Suppose that you wish to by a car in the sport/compact car
class and because you can not make up your mind you are going to just select a
model at random and hope for the best. You look up in consumer reports magazine
all the cars in these two classes and discover that there are 93 cars in these
classes. You are wondering what the probability is that you select one of these
cars at random and get one with an average city mpg greater than or equal to 32
mpg
You decide to construct a histogram (using STATDISK with 10
classes) and get the following result
The histogram is approximately bell-shaped. So we might
consider using a normal distribution. What is the probability that if we select
a car it's average mpg exceeds 32 mpg?
In the last section we learned how to answer questions like
this when we were dealing with a normal distribution with mean 0 and standard
deviation 1. But this distribution clearly does NOT have a mean of 0 and a
standard deviation of 1. (it's mean is 22.4 and its standard deviation is 5.62).
How can we do a problem like this?
In the above example we want to find the probability that if
we select one car at random we get one with an average city mpg that is greater
than 32 mpg. If we assume that the average city mpg of cars is normally
distributed with a mean of 22.4 and a standard deviation of 5.62, to answer the
question concerning probability we need to answer the question what is the area
under the normal curve with mean 22.4 and standard deviation 5.62 that is above
32? To do this we will utilize the z-score. Recall a z-score is
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So for our problem to find 
, we need to calculate the z-score corresponding to 32 in our
normal distribution with mean 22.4 and standard deviation 5.62
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What the z-score tells us is that the area above the point
32 on our normal curve (with mean 22.4 and standard deviation 5.62) is
equivalent to the area above the Standard normal distribution (with mean 0 and
standard deviation 1) above the point 1.708. This means if we find the area
above 1.708 using the techniques of the last section , we will have found the
area above 32 and answered our probability question
So what is the area above 1.708 on the standard normal
curve?
Using Table A2, we find the area between 0 and 1.71 is
0.4564, so the area above 1.71 is 0.5 - 0.4564 = .0436
On the calculator you can take a shortcut you do not need to
consider the standard normal distribution at all! Just type in 2nd VARS to get
the distr menu and then select 2 for normalcdf(, type in 32 a comma and
2nd comma 99 and another comma and then
type in 22.4 another comma and 5.62 and a final parenthesis. Your screen should
look like this
normalcdf(32,E99,22.4,5.62)
When you hit enter you will get the probability you are
above 32 in the normal distribution with mean 22.4 and standard deviation 5.62.
So with the TI-83 you do not need the z-score at all!
Another example:
The garment industry is interested in the average waist size
of adult men. They do some research and determine that waist sizes are normally
distributed with a mean of 35 and a standard deviation of 2.3 inches. What is
the probability that if an adult male is selected at random they have a waist
size between 32 inches and 39 inches.
To do this problem via the table we need to find the
z-scores for both 32 and 38 : Here are the calculations:
So to find 
, we use the table to find the area between 0 and 1.31 (this
yields .4049) and the table to find the area between 0 and 1.74 (this yields
.4591). Summing these two values gives .8640. SO the probability of getting a
man with waist size between 32 and 39 inches is 0.8640
It is especially easy with the TI-83 - type in
normalcdf(32,38,35,2.3) and hit enter to get an identical answer!
Finding z-scores
given probabilities with nonstandard normal distributions
In some cases we wish to find the score representing a particular
percentile, decile, or quartile instead of the probability corresponding to a
particular score. For example:
Example 1: Consider the men's waist size
example above. What score represents the 70th percentile?
Solution: The 70th percentile is the point with 70% of the
cores below it or the point with 70% of the area under the curve below it.
Since the area under all density functions is 1, we want the point with area
0.7 below it. Since table A-2 only gives the positive portion of the standard
normal curve, we want the curve with an area between 0 and the point being 0.2.
Using the body of table 2 we get the closest point to be 0.52, so the point on
the standard normal curve that represents the 70th percentile is about 0.52. To
convert to our non standard normal distribution we need a little algebra. Consider
the formula for the z-score
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So for our problem: 
. So 36.242 is about the 70th percentile
If you have a TI-83 it is really easy
2nd VARS for the distr menu then 3 for invNorm( now enter .7
and a comma and 35 and a comma and 2.3 and the final parenthesis. Your screen
should look as follows:
invNorm(0.7,35,2.3)
You get an answer of 36.206
Your answers using the calculator will be more accurate than
those using the table. Be aware of this if you are checking your answers in the
back of the book
Example 2:
According to the Opinion Research Corporation, men spend an
average of 11.4 minutes in the shower with a standard deviation of 1.8 minutes.
If we assume these values are normally distributed, find the first quartile
Solution via table: Let's first find the 25th percentile on
the standard normal distribution. This is the point with 25% of the area below
it. By symmetry we will use the point with 75% of the area below it. Since the
table only gives positive values, we look for the point closest to 0.25 in the
body of the table. You will find this to be 0.67. But this is the 75th
percentile, so the 25th percentile is about -0.67. Now using the algebraic
formula above we can find the score of the 25th percentile in our distribution:
So 10.194 is approximately the first quartile
On the TI-83: use invNorm(0.25,11.4,1.8) and press enter to
get a more accurate answer of 10.185918