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7.4 Normal Distributions Part II

7.4 Normal Distributions Part II. p. 264. 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. From Yesterday’s notes. GUIDED PRACTICE. P ( > ). x.

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7.4 Normal Distributions Part II

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  1. 7.4 Normal DistributionsPart II p. 264

  2. 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 From Yesterday’s notes GUIDED PRACTICE

  3. P(> ) x 2. x ANSWER 0.5 From Yesterday’s notes GUIDED PRACTICE

  4. P(<< + 2σ ) x 3. x x ANSWER 0.475 From yesterday’s notes GUIDED PRACTICE

  5. P( – σ<x<) x x ANSWER 0.34 From yesterday’s notes GUIDED PRACTICE 4.

  6. P(x ≤ – 3σ) 5. x ANSWER 0.0015 From yesterday’s notes GUIDED PRACTICE

  7. P(x > + σ) 6. x ANSWER 0.16 From yesterday’s notes GUIDED PRACTICE

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

  9. Formula for SND 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.

  10. 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.

  11. 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.

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

  13. 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.

  14. 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.

  15. EXAMPLE 4 Use a z-score and the standard normal table 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. Find each person’s z-score Matt – completed course A in 59 seconds John – completed course B in 1 minute, 31 seconds

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

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