15.4 The Normal Distribution
Objectives:
• Understand the basic properties of the normal
curve.
• Relate the area under a normal curve to
z-scores.
• Make conversions between raw scores and
z-scores.
• Use the normal distribution to solve applied
problems.
The Normal Distribution
The normal distribution describes many real-life
data sets. The histogram shown gives an idea
of the shape of a normal distribution.
Normal Curve properties
We represent the mean by μ and the
standard deviation by σ.
Normal Curve properties cont. (2)
The “68 – 95 – 99.7 rule” for a normal distribution.
The Normal Distribution
Applications
Example: Suppose that the distribution of scores
for 1,000 students who take a standardized
intelligence test is a normal distribution. If the
distribution’s mean is 450 and its standard
deviation is 25,
a) how many scores do we expect to fall
between 425 and 475?
b) how many scores do we expect to fall above
500?
The Normal Distribution Applications cont. (2)
a) Solution: 425 and 475 are each 1 standard deviation from the
mean. Approximately 68% of the scores lie within 1 standard
deviation of the mean.
So we expect about 0.68 × 1,000 = 680 scores are in the range
from 425 to 475.
The Normal Distribution Applications cont. (3)
b)
Solution:
We know 5% of the scores lie more than 2 standard
deviations above or below the mean, so (by symmetry) we expect to
have 0.05 ÷ 2 = 0.025 of the scores to be more than 2 standard
deviations above the mean. For this distribution,
μ + 2σ = 450 + 2(25) = 500.
2.5% of 1000 is 25 so we can expect 0.025 * 1,000 = 25 of the scores to be
above 500.
z-Scores
• Every raw score in a normal distribution has its own
Z-score.
• The Z-score for a particular raw score is the number
of standard deviations that the raw score is from
the mean.
• For example, for a normal distribution with mean
450 and standard deviation 25, the value 500 is 2
standard deviations above the mean; that is, the
value 500 corresponds to a z-score of 2.
Standard Normal Distribution
The standard normal distribution has a mean of
μ = 0 and a standard deviation of σ = 1.
There are tables that give the area (which represents
the percentage of data) under this curve between
the mean and a particular z-score.
Using z-Score tables
Example:
Use a table to find the percentage of the data
(area under the curve) that lie in the following regions for a
standard normal distribution:
a) between z = 0 and z = 1.3
b) between z = 1.5 and z = 2.1
c) between z = 0 and z = –1.83
Solution (a): The area under the curve between z = 0 and z = 1.3
is shown. Using a table we find this area for the z-score 1.30.
We find that A is 0.403. So we expect 40.3%, of the data to
fall between 0 and 1.3 standard deviations above the mean.
Using z-Score tables cont. (2)
Solution (b): The area under the curve between z = 1.5 and z =
2.1 is shown. We first find the area from z = 0 to z = 2.1 and
then subtract the area from z = 0 to z = 1.5. Using a table we
get A = 0.482 when z = 2.1, and A = 0.433 when z = 1.5. The
area is 0.482 – 0.433 = 0.049 or 4.9%
Using z-Score tables cont. (3)
Solution (c): Due to the symmetry of the normal distribution,
the area between z = 0 and z = –1.83 is the same as the area
between z = 0 and z = 1.83.
Using a table, we see that A = 0.466 when z = 1.83.
Therefore, 46.6% of the data values lie between 0 and –1.83.
Using z-Scores on a calculator
The equation for the standard normal distribution is
y
e
 x2
2
2
1) Enter the function in
Y=
2) Set the WINDOW as
Xmin = -5
Xmax = 5
Ymin = -0.2
Ymax = 0.5
3) To calculate the area under the normal distribution curve find the associated
percentages, go to
2nd TRACE
choose 7: f(x)dx
and then enter the z-values.
Using z-Scores on a calculator cont. (2)
Example: Use a calculator to verify the percentage of
data (area under the curve) that lie in the following
regions for a standard normal distribution:
a)
between z = 0 and z = 1.3
b)
between z = 1.5 and z = 2.1
c)
between z = 0 and z = –1.83
a) 40.3%
b) 4.9%
c) 46.6%
Converting Raw Scores to z-Scores
1) See how far the raw score is from the mean.
i.e. Take the difference
x-μ
2) See how many standard deviations are in that
distance.
i.e. Divide that difference by σ
Converting Raw Scores to z-Scores cont. (2)
Example: Suppose the mean of a normal distribution
is 20 and its standard deviation is 3.
a) Find the z-score corresponding to the raw score 25.
b) Find the z-score corresponding to the raw score 16.
Applications
• Example: Suppose you take a standardized test. Assume that
the distribution of scores is normal and you received a score
of 72 on the test, which had a mean of μ = 65 and a standard
deviation of σ = 4. What percentage of those who took this
test had a score below yours?
• Example: Consider the following information:
1911:
Ty Cobb hit .420.
Mean average was .266 with standard deviation .0371.
1941:
Ted Williams hit .406.
Mean average was .267 with standard deviation .0326.
1980:
George Brett hit .390.
Mean average was .261 with standard deviation .0317.
Assuming normal distributions, use z-scores to determine which of
the three batters was ranked the highest in relationship to his
contemporaries.
Applications cont. (2)
A manufacturer plans to offer a warranty on an electronic device. Quality control
engineers found that the device has a mean time to failure of 3,000 hours with
a standard deviation of 500 hours. Assume that the typical purchaser will use
the device for 4 hours per day. If the manufacturer does not want more than
5% to be returned as defective within the warranty period, how long should
the warranty period be to guarantee this?
Solution:
We need to find a z-score
such that at least 95% of the area is beyond
this point. This score is to the left of the
mean and is negative.
By symmetry we find the z-score such that 95% of the area is below this score.
Applications cont. (3)
50% of the entire area lies below the mean, so our problem reduces to
finding a z-score greater than 0 such that 45% of the area lies between the
mean and that z-score.
If A = 0.450, the corresponding z-score is 1.64. 95% of the area
underneath the standard normal curve falls below z = 1.64. By symmetry,
95% of the values lie above –1.64.
Since
, we obtain
Then solving for x gives
Owners use the device about 4 hours per day, so we divide 2,180 by 4 to get
545 days. This is approximately 18 months if we use 31 days per month.
The warranty should be for roughly 18 months.