Download f(x) 1 2πσ e Normal distribution of grades with µ=70 and σ=10, and

Normal Distribution (Part I)
The normal distribution is a rather special continuous density function that is observed frequently
in the natural world. Examples of continuous random variables that likely follow a normal
distribution include:
X = cholesterol levels of a healthy male population
X = head circumference, in centimeters, of adult females
X = body length of adult King salmon
Characteristics of a Normal Random Variable
The probability density function f(x) of a normal random variable depends only on the mean µ
(“mu”) and the standard deviation ) (“sigma”). Its formula
f(x) 1
2%)
e
1 xµ
2
2
)
suggests that there are an infinite number of possible normal density functions, whose shapes
each depend on µ and ). The mean µ indicates the LOCATION (or “center”) of the distribution,
while the standard deviation ) indicates the SHAPE (or “spread”) of the distribution. Because the
shape of the normal distribution is symmetric and looks like a bell, the distribution is also often
called a BELL CURVE.
0.08
Height of curve
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00
40
50
60
70
80
90
100
Grades
Normal distribution of grades with µ=70 and )=10, and
normal distribution of grades with µ=70 and )=5.
Handout 09
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Of course, the normal distribution follows the basic properties of all continuous probability
density functions, namely:
i) f(x) 0,
ii) we find probabilities by finding the area under the curve, and
iii) the area under the whole curve is 1.
Example:
Consider the population of elite male distance runners. It is known that their weight follows
a normal distribution with an average weight of µ=138 pounds and a standard deviation of
)=8 pounds. That is, their distribution of weights looks like:
0.05
Height of curve
0.04
0.03
0.02
0.01
0.00
106
114
122
130
138
146
154
162
170
Weight
Then, to find the probability that a randomly selected elite male distance runner weighs less
than 128 pounds, i.e. F(128) = P(X 128), we need to find the area under the curve. If we
had a cumulative distribution table for a normal random variable with µ=138 and )=8, we
could just easily look up F(128). But, this is an unrealistic proposition: since there are an
infinite number of normal distributions, we would need an infinite number of cumulative
distribution tables.
To solve this problem, we “STANDARDIZE” by picking one normal distribution, namely the
one where µ=0 and )=1. We denote this random variable by the capital letter Z, and call it
the STANDARD NORMAL RANDOM VARIABLE. A cumulative distribution table for Z has been
created. (It is Table III in our text.) To use the table for Z, you must first transform your
random variable X by using:
Xµ
Z )
Note that Z is merely the number of standard deviations that X falls above or below the
mean.
Handout 09
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So, here, we transform X = 128 by:
Z Xµ
128 138
1.25
)
8
That is, FX(128) = P(X 128) is equivalent to FZ(-1.25) = P(Z -1.25). Then, we look up
the cumulative probability corresponding to Z = -1.25 on a standard normal table as follows:
1.
2.
3.
4.
Carry out all calculations of Z to two decimal places, that is X.XX
Find the first 2 digits of Z in the column headed by z.
Find the third digit of Z in the first bolded row.
P(Z z) = the probability found at the intersection of the row and column found in steps
2 and 3 above.
So, P(X 128) = P(Z -1.25) = 0.1056. That is, 10.56% of all male elite distance runners
weigh less than 128 pounds.
More examples:
Find the probability that a randomly selected male elite distance runner weighs between
130.8 and 152.3 pounds.
0.05
Height of curve
0.04
0.03
0.02
0.01
0.00
106
114
122
130
138
146
154
162
170
Weight
Handout 09
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Find the probability that a randomly selected male elite distance runner weighs more than
163 pounds.
0.05
Height of curve
0.04
0.03
0.02
0.01
0.00
106
114
122
130
138
146
154
162
170
Weight
Normal Probability Rule
The probability that a normal random variable X lies within ....
...one standard deviation of its mean is 0.68
...two standard deviations of its mean is 0.95 (called “2-SIGMA LIMITS”)
...three standard deviations of its mean is 0.99
Handout 09
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