2. The mean high temperature in July for a town in the Central Valley is 106. If the standard deviation is 2 degrees, what is the probability that the temperature will not reach 100 on a given day?

First we change the temperature to a z-score, i.e., the number of standard deviations which this score is from the mean, using the formula

where

x = 100

so

= -6/2 = -3

and we see that a temperature of 100 is 3 standard deviations below the mean. To find the probability that a temperature could be that low, it is important to keep a picture of the bell shaped curve firmly in mind.

The scale on the horizontal axis is z scores. We see that a temperature which is 3 standard deviations below the mean is getting fairly far out into the left tail of the curve. We thus see that a temperature which is 3 standard deviations below the mean is pretty unlikely. In general, t he probability that a z score is in a certain range is given by the area under the curve over that range. Since probabilities are numbers between 0 and 1, the probability is actually the percentage of the total area under the bell shaped curve which is over that inteval.

Finding the area under a curve is a problem from itegral calculus. Getting the area over a certain interval under the bell shaped curve is a diabolically complicated problem. First you need the formula for the curve. That is a tractable calculus problem. For a discussion of the formula, click here. The trouble is that this formula does not submit to the techniques of Calculua I. However, there are methods, involving infinite series, which will enable us to find the area over an interval under the bell shaped curve, but it doesn't get covered until Calculus II.

 

However, there are methods, and in the days before the invention of calculators, they would prepare tables. This semester we are using Sullivan's Finite Math book. To find out what the area is, we go to the table on the inside of the back cover of the text. In that book, the areas which correspond to the z scores are the areas between the mean and the z score.

We look up the area which corresponds to a z-score of -3 in the table, we get

..4987

Since the bell shaped curve is symmetric, the area between the mean and 3 standard deviations below is the same as the area between the mean and 3 standard deviatioins above, which is the .4987. So the picture looks like

 

.4987 is the probability that the rtemperature will be between 100 and 106. We use the fact that the probability that the temperature will be below 106 is 1/2 or .5, and we see that we need to subtract the .4987 fro ,5.


.5000
.4987
.0013

and the probability that the temperature will be less than 100 is .0013.

In general, to find the area you want, keep a picture of the bell shaped curve firmly in mind, and figure out how to use the area you are given to figure out the area you want.

If you have a TI -83 or a TI-84+ gaphing calculator, it is possible to get these areas from your calculator. Over the VARS button it says DISTR. Access that by hitting the 2nd button before hitting the VARS.The top menu will be

DISTR     DRAW

If you haven't done anything, the DISTR will be selected. If it isn't, select it. Going down is another menu. Select

2: normalcdf (

You will then see a

normalcdf (

in your display. Punch in the endpoints of the interval separated by a comma (there is a "," button on your calculator) and close off the interval with a right parentheses. Our interval goes from negative infinity to -3. Actually, there is no infinity button on your calculator, but it will suffice if you approximate negative infinity with -99. 99 standard deviations is so far from the mean that the curve is so close to the x axis that the area in the tail that is farther than 99 standard deviations from the mean, while it is a positive real number, has so many zeroes between the decimal point and the first nonzero digit that it is quite negligible.

normalcdf (-99, -3)

= .001349967 . . .

which actually rounds off to our previous answer of .0013.

You may wonder if 99 standard deviations is far enough away. We could find the area between the mean and 3 standard deviations from the mean

normalcdf (-3, 0)

= .498650033

If we subtrct this from .5

.500000000
.498650033
.001349967

and we get the same thing we got before. You can see that going 99 standard deviations gives us accuracy to more places than we are interested in.