Transcript Chapter 5

Chapter 5
Normal Probability Distributions
Chapter 5
Normal Probability Distributions
Section 5-1 – Introduction to Normal Distributions and the Standard
Normal Distribution
A. The normal distribution is the most important of the continuous
probability distributions.
1. Definition: A normal distribution is a continuous probability
distribution for a random variable x.
a. The graph of a normal distribution is called a normal curve.
2. A normal distribution has the following properties:
a. The mean, median, and mode are equal (or VERY close to
equal).
b. The normal curve is bell-shaped and is symmetric about the
mean.
c. The total area under the normal curve is equal to one.
d. The normal curve approaches, but never touches, the x-axis
as it gets further away from the mean.
Chapter 5
Normal Probability Distributions
Section 5-1 – Introduction to Normal Distributions and the Standard
Normal Distribution
e. The graph curves downward within one standard deviation of
the mean, and it curves upward outside of one standard
deviation from the mean.
1) The points where the curve changes from curving
upward to curving downward are called inflection
points.
B. We know that a discrete probability can be graphed with a
histogram (although we didn’t emphasize this in Chapter 4).
1. For a continuous probability distribution, you can use a
probability density function (pdf).
a. A probability density function has two requirements:
1) The total area under the curve has to equal one.
2) The function can never be negative.
Chapter 5
Normal Probability Distributions
Section 5-1 – Introduction to Normal Distributions and the Standard
Normal Distribution
b. We can graph a normal curve with the mean of μ and
standard deviation of σ by using the normal probability
2
2
1
density function: y=
𝑒 − 𝑥−𝜇 /2𝜎
σ 2π
1) Notice that this entire equation depends completely on
what μ and σ are, since e and π are constants.
2. A normal distribution can have ANY mean and ANY POSITIVE
standard deviation.
a. These two parameters completely determine the shape of the
normal curve.
1) The mean gives the axis of symmetry.
2) The standard deviation describes how spread out (or
bunched up) the data is.
Chapter 5
Normal Probability Distributions
Section 5-1 – Introduction to Normal Distributions and the Standard
Normal Distribution
C. The Standard Normal Curve
1. There are infinitely many normal distributions, because there are
infinitely many possible combinations of means and standard
deviations.
a. The standard normal distribution has a mean of zero and a
standard deviation of 1.
1) The horizontal scale of the graph of the standard normal
distribution corresponds to z-scores.
a) Remember that z-scores are measures of position that
indicate the number of standard deviations values lie
away from the mean.
1. z = x minus the mean over the standard deviation, or
𝑥−𝜇
𝑧=
𝜎
Chapter 5
Normal Probability Distributions
Section 5-1 – Introduction to Normal Distributions and the Standard
Normal Distribution
2. The standard normal distribution has the following properties:
a. The cumulative area under the curve is close to 0 for z-scores
close to -3.49.
b. The cumulative area increases as the z- scores increase.
c. The cumulative area for z = 0 is 0.5000.
d. The cumulative area is close to 1 for z-scores close to 3.49.
3. To find the corresponding area under the curve for any given (or
calculated) z-score, there are two main methods.
a. The easiest, and the one I suggest, is to use the TI-84
calculator.
1) 2nd VARS normalcdf (lower boundary, upper boundary)
a) If you want the area to the left of a z-score, use -10,000
as your lower boundary and the z-score you are
interested in as your upper boundary.
Chapter 5
Normal Probability Distributions
Section 5-1 – Introduction to Normal Distributions and the Standard
Normal Distribution
b) If you want the area to the right of a z-score, use zscore you are interested in as your lower boundary and
use 10,000 as your upper boundary.
c) If you want the area between two z-scores, use them
both (smaller one as lower, larger one as upper).
b. The other way, which will be demonstrated in class, is to use
the z-score chart.
1) You should have done this in Algebra 2, so it should be a
quick review for us.
Chapter 5
Normal Probability Distributions
Section 5-1 – Introduction to Normal Distributions and the Standard
Normal Distribution
4. Remember that in Section 2.4 we learned from the Empirical
Rule that values lying more than 2 standard deviations from the
mean are considered to be unusual.
a. We also learned that values lying more than 3 standard
deviations from mean are very unusual, or outliers.
b. In terms of z-scores, this means that a z-score of less than -2
or greater than 2 means an unusual event.
1) A z-score of less than -3 or greater than 3 means a very
unusual event. (Outlier)
Section 5-1
Normal Probability Distributions
Examples
Page 249 Selected Even Problems, Using the TI-83
#22
Find the area to the left of z = 0.08.
2nd VARS normalcdf(-10000,.08) = .5319
#32
Find the area to the right of z = 2.51
2nd VARS normalcdf(2.51,10000) = .0060
#36
Find the area between z = -0.51 and z = 0
2nd VARS normalcdf(-0.51,0) = .1950
#40
Find the area to the left of z = -1.96 or to the right of z = 1.96
Remember that we ADD probabilities for OR questions.
