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7.4 Normal Distributions EXAMPLE 1 Find a normal probability 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 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? b. Readings higher than 200 are considered undesirable. About what percent of the readings are undesirable? 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 b. Interpret normally distribute data 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 x and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution. 1. P( x ≤ x ) ANSWER 0.5 GUIDED PRACTICE 2. P( x > x ) ANSWER 0.5 for Examples 1 and 2 GUIDED PRACTICE for Examples 1 and 2 3. P( x < x < x + 2σ ) ANSWER 0.475 GUIDED PRACTICE 4. P( x – σ < x < x ) ANSWER 0.34 for Examples 1 and 2 GUIDED PRACTICE 5. P(x ≤ x – 3σ) ANSWER 0.0015 for Examples 1 and 2 GUIDED PRACTICE 6. P(x > x + σ) ANSWER 0.16 for Examples 1 and 2 GUIDED PRACTICE for Examples 1 and 2 7. WHAT IF? In Example 2, what percent of the women have readings between 172 and 200? ANSWER 47.5% 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. 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. z = x – x = 50 – 73 14.1 –1.6 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 8. for Example 3 WHAT IF? In Example 3, find the probability that at most 90 seals were observed during a survey. ANSWER 0.8849 GUIDED PRACTICE 9. for Example 3 REASONING: Explain why it makes sense that P(z < 0) = 0.5. ANSWER 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. Daily Homework Quiz 1. A normal distribution has mean x and standard deviation . For a randomly selected x-value from the distribution, find P(x x – 2). ANSWER 2. For use after Lesson 11.3 0.025 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