To find the area under the standard normal curve to the left of a given value of z, we look up the ones and tenths values of z in the column at the right and the hundredths value in the row
at the top of the table. The area probability is then the value in the table at that column and row.
Example:
If z is a continuous random variable with the standard normal distribution, then by using the Standard Normal Table we can see that:
.47 .3192 P z
and
1.28 .8997 P z
.47 1 .3192 .6808
P z
and
1.28 1 .8997 .1003 P z
.47 1.28
.8997 .3192 .5805 P
z
Exercises:
Compute the following probabilities for a random variable z with the standard normal distribution.
0.88 P z
1.96 P z
.32 1.10
P z
1.22 or 1.22
P z z
E. Calculating Probabilities for a Normal Random Variable
If x is a normally distributed random variable with mean
and standard deviation
, then
the z-scores of the values for the random variable have the standard normal distribution. That is the random variable z defined by:
x z
is normally distributed with
and 1
. Therefore, any interval for the variable x can be written as an interval for the z-score z and
then the probability found by using the Standard Normal Table. Example:
The height h in inches of a randomly selected woman is approximately normally distributed with a mean of
63.5
and a standard deviation of
2.75
inches. To
calculate the probability that a woman is less than 63 inches tall, we first find the z-score for 63 inches:
63 63.5 0.5
0.18 2.75
2.75 z
Thus
63 0.18
P h P z
. Using the Standard Normal Table, we see that
0.18 .4286 P z
. So the probability that a randomly chosen woman’s height is less than 63 inches is also .4286 or equivalently,
42.86 of women are less than 63 inches tall. Exercises:
Using the information from the women’s height example above, Calculate:
65 P h
60 70
P h
63.5 72
P h
F. Calculating Values Using the Standard Normal Table
The Standard Normal Table can be used to find percentiles for variables which are normally distributed.
Example:
To find the score which marks the 80
th
percentile for SAT Math Scores, we use the fact that SAT Math scores s are approximately normally distributed with
514
and
113
. From the Standard Normal Table, the z-score for which closest to 80 percent of
values lie to the left is 0.84 which corresponds to a probability of .7995. The SAT score which corresponds to a z-score of 0.84 can be found by solving
514 0.84
113 s
for s. This yields 608.92
s
. So a score of 609 is better than 80 of all other test scores.
Exercises:
For the normal distribution above: Find
35
P . If a person scores in the top 5 of test scores, what is the minimum score they could have
received?
If a person scores in the bottom 10 of test scores, what is the maximum score they could have received?
G. The Central Limit Theorem
