7.4 Normal Distributions. EXAMPLE 1 Find a normal probability SOLUTION The probability that a randomly selected x -value lies between – 2σ and is.

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Presentation transcript:

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. x x P ( – 2σ ≤ x ≤ )xx 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). = = 0.475

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% %, 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 1.1. P ( ≤ )xx 0.5 ANSWER

GUIDED PRACTICE for Examples 1 and P( > )xx 0.5 ANSWER

GUIDED PRACTICE for Examples 1 and P( < < + 2σ )xxx ANSWER

GUIDED PRACTICE for Examples 1 and P( – σ < x < )xx 0.34 ANSWER

GUIDED PRACTICE for Examples 1 and P (x ≤ – 3σ) x ANSWER

GUIDED PRACTICE for Examples 1 and P (x > + σ) x 0.16 ANSWER

GUIDED PRACTICE for Examples 1 and 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. –1.6 z = x – x 50 – = STEP 2 Use: the table to find P(x < 50) P(z < – 1.6). The table shows that P(z < – 1.6) = So, the probability that at most 50 seals were observed during a survey is about

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 ANSWER

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