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7.4 Normal Distributions. 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 ). x. x. x. x. - PowerPoint PPT Presentation
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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).