6.3 Use Normal Distributions

Page 399What is a normal distribution?

What is standard normal distribution?What does the z-score represent?

Normal Distribution and Normal CurveNormal 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 CurveA normal distribution with mean 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

SOLUTIONThe probability that a randomly selected x-value lies between – 2σ and is the shaded area under the normal curve shown.

xx

P( – 2σ ≤ x ≤ )x x

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).

= 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.a. About what percent of the women have readings

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 women have readings between 158 and 186 (34% + 34% = 68%).

Mg/dl = milligrams per deciliter

Readings less than 158 are considered desirable. About what percent of the readings are undesirable?

b.

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. 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 and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution.

x

1. P( ≤ )x xP( ≤ ) =x x P( – 3σ) + x P( – 2σ) +x

P( – σ) x

= 0.0015 + 0.0235 + 0.135 + 0.34

0.5ANSWER

2. P( > )x x

0.5ANSWER

P( > ) =x x P( + σ) + x P( + 2σ) +xP( + 3σ) x

= 0.34 + 0.135 + 0.0235 + 0.0015

A normal distribution has mean and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution.

x

A normal distribution has mean and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution.

x

3. P( < < + 2σ ) x x x

0.475ANSWER

= 0.34 + 0.135

P( + σ) + xP( + 2σ)x

P( < < + 2σ )=x x x

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 -values from a normal distribution with mean and standard deviation σ into -values having a standard normal distribution.

Formula: Subtract the mean from the given -value, then divide by the standard deviation.

-scoreThe z-value for a particular -value is called the -score for the -value and is the number of standard deviations the -value lies above or below the mean .

Standard Normal Table

If is a randomly selected value from a standard normal distribution, the table below can be used to find the probability that is less than or equal to some given value.

See page 401In the table, the value .000+ means “slightly more than 0” and the value 1.0000− means “slightly less than 1”

Scientists conducted aerial surveys of a seal sanctuary and recorded the number 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.

Biology

SOLUTION

STEP 1 Find: the z-score corresponding to an x-value of 50.

–1.6z= x – x 50 – 7314.1=

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 survey.

0.8849ANSWER

1.2z= x – x 90 – 7314.1=

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.

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.

ANSWER

• The bell shaped normal curve models a normal distribution

• In a normal distribution with mean 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.

• The formula can be used to transform x-values from a normal distribution with mean 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