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```6.3 Use Normal Distributions
Page 399
What is a normal distribution?
What is standard normal distribution?
What does the z-score represent?
Normal Distribution and Normal Curve
Normal distribution is one type of probability
distribution. It is modeled by a bell shaped
curve called a normal curve.
A normal curve is symmetric about the mean.
Areas Under a Normal Curve
A normal distribution with mean 𝑥 and standard
deviation σ has the following properties:
• The total area under the related normal curve is 1.
• About 68% of the area lies within 1 standard deviation
of the mean.
• About 95% of the area lies within 2 standard
deviations of the mean.
• About 99.7% of the area lies within 3 standard
deviations of the mean.
See page 399 Key Concept
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
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.
between 158 and 186?
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
158 and 186 (34% + 34% = 68%).
Mg/dl = milligrams per deciliter
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.
b. Readings less than 158 are considered
desirable. About what percent of the
b. A reading of 158 is one standard deviation to
the left of the mean, as shown. So, the percent of
readings that are desirable is 0.15% + 2.3% + 13.5%,
or 16%.
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 )
P( x ≤ x ) = P( x – 3σ) +P( x – 2σ) +
P( x – σ)
= 0.0015 + 0.0235 + 0.135 + 0.34
0.5
A normal distribution has meanx and
standard deviation σ. Find the indicated
probability for a randomly selected x-value
from the distribution.
2. P( x > x )
P( x >x ) = P( x + σ) + P( x + 2σ) +
P( x + 3σ)
= 0.34 + 0.135 + 0.0235 + 0.0015
0.5
A normal distribution has meanx and
standard deviation σ. Find the indicated
probability for a randomly selected x-value
from the distribution.
3. P( x < x < x + 2σ )
P( x < x< x + 2σ )= P( x+ σ) +
P( x + 2σ)
= 0.34 + 0.135
0.475
Standard Normal Distribution
The standard normal distribution is the normal
distribution with mean 0 and standard deviation
1. The formula below can be used to transform
𝑥-values from a normal distribution with mean
𝑥 and standard deviation σ into 𝑧-values having
a standard normal distribution.
Formula: 𝑧 =
𝑥−𝑥
𝜎
Subtract the mean from the given 𝑥-value, then
divide by the standard deviation.
𝑍-score
The z-value for a particular 𝑥-value is called the
𝑧-score for the 𝑥-value and is the number of
standard deviations the 𝑥-value lies above or
below the mean 𝑥.
Standard Normal Table
If 𝑧 is a randomly selected value from a standard
normal distribution, the table below can be used to
find the probability that 𝑧 is less than or equal to some
given value.
See page 401
In the table, the value .000+ means “slightly more than
0” and the value 1.0000− means “slightly less than 1”
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.
SOLUTION
STEP 1 Find: the z-score corresponding to an xvalue of 50.
z = x – x = 50 – 73 –1.6
14.1
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.
8.
WHAT IF? In Example 3, find the probability
that at most 90 seals were observed during a
– 73 1.2
survey.
z = x – x = 9014.1
Use: the table to find P(x < 90) P(z < 1.2).
The table shows that P(z < 1.2) = 0.8849. So, the
probability that at most 90 seals were observed
during a survey is about 0.8849.
0.8849
9.
REASONING: Explain why it makes sense that
P(z < 0) = 0.5.
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.
• The bell shaped normal curve models a normal
distribution
• In a normal distribution with mean 𝑥 and standard
deviation σ, the total area under the curve is 1.
About 68% of the area lies within 1 standard
deviation of the mean, about 95% of the area lies
within 2 standard deviations of the mean, and about
99.7% of the area lies within 3 standard deviations of
the mean.
𝑥−𝑥
𝜎
• The formula z =
can be used to transform xvalues from a normal distribution with mean 𝑥 and
standard deviation σ into z-values having a standard
normal distribution with mean 0 and standard
deviation 1.
6.3 Assignment
Page 402, 3-14, 19-24
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