 8?
h. What is the z-score for 288 days for a sample where n = 8?
i. What is the proportion that corresponds to 288 days for a sample where n = 8?
j. What proportion of pregnancies last between 273 to 288 days for a sample where n = 8?

Consider the following data set.

17, 19, 19, 18, 19, 19, 15, 20, 20, 20, 20, 14, 18, 20, 19, 20, 18, 19, 20, 16

Find the frequencies and relative frequencies for each value.

Answers in the comments.

Class notes for 3/9

Not all data sets are normally distributed, but it has been proven that the set of averages of subsets of a fixed size of a data set are normally distributed around the average of the whole data set. In the language we have used in class, this means that if we take a sample and get an average x-bar, it should be relatively close to the average of the population mux. This is called the Central Limit Theorem.


It's a little hard to see in the picture here, but the equation reads sigmax-bar = sigmax/sqrt(n). As n gets larger, sigmax-bar gets smaller.


To find z(x-bar), we subtract mux from x-bar and divide by sigmax-bar. On a calculator it's easiest to type in (x-bar-mux)/sigmax*sqrt(n).

Here is an example of the difference between z(x) and z(x-bar). We know that for IQ scores, the data set is normally distributed, mux = 100 and sigmax = 15. This means an IQ of 115, has a z-score of (115-100)/15 = 1. The z-score of 1 corresponds to the proportion .8413, which says that the percentage of people with IQs below 115 is 84.13% and the percentage with IQs above 115 is (100-84.13)% = 15.87%.

The Central Limit Theorem z-score answers a different question. What if we have a group of 8 people whose average IQ is 115. How often does that happen? Now, the formula changes to (115-100)/15*sqrt(8) = 2.828..., which rounds to 2.83. The proportion that corresponds to a z-score of 2.83 is .9977, which means that about 99.77% of all groups of eight people have average IQs under 115, while only (100-99.77)% = 0.23% of groups of eight have average IQs at 115 or over.

A standard usage of the Central Limit Theorem is to take a data set and see if the result is unusual or not. This is done by the following procedure.

1. Choose an outlying value, either a z-score or the percentage that corresponds to it.
2. Take a data sample from a population where you know the average and standard deviation already. If the Central Limit Theorem z-score gives us a value beyond the outlying value, we flag the sample we took as an outlying sample.