7.4 Normal Distributions

EXAMPLE 1 Find a normal probability

SOLUTION

The 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

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?

Readings higher than 200 are considered undesirable. About what percent of the readings are undesirable?

b.

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 Interpret normally distribute data

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%.

GUIDED PRACTICE for Examples 1 and 2

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 x

0.5ANSWER

GUIDED PRACTICE for Examples 1 and 2

2. P( > )x x

0.5ANSWER

GUIDED PRACTICE for Examples 1 and 2

3. P( < < + 2σ ) x x x

0.475ANSWER

GUIDED PRACTICE for Examples 1 and 2

4. P( – σ < x < ) x x

0.34ANSWER

GUIDED PRACTICE for Examples 1 and 2

5. P(x ≤ – 3σ)x

0.0015ANSWER

GUIDED PRACTICE for Examples 1 and 2

6. P(x > + σ)x

0.16ANSWER

GUIDED PRACTICE for Examples 1 and 2

7. WHAT IF? In Example 2, what percent of the women have readings between 172 and 200?

47.5%ANSWER

EXAMPLE 3 Use a z-score and the standard normal table

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.

Biology

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.

–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.

EXAMPLE 3 Use a z-score and the standard normal table

GUIDED PRACTICE for Example 3

8. WHAT IF? In Example 3, find the probability that at most 90 seals were observed during a survey.

0.8849ANSWER

GUIDED PRACTICE for Example 3

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

Daily Homework Quiz For use after Lesson 11.3

0.025ANSWER

2. 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

1. A normal distribution has mean x and standard deviation . For a randomly selected x-value from the distribution, find P(x x – 2).