Transcript 5.3 Applications of Normal Distributions

5.3 Applications of Normal
Distributions
Statistics
Mrs. Spitz
Fall 2008
Objectives/Assignment
• How to compare data from two normal
distributions
• How to find probabilities for normally
distributed variables using a table and
using technology
• How to find a specific data entry of a
normal distribution given the probability.
Assignment: pp. 217-219 #1-16
Comparing Normal Distributions
• In 5.2, you learned how to transform a
normally distributed variable, x to a zscore using the equation
value  mean x  
z

St.Dev

continued
• You also learned to find areas under the
standard normal curve. In this section,
you will learn how to apply those skills.
For instance, transforming x-values to zscores allows you to compare variables in
two normal distributions as shown in the
following example.
Ex. 1: Comparing scores from two
distributions
• The Graduate Record Exam (GRE) and
the Miller Analogy Test (MAT) are tests
that graduate schools use to evaluate
applicants. GRE scores are normally
distributed with = 1500 and =300, while
MAT scores are normally distributed with
=50 and =5. You decide to take both
tests. You score 1875 on the GRE and a
57 on the MAT. On which test did you
score better? Explain
Solution: You can transform each score to a standard z-score to
determine which is better. The area to the left of the z-score is the
percentile.
x
1875  1500
z

 1.25

300
x   57  50
z

 1.4

5
The area to the left of z=1.25 is .8944 and the area to the left of z=1.4 is
.9222. The MAT z-score is in the 92nd percentile which is greater than the
GRE z-score (89th percentile). So you scored better on the MAT.
Probability and Normal Distributions
•
If a random variable x is normally
distributed, you can find the
probability that x will fall in a given
interval by calculating the ara
under the normal curve for the
given interval.
• To find the area under
any normal curve, first
convert each data value
to a z-score. Then use
the standard normal
distribution. For instance,
consider a normal curve
with = 500 and = 100,
as shown at the upper
left. The value of x one
standard deviation above
the mean is  +  = 500 +
100 = 600.
Probability and Normal Distributions
• Now consider the
standard normal curve
shown at the lower left.
The v alue of z one
standard deviation above
the mean is  + =0 + 1=
1. Because a z-score of
1 corresponds to an xvalue of 600 and areas
are not changed with a
transformation to a
standard normal curve,
the shaded areas in the
graph are equal
Ex. 2: Finding probabilities for Normal
Distributions
•
A survey indicates that people use
their computers an average of 2.4
years before upgrading to a new
machine. The standard deviation
is .5 year. If a computer owner is
selected at random, find the
probability that he or she will use it
for less than two years before
upgrading. Assume that the
variable x is normally distributed
SOLUTION: The graph shows a normal curve with = 2.5 and  = .5 and a
shaded area for x less than 2. The z-score that corresponds to two years
is:
z
x

2  2.4

 0.8
0.5
Ex. 2: Finding probabilities for Normal
Distributions
•
Using the Standard Normal Table,
P(z x -0.8) = 0.2119. The
probability that the computer will
be upgraded in less than 2 years
is 0.2119. So, 21.19% of new
owners will upgrade in less than
two years.
SOLUTION: The graph shows a normal curve with = 2.5 and  = .5 and a
shaded area for x less than 2. The z-score that corresponds to two years
is:
z
x

2  2.4

 0.8
0.5
Ex. 3: Finding probabilities for Normal
Distributions
•
A survey indicates that for each trip to
the supermarket, a shopper spends an
average of  = 45 minutes with a
standard deviation of  = 12 minutes.
The length of time spent in the store is
normally distributed and is represented
by the variable, x. A shopper enters
the store. Find the probability that the
shopper will be in the store for the
lengths of time 1) between 24 and 54
minutes and 2) more than 39 minutes.
SOLUTION: The graph shows a normal curve with = 45 minutes and  =
12 minutes. The area for x between 24 and 54 minutes is shaded. The zscore that corresponds to 24 minutes and to 54 minutes are:
z
x

24  45

 1.75
12
z
x

54  45

 0.75
12
Ex. 3: Finding probabilities for Normal
Distributions (continued)
•
So the probability that a shopper will be in the store between 24 and 54
minutes is:
P(24  x  54)  P(1.75  z  0.75)
 P( z  0.75)  P( z  1.75)
 0.7734  0.0401
 0.7333
Another way of interpreting this probability is to say that 77.33% of the
shoppers will be in the store between 24 and 54 minutes
Ex. 3: Finding probabilities for Normal
Distributions
•
The graph at the right shows a
normal curve with  = 45 minutes
and  = 12 minutes. The area for
x greater than 39 minutes is
shaded. The z-score that
corresponds to 39 minutes is:
z
x

39  45

 0.5
12
So, the probability that a shopper will be in the store more than 39 minutes
is:
P( x  39)  1  P( z  0.5)
 1  0.3085
 0.6915
Ex. 4: Using technology to find Normal Distributions
Another way to find normal distributions is using a calculator
or computer. You can find normal probabilities with a TI-83 or
TI-84 or using Excel.
• Cholesterol levels of American men are normal
distributed with a mean of 215 and a standard deviation
of 25. If you randomly select an American man, what is
the probability that his cholesterol level is less than 175?
Use a calculator or computer to find the probability.
SOLUTION: The TI-83 has a feature that allows you to find
normal probabilities without first converting to a
standard z-score. You must specify the mean and
standard deviation of the population as well as the xvalue(s) that determine the interval.
Ex. 4: Using technology to find Normal Distributions
2nd VARS - #2
Type in 0 followed by
175 (what you are
looking for) 215 (the
mean) and 25 (Standard
Deviation.
SOLUTION: From the display, you can see that the
probability that his cholesterol is less than 175 is about
0.055, or 5.5%.
Ex. 5: Finding a Specific Data Value
You can also use the normal distribution to find a specific
data value (x-value) for a given probability shown in ex. 5.
Scores for a civil service
exam are normally
distributed with a mean
of 75 and a standard
deviation of 6.5. To be
eligible for civil service
employment, you must
score in the top 5%.
What is the lowest score
you can earn and still be
eligible for employment?
Ex. 5: Finding a Specific Data Value - continued
Exam scores in the top
5% correspond to the
shaded region shown.
An exam score in the top
5% is any score above
the 95th percentile. To
find the score that
represents the 95th
percentile, you must first
find the z-scores that
correspond to a
cumulative area of 0.95.
From the Standard
Normal Table, you can
find that the areas closest to
0.95 are 0.9495 (z=1.64) and
.9508 (z=1.65). Because
0.95 is halway between the
two areas in the table, use
the z-score that is halfway
between 1.64 and 1.65. That
is z=1.645.
Ex. 5: Finding a Specific Data Value - continued
Using the equation,
x =  + z
x = 75 + 1.645(6.5)
x ≈85.69
So, the lowest score you can earn and still be eligible for
employment is 86.