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7.4 Use Normal
Distributions
p. 266
Normal Distribution
A bell-shaped curve is called a normal curve. It is
symmetric about the mean. The percentage of the
area in each standard deviation is shown above.
Standard Normal Distribution
Mean is 0 and standard deviation is 1. The formula is
used to find the z scores. (# of standard deviations
away from the mean)
z
x X

EXAMPLE 1
Find a normal probability
A normal distribution has mean x and standard
deviation σ. For a randomly selected x-value from the
distribution, find P(x – 2σ ≤ x ≤ x).
SOLUTION
The probability that a randomly
selected x-value lies between x – 2σ
and x is the shaded area under the
normal curve shown.
P( x – 2σ ≤ x ≤ x ) = 0.135 + 0.34 = 0.475
EXAMPLE 2
a.
Interpret normally distribute data
The blood cholesterol readings for a group of women
are normally distributed with a mean of 172 mg/dl and
a standard deviation of 14 mg/dl.
About what percent of the women have readings between
158 and 186? (Hint: Find out how far away 158 and 186 are
from the mean.)
a. The readings of 158 and 186 represent one standard
deviation on either side of the mean, as shown below.
So, 68% of the women have readings between 158 and
186.
EXAMPLE 2
Interpret normally distribute data
b. Readings higher than 200 are considered
undesirable. About what percent of the readings
are undesirable?
b.
A reading of 200 is two standard deviations to the
right of the mean, as shown. So, the percent of
readings that are undesirable is 2.35% + 0.15%, or
2.5%.
for Examples 1 and 2
GUIDED PRACTICE
x standard deviation σ.
A normal distribution has mean and
Find the indicated probability for a randomly selected xvalue from the distribution.
1. P( x ≤ x )
2. P( x > x )
0.5
0.5
3. P( x < x < x + 2σ )
ANSWER
0.475
4. P( x – σ < x < x )
ANSWER
0.34
GUIDED PRACTICE
5. P(x ≤ x – 3σ)
ANSWER
for Examples 1 and 2
6. P(x > x + σ)
0.0015
ANSWER
0.16
EXAMPLE 3
Use a z-score and the standard normal table
Biology
Scientists conducted aerial surveys of a seal sanctuary
and recorded the number x of seals they observed
during each survey. The numbers of seals observed
were normally distributed with a mean of 73 seals and a
standard deviation of 14.1 seals. Find the probability
that at most 50 seals were observed during a survey.
EXAMPLE 3
Use a z-score and the standard normal table
SOLUTION
STEP 1 Find: the z-score corresponding to an x-value
of 50.
z = x – x = 50 – 73
14.1
–1.6
STEP 2 Use: the table to find P(x < 50)
P(z < – 1.6).
The table shows that P(z < – 1.6) = 0.0548. So,
the probability that at most 50 seals were
observed during a survey is about 0.0548.
GUIDED PRACTICE
9.
for Example 3
REASONING: Explain why it makes sense that
P(z < 0) = 0.5.
ANSWER
A z-score of 0 indicates that the z-score and the
mean are the same. Therefore, the area under
the normal curve is divided into two equal parts
with the mean and the z-score being equal to 0.5.