1. 7.4 Normal Distributions Part II p. 264

2. GUIDED PRACTICE From Yesterday’s notes 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

3. GUIDED PRACTICE From Yesterday’s notes 2.2. P( > )xx 0.5 ANSWER

4. GUIDED PRACTICE From yesterday’s notes 3.3. P( < < + 2σ )xxx 0.475 ANSWER

5. GUIDED PRACTICE From yesterday’s notes 4.4. P( – σ < x < )xx 0.34 ANSWER

6. GUIDED PRACTICE From yesterday’s notes 5.5. P (x ≤ – 3σ) x 0.0015 ANSWER

7. GUIDED PRACTICE From yesterday’s notes 6.6. P (x > + σ) x 0.16 ANSWER

8. VOCABULARY Z-Score – the number (z) of standard deviations that a data value lies above or below the mean of the data set. Z-Score – the number (z) of standard deviations that a data value lies above or below the mean of the data set.

10. The formula below can be used to transform x-values from a normal distribution with mean and standard deviation into z-values having a standard normal distribution.

11. 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

12. 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 – 73 14.1 = 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.

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

14. GUIDED PRACTICE for Example 3 8. WHAT IF? In Example 3, find the probability that at most 90 seals were observed during a survey. 0.8849 ANSWER

15. 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

16. EXAMPLE 4 Use a z-score and the standard normal table Find each person’s z-score Matt – completed course A in 59 seconds John – completed course B in 1 minute, 31 seconds OBSTACLE COURSE Two different obstacle courses were set up for gym class. The times to complete Course A are normally distributed with a mean of 54 seconds and a standard deviation of 6.1 seconds. The times to complete Course B are normally distributed with a mean of 1 minute, 25 seconds and a standard deviation of 8.7 seconds.

17. EXAMPLE 4 Use a z-score and the standard normal table SOLUTION MATT Find: the z-score corresponding to an x-value of 59. 0.82 z = x – x 59 – 54 6.1 = JOHN Find: the z-score corresponding to an x-value of 91. 0.69 z = x – x 91 – 85 8.7 = MATT = 0.7881 or 78.8 % JOHN = 0.7580 or 75.8 %