You manufacture thermometers and these thermometers should measure at the freezing point of water. You test a large sample of these thermometers and discover that the distribution of actual temperatures measured at the freezing point of water follows a normal distribution with a mean of 0 degrees and a standard deviation of 1 degree.

What is the probability that if you select one thermometer at random, its measurement at the freezing point of water will be between 0 and 1.32 degrees?

Hit 2^nd then the VARS key to get the distribution menu - your screen should look like this:

Now hit 2 for normalcdf (NOTE THIS MEANS NORMAL CUMULATIVE PROBABILITY DENSITY FUNCTION) - You should see this on your screen

Now type in the lowerbound (In this case 0) followed by a comma, followed by the upperbound (in this case 1.32) - you should see this on the screen

Hit enter for your answer - compare this answer to the one received from the table

 How about another problem - What is the probability if a thermometer is randomly selected it registers -2.57 to 0 at the freezing point of water?

Remember when you used the table you had to use the fact of symmetry for the normal curve to get the answer - NO NEED FOR THAT HERE - since -2.57 is the lower bound and 0 is the upper bound - we can just type it in directly!

Again Hit 2^nd, VARS, 2 for normalcdf

Now enter -2.57, a comma and 0, your screen should look as follows:

Hit enter for your answer - compare with the problem you did using the table!

 Another problem - What is the probability a thermometer selected registers greater than 1.53 degrees at the freezing point of water?

When you used the table you had to find the probability and subtract from .5 on this one, NO NEED FOR THIS HERE - try this

Hit 2^nd, VARS, 2 for normalcdf

Now enter 1.53, a comma and hit 2^nd comma for the E key and type in 99 - your screen should look as follows

The E99 stands for infinity here (actually just the largest value the calculator can handle), hit enter to get the answer - you should have the following answer - how does this compare to your table answer?

Can you guess what to do? 2^nd, VARS, 2 for normalcdf, now hit the (-) key and 2^nd comma 99 to get this on your screen

Enter a comma and type in -2.01 and hit enter to get this answer

-E99 stands for negative infinity

Recall to do these problems with the Table you needed to find a z-score and then use the standard normal table. The TI-83 can do these problems directly without calculating the z-score and it gives you a more accurate answer than the table gives you - here is how

Remember this problem

The highway miles per gallon of sporty and compact cars were determined to be normally distributed with a mean of 33.78 and a standard deviation of 9.94. If a car is selected at random what is the probability that it will have highway miles per gallon greater than 42 miles per gallon?

You could still calculate the z-score and use the standard normal distribution - but you don't need to do so!

Hit 2^nd, VARS, 2 for normalcdf, enter 42, a comma and E99, now enter a comma and enter the mean (33.78) and the standard deviation (9.94) Your screen should look as follows:

Hit enter to get your answer - how does the screen below compare to the table method?

 SUMMARY - TO GET A PROBABILITY CORRESPONDING TO A Z SCORE

1. Hit 2^nd enter to get the Distribution menu, hit 2 for normalcdf
2. Enter the lowerbound followed by a comma and the upperbound use -E99 for negative infinity and E99 for positive infinity if needed
3. If the distribution you are interested in is NOT standard normal, enter the mean, a comma and the standard deviation
4. Hit enter to get your answer