1 / 17

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.

glen
Download Presentation

7.4 Normal Distributions Part II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 %

More Related