2nd VARS normalcdf(-10000,-1.96) = .025
2nd VARS normalcdf(1.96,10,000) = .025
.025 + .025 = .0500
Page 250, # 42
You are performing a study about the height of 20-29 year old men. A previous
study found the height to be normally distributed, with a mean of 69.6 inches and
a standard deviation of 3.0 inches. You randomly sample 30 men and find their
heights (in inches) to be as follows:
72.1
71.2
67.9
67.3
69.5
68.6
68.8
69.4
73.5
67.1
69.2
75.7
71.1
69.6
70.7
66.9
71.4
62.9
69.2
64.9
68.2
65.2
69.7
72.2
67.5
66.6
66.5
64.2
65.4
70.0
A) Draw a frequency histogram to display these data points using seven classes.
Is it reasonable to assume that the heights are normally distributed? Why?
B) Find the mean and standard deviation of your sample.
C) Compare the mean and standard deviation of your sample with those in the
previous study. Discuss the differences.
Page 250, # 42
You are performing a study about the height of 20-29 year old men. A previous
study found the height to be normally distributed, with a mean of 69.6 inches and
a standard deviation of 3.0 inches. You randomly sample 30 men and find their
heights (in inches) to be as follows:
72.1
71.2
67.9
67.3
69.5
68.6
68.8
69.4
73.5
67.1
69.2
75.7
71.1
69.6
70.7
66.9
71.4
62.9
69.2
64.9
68.2
65.2
69.7
72.2
67.5
66.6
66.5
64.2
65.4
70.0
Entering the 30 data points into the TI-83, using STAT and Edit, we can
calculate the mean, standard deviation, and median.
STAT, Calc, 1-Var Stats gives us what we need.
The mean is 68.75, the standard deviation is 2.85, and the median is
69.00.
72.1
71.2
67.9
67.3
69.5
68.6
68.8
69.4
73.5
67.1
69.2
75.7
71.1
69.6
70.7
66.9
71.4
62.9
69.2
64.9
68.2
65.2
69.7
72.2
67.5
66.6
66.5
64.2
65.4
70.0
Rel.
Freq.
Cum.
Freq.
Max Value:
75.7
Min Value:
62.9
Range: 75.7 – 62.9 =
First Lower Limit is the
12.8
Minimum Value!!!
Class Width: 12.8/7 =
Add Class Width 2
Down
Remember
to ROUND UP!!
LL
UL
62.9
64.8
64.9
66.8
66.9
68.8
68.9
70.8
70.9
72.8
72.9
74.8
74.9
76.8
LB
UB
MdPt
Freq.
First Upper Limit is one unit less than the 2nd
Lower Limit (Remember, our units are tenths,
not whole numbers).
Add Class Width Down
72.1
71.2
67.9
67.3
69.5
68.6
68.8
69.4
73.5
67.1
69.2
75.7
71.1
69.6
70.7
66.9
71.4
62.9
69.2
64.9
68.2
65.2
69.7
72.2
67.5
66.6
66.5
64.2
65.4
70.0
Subtract one-half unit
from lower limits to
get lower boundaries.
REMEMBER that our
units are tenths!!
One-half of a tenth is
5 hundredths (.05)
Add one-half unit to
upper limits to get
upper boundaries
LL
UL
LB
UB
MdPt
62.9
64.8
62.85 64.85
63.85
64.9
66.8
64.85 66.85
65.85
66.9
68.8
66.85 68.85
67.85
68.9
70.8
68.85 70.85
69.85
70.9
72.8
70.85 72.85
71.85
72.9
74.9
74.8
76.8
72.85 74.85
74.85 76.85
73.85
75.85
Freq.
Rel.
Freq.
Cum.
Freq.
Find the mean of
the limits (or
boundaries) to find
the midpoint of
each class.
72.1
71.2
67.9
67.3
69.5
68.6
68.8
69.4
73.5
67.1
69.2
75.7
71.1
69.6
70.7
66.9
71.4
62.9
69.2
64.9
68.2
65.2
69.7
72.2
67.5
66.6
66.5
64.2
65.4
70.0
Count how many data
points fit in each class and
enter that into the
Frequency column
LL
UL
62.9
64.8
64.9
LB
UB
MdPt
Freq.
62.85 64.85
63.85
2
66.8
64.85 66.85
65.85
5
66.9
68.8
66.85 68.85
67.85
8
68.9
70.8
68.85 70.85
69.85
8
70.9
72.8
70.85 72.85
71.85
5
72.9
74.9
74.8
76.8
72.85 74.85
74.85 76.85
73.85
75.85
1
1
Draw the histogram using the
frequencies obtained from the table
we just did.
Looking at the histogram drawn from
the frequency table, it is easy to see
that the data is almost perfectly bellshaped, symmetrical and centered
about the mean.
8
8
5
5
2
1
62.85
64.85
66.85 68.85
70.85
72.85
1
74.85 76.85
The mean, median, and mode are also very closely bunched together.
Mean is 68.75, median is 69.00 and the mode is 69.2
For these reasons, it is reasonable to assume that the heights are normally
distributed.
The last part of the question was to compare the mean and standard deviation of
your sample with those in the previous study. Discuss the differences.
Our mean and standard deviation were 68.75 and 2.85.
The previous study had a mean of 69.6 and a standard deviation of 3.0.
This means that our sample of men was shorter than the previous study, but that
they were also more closely bunched together in height.
Your assignments are:
Classwork:
Pages 248-250, #1-8 All, and #9-41 Odd
Homework:
Pages 250-252, #43-61 Odd