Chapter 7 The Normal Probability Distribution

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Chapter 7The Normal Probability Distribution

• 7.1
• Properties of the Normal Distribution

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EXAMPLE Illustrating the Uniform
Distribution Suppose that United Parcel Service
is supposed to deliver a package to your front
door and the arrival time is somewhere between 10
am and 11 am. Let the random variable X represent
the time from10 am when the delivery is supposed
to take place. The delivery could be at 10 am (x
0) or at 11 am (x 60) with all 1-minute
interval of times between x 0 and x 60
equally likely. That is to say your package is
just as likely to arrive between 1015 and 1016
as it is to arrive between 1040 and 1041. The
random variable X can be any value in the
interval from 0 to 60, that is, 0 lt X lt 60.
Because any two intervals of equal length between
0 and 60, inclusive, are equally likely, the
random variable X is said to follow a uniform
probability distribution.
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Probability Density Function A probability
density function is an equation that is used to
compute probabilities of continuous random
variables that must satisfy the following two
properties. Let f(x) be a probability density
function. 1. The area under the graph of the
equation over all possible values of the random
variable must equal one,that is 2. The graph of
the equation must be greater than or equal to
zero for all possible values of the random
variable. That is, the graph of the equation
must lie on or above the horizontal axis for all
possible values of the random variable, i.e.
f(x)gt0
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The area under the graph of a density function
over some interval represents the probability of
observing a value of the random variable in that
interval.
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EXAMPLE Area as a Probability Referring to the
earlier example, what is the probability that
your package arrives between 1010 am and 1020
am?
(1020-1010)/6010/601/6
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Relative frequency histograms that are symmetric
and bell-shaped are said to have the shape of a
normal curve.
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If a continuous random variable is normally
distributed or has a normal probability
distribution, then a relative frequency histogram
of the random variable has the shape of a normal
curve (bell-shaped and symmetric).
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Properties of the Normal Density Curve 7. The
Empirical Rule About 68 of the area under the
graph is within one standard deviation of the
mean about 95 of the area under the graph is
within two standard deviations of the mean about
99.7 of the area under the graph is within three
standard deviations of the mean.
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EXAMPLE A Normal Random Variable The following
data represent the heights (in inches) of a
random sample of 50 two-year old males. (a)
Create a relative frequency distribution with the
lower class limit of the first class equal to
31.5 and a class width of 1. (b) Draw a histogram
of the data. (c ) Do you think that the variable
height of 2-year old males is normally
distributed?
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36.0 36.2 34.8 36.0 34.6 38.4 35.4 36.8 34.7 33.4
37.4 38.2 31.5 37.7 36.9 34.0 34.4 35.7 37.9 39.3
34.0 36.9 35.1 37.0 33.2 36.1 35.2 35.6 33.0 36.8
33.5 35.0 35.1 35.2 34.4 36.7 36.0 36.0 35.7 35.7
38.3 33.6 39.8 37.0 37.2 34.8 35.7 38.9 37.2 39.3
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In the next slide, we have a normal density curve
drawn over the histogram. How does the area of
the rectangle corresponding to a height between
34.5 and 35.5 inches relate to the area under the
curve between these two heights?
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EXAMPLE Interpreting the Area Under a Normal
Curve The weights of pennies minted after
1982 are approximately normally distributed with
mean 2.46 grams and standard deviation 0.02
grams. (a) Draw a normal curve with the
parameters labeled. (b) Shade the region under
the normal curve between 2.44 and 2.49 grams. (c)
Suppose the area under the normal curve for the
shaded region is 0.7745. Provide two
interpretations for this area.
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EXAMPLE Relation Between a Normal Random
Variable and a Standard Normal Random
Variable The weights of pennies minted after
1982 are approximately normally distributed with
mean 2.46 grams and standard deviation 0.02
grams. Draw a graph that demonstrates the area
under the normal curve between 2.44 and 2.49
grams is equal to the area under the standard
normal curve between the Z-scores of 2.44 and
2.49 grams.
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Chapter 7The Normal Probability Distribution

• 7.2
• The Standard Normal Distribution

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Properties of the Normal Density Curve 7. The
Empirical Rule About 68 of the area under the
graph is between -1 and 1 about 95 of the area
under the graph is between -2 and 2 about 99.7
of the area under the graph is between -3 and 3.
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The table gives the area under the standard
normal curve for values to the left of a
specified Z-score, zo, as shown in the figure.
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EXAMPLE Finding the Area Under the
Standard Normal Curve Find the area under the
standard normal curve to the left of Z -0.38.
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Area under the normal curve to the right of zo
1 Area to the left of zo
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EXAMPLE Finding the Area Under the
Standard Normal Curve Find the area under the
standard normal curve to the right of Z 1.25.
Look in the Normal distribution table
P(xgt1.25) 1-P(xlt1.25)1-0.89440.1056
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EXAMPLE Finding the Area Under the
Standard Normal Curve Find the area under the
standard normal curve between Z -1.02 and Z
2.94.
P(-1.02ltxlt2.94)P(xlt2.94)-p(xlt-1.02)
0.9984-0.1539 0.8445
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EXAMPLE Finding a Z-score from a Specified
Area to the Left Find the Z-score such that the
area to the left of the Z-score is 0.68.
i.e., find Z such that P(xltZ)0.68 , Z 0.46
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EXAMPLE Finding a Z-score from a Specified
Area to the Right Find the Z-score such that the
area to the right of the Z-score is 0.3021.
P(xgtZ) 0.3021, so P(xltZ) 0.6979 z0.52
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EXAMPLE Finding a Z-score
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EXAMPLE Finding the Value of z? Find the value
of z0.25
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Notation for the Probability of a Standard Normal
Random Variable P(a lt Z lt b) represents the
probability a
standard normal random variable is
between a and b P(Z gt a) represents
the probability a
standard normal random variable is
greater than a. P(Z lt a) represents
the probability a standard normal random
variable is less than a.
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EXAMPLE Finding Probabilities of Standard
Normal Random Variables Find each of the
following probabilities (a) P(Z lt -0.23) (b) P(Z
gt 1.93) (c) P(0.65 lt Z lt 2.10)
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