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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.
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A normal distribution has mean xand standard deviation σ. For a randomly selected x-value from the distribution, find P(x – 2σ ≤ x ≤ x). x x x x The probability that a randomly selected x-value lies between –2σ and is the shaded area under the normal curve shown. P(–2σ ≤ x ≤ ) EXAMPLE 1 Find a normal probability SOLUTION = 0.135 + 0.34 = 0.475
Readings higher than 200 are considered undesirable. About what percent of the readings are undesirable? b. 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?
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 SOLUTION
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%. EXAMPLE 2 Interpret normally distribute data
A normal distribution has mean and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution. x x P(≤ ) x 1. ANSWER 0.5 for Examples 1 and 2 GUIDED PRACTICE
P(> ) x 2. x ANSWER 0.5 for Examples 1 and 2 GUIDED PRACTICE
P(<< + 2σ ) x 3. x x ANSWER 0.475 for Examples 1 and 2 GUIDED PRACTICE
P( – σ<x<) x x ANSWER 0.34 for Examples 1 and 2 GUIDED PRACTICE 4.
P(x ≤ – 3σ) 5. x ANSWER 0.0015 for Examples 1 and 2 GUIDED PRACTICE
P(x > + σ) 6. x ANSWER 0.16 for Examples 1 and 2 GUIDED PRACTICE
WHAT IF?In Example 2, what percent of the women have readings between 172 and 200? 7. ANSWER 47.5% for Examples 1 and 2 GUIDED PRACTICE
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.
x 50 – 73 –1.6 z = = 14.1 Use: the table to find P(x <50) P(z <– 1.6). x – EXAMPLE 3 Use a z-score and the standard normal table SOLUTION Find: the z-score corresponding to an x-value of 50. STEP 1 STEP 2 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
ANSWER 0.8849 for Example 3 GUIDED PRACTICE 8. WHAT IF? In Example 3, find the probability that at most 90 seals were observed during a survey.
ANSWER Az-scoreof 0 indicates that thez-score and the mean are the same. Therefore, the area under the normal curve is divided into two equal parts with the mean and thez-score being equal to 0.5. for Example 3 GUIDED PRACTICE 9. REASONING: Explain why it makes sense that P(z< 0) = 0.5.
1. A normal distribution has mean xand standard deviation . For a randomly selected x-value from the distribution, find P(xx – 2). ANSWER 0.025 ANSWER 0.9953 Daily Homework Quiz For use after Lesson 11.3 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.