```Question 486870
PROBLEM

Assume that the population of heights of male college students is approximately normally distributed with mean m of 69 inches and standard deviation s of 4.35 inches. Show all work.

SOLUTION

we'll do this the easy way and then we'll do it the hard way.

the easy way uses the following graphics calculator.

<a href = "http://davidmlane.com/hyperstat/z_table.html" target = "_blank">http://davidmlane.com/hyperstat/z_table.html</a>

use the top graph.
enter 69 for the mean.
enter 4.35 for the standard deviation (Sd)

select "Above".
enter 72.
It will tell you that shaded area is equal to .245206
This means that .245206 proportion of the population is more than 72 inches in height.
It will show you where the shaded area is under the normal distribution curve.

select "Below"
enter 72
it will tell you that shaded area is equal to .754793
This means that .754795 proportion of the population is less than 72 inches in height.
It will show you where the shaded area is under the normal distribution curve.

This graph provides a visual representation of the area under the normal distribution curve that is being calculated.

If you wanted to know the proportion of the population that was between 69 and 72 inches in height, you would select Between and enter 68 in the first box and 72 in the second box.  It would show you that the shaded area is equal to .345704 proportion of the population and would show you where that area was on the graph of the normal distribution curve.

That's the easy way.

The hard way is to determine what the z-score is and then look up the value in the z-table.

For this exercise we will be using the following table.

<a href = "http://www.regentsprep.org/Regents/math/algtrig/ATS7/ZChart.htm" target = "_blank">http://www.regentsprep.org/Regents/math/algtrig/ATS7/ZChart.htm</a>

First you have to find out what the z-score is.

You do that with the following formula:

z-score = (data value minus mean) / standard deviation.

for your problem, since your mean is 69 and your data value is 72 and your standard deviation is 4.35, this translates to:

z-score = (72 - 69) / (4.35)

this comes out to be equal to be:

z-score = .689655172

this means that a score of 72 on the normal distribution curve is .689655172 standard deviations above the mean.

you want to know the proportion of the population that is taller than 72 inches in height.

you need to read the instructions on the z-score table you are using.

this particular table didn't come with instructions, but it's clear from looking at it that the values in the table tell you the proportion of the population that is to the left of the value indicated.

we will look up a z-score of .689655172.

the table only goes to 2 decimal places, so we round our number to .69.

we could try to get more accurate through interpolation, but it's not going to buy us much and, at this point in time, I just want to show you how to use the table.

the z-score we are looking for is .69.

the number is actually 0.69

we look for 0.6 in the left column of each row until we get to the row that contains .6 in the first column.

then we look for .09 in the top row of each column until we get to the column that contains .09 in the first row.

the value we are looking for is in the intersection of the specified row and column.

that value is equal to .7549.

it is found in the intersection of the 8th row and the 11th column in the section of the table that contains positive z-scores.

the 8th row contains a 0.6 in the first column.
the 11th column contains a .09 in the first row.
add 0.6 and .09 and you get 0.69 which is the z-score we want.
the value in the intersection is equal to .7549 which means that .7549 proportion of the population is to the left of a z-score of 0.69.

this also means that 1 - .7549 = .2451 proportion of the population is to the right of a z-score of 0.69

since a z-score of .69 was derived from a raw score of 72, this means that:

.7549 of the population is less than 72 inches in height.
.2451` of the population is greater than 72 inches in height.

we could have gotten more accurate through interpolation.

if you need to be more accurate and have to use interpolation and need to know how to do it, let me know and i'll work you through it.

otherwise, just round your z-score to the accuracy of the table you are using unless you are to be more accurate than that.

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