The Z distribution is a normal distribution that looks
much like the graph below. The figures in the table show the
probability that a random variable x will fall within the
range to the left of the Z score (from negative infinity up
to the Z score.) Consider the diagram below for a more
intuitive understanding of this concept. If you looked up a
Z score of two in a Z distribution table, you would
find that the probability listed in the table represents the pink
shaded area under the curve in the illustration. If you are
interested in the probability of the clear area under the curve—P(Z >
2)—you would have to subtract the probability
found in the table from one to determine the probability of Z
greater than two.
As you might expect, a Z score of zero, when looked up
in the table, yields a probability of .5. This makes sense if you
remember the definition of a normal distribution having half of its
values distributed on either side of the mean.
Returning to the example of apparel shipments, you can look up
the Z score of 2.00 in the table to find the probability that the
shipment will arrive within 30 days. The probability can be found by
looking first down the lefthand column of the table for the number
2.0. Since the number after the decimal point is zero, use the first
column titled .00 for the probability. The probability that
Z will be less than or equal to 2.0 is .9772. (This number
is highlighted in this Z table.) This
means that the probability that the shipment will arrive within 30
days is .9772.
With any normally distributed random variable, you can find the
probability of the variable x falling between two values: calculate the Z score of each value, look up their
probabilities in a Z distribution table, and then
subtract the smaller probability from the larger one. The difference
is the probability of the random variable x falling between
the two values.
For a demonstration of finding the probability of an interval of
values, consider again the apparel manufacturer example from above.
Calculate the probability of an apparel shipment arriving within the
range of 22 to 26 days, using above data for
m_{x} and
s_{x} of 24 days and 3 days,
respectively.
To solve this problem, first standardize the
variables at each end of the interval in question. Calculate the Z value using the following formula for
Z.
First, for the lower end of the interval of 22 days, calculate the Z
value.
Then, for the upper end of the interval of 26 days, calculate the Z value.
Now look up the probability of each standardized random variable
Z in a Z distribution table. The probability
associated with the standardized random variable –.67 is .2514 (a
value which is determined by calculating 1 – .7486). The probability
associated with the standardized random variable .67 is .7486.
To find the probability of an interval, subtract the lower end of
the interval from the upper end of the interval.
The probability of an apparel shipment arriving in 22 to 26 days
is .4972.
1. What is the assumed distribution behind a standardized random
variable, Z?
Solution 1
2. Why is the concept of a standardized variable important?
Solution 2
3. What is the probability that a standardized random variable
Z is less than or equal to 1.96—P(Z <
1.96)?
Solution 3
4. What is the probability that a Z score is greater
than or equal to 1.5 and less than or equal to 2.0— P(1.5 < Z
< 2.0)?
Solution 4
5. If the population of monthly cell phone bills of a group of students
has a mean of $43.00 (m = $43.00) and a
standard deviation of $5.70 (s = $5.70),
what is the probability of selecting a student at random whose
monthly bill is $36.50 or less?
Solution 5
6. The manufacturer of a box of cookies states on the outside of
each box that the product contents weigh 16 ounces. The population of
boxes of cookies has a mean of 16.25 ounces
(m = 16.25 ounces) and a standard deviation
of .30 ounces (s = .30 ounces).
The manufacturer would like to ensure that the contents of each
box weigh between 15.90 and 16.90 ounces. Less than 15.90 ounces may
lead to customer satisfaction issues, while more than 16.90 ounces
results in too much product being given away in each box. What is the
probability of selecting a box of cookies with product content
weighing between 15.90 and 16.90 ounces?
Solution 6
