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7.4 Normal Distributions
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
Interpret normally distribute data
Health
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.
a.
About what percent of the women have readings
between 158 and 186?
b. Readings higher than 200 are considered
undesirable. About what percent of the readings
are undesirable?
EXAMPLE 2
Interpret normally distribute data
SOLUTION
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
b.
Interpret normally distribute data
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%.
GUIDED PRACTICE
for Examples 1 and 2
A normal distribution has mean x and standard
deviation σ. Find the indicated probability for a
randomly selected x-value from the distribution.
1. P( x ≤ x )
ANSWER
0.5
GUIDED PRACTICE
2. P( x > x )
ANSWER
0.5
for Examples 1 and 2
GUIDED PRACTICE
for Examples 1 and 2
3. P( x < x < x + 2σ )
ANSWER
0.475
GUIDED PRACTICE
4. P( x – σ < x < x )
ANSWER
0.34
for Examples 1 and 2
GUIDED PRACTICE
5. P(x ≤ x – 3σ)
ANSWER
0.0015
for Examples 1 and 2
GUIDED PRACTICE
6. P(x > x + σ)
ANSWER
0.16
for Examples 1 and 2
GUIDED PRACTICE
for Examples 1 and 2
7. WHAT IF? In Example 2, what percent of the
women have readings between 172 and 200?
ANSWER
47.5%
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.
EXAMPLE 3
Use a z-score and the standard normal table
GUIDED PRACTICE
8.
for Example 3
WHAT IF? In Example 3, find the probability that at
most 90 seals were observed during a survey.
ANSWER
0.8849
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.
Daily Homework Quiz
1.
A normal distribution has mean x and standard
deviation . For a randomly selected x-value
from the distribution, find P(x  x – 2).
ANSWER
2.
For use after Lesson 11.3
0.025
The average donation during a fund drive was $75.
The donations were normally distributed with a
standard deviation of $15. Use a standard normal
table to find the probability that a donation is at
most $115.
ANSWER
0.9953