1 / 11

7.4 Use Normal Distributions

7.4 Use Normal Distributions. p. 266. Normal Distribution. A bell-shaped curve is called a normal curve. It is symmetric about the mean. The percentage of the area in each standard deviation is shown above. Standard Normal Distribution.

holly-moore
Download Presentation

7.4 Use Normal Distributions

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 Use Normal Distributions p. 266

  2. Normal Distribution A bell-shaped curve is called a normal curve. It is symmetric about the mean. The percentage of the area in each standard deviation is shown above.

  3. Standard Normal Distribution Mean is 0 and standard deviation is 1. The formula is used to find the z scores. (# of standard deviations away from the mean)

  4. A normal distribution has mean xand standard deviation σ. For a randomly selected x-value from the distribution, find P(x – 2σ ≤ x ≤ x). x x x x The probability that a randomly selected x-value lies between –2σ and is the shaded area under the normal curve shown. P(–2σ ≤ x ≤ ) EXAMPLE 1 Find a normal probability SOLUTION = 0.135 + 0.34 = 0.475

  5. EXAMPLE 2 Interpret normally distribute data 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? (Hint: Find out how far away 158 and 186 are from the mean.) 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.

  6. Readings higher than 200 are considered undesirable. About what percent of the readings are undesirable? b. 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% + 0.15%, or 2.5%.

  7. A normal distribution has mean and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution. P( – σ<x<) P(<< + 2σ ) P(> ) x x 2. 3. x x x x x x x P(≤ ) x 1. ANSWER ANSWER 0.475 0.34 for Examples 1 and 2 GUIDED PRACTICE 0.5 0.5 4.

  8. P(x ≤ – 3σ) P(x > + σ) 5. 6. x x ANSWER 0.16 ANSWER 0.0015 for Examples 1 and 2 GUIDED PRACTICE

  9. 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.

  10. 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.

  11. 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.

More Related