1 / 7

§ 5.3

§ 5.3. Normal Distributions: Finding Values. μ = 10 σ = 5. P ( x < 15). x. μ = 10. 15. Probability and Normal Distributions.

aaron-martinez
Télécharger la présentation

§ 5.3

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. § 5.3 Normal Distributions: Finding Values

  2. μ = 10 σ = 5 P(x < 15) x μ =10 15 Probability and Normal Distributions If a random variable, x, is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval.

  3. Normal Distribution Standard Normal Distribution μ = 10 σ = 5 μ = 0 σ = 1 P(z < 1) P(x < 15) z x μ =10 15 μ =0 1 Probability and Normal Distributions Same area P(x < 15) = P(z < 1) = Shaded area under the curve = 0.8413

  4. μ = 78 σ = 8 P(x < 90) x μ =78 90 z μ =0 ? Probability and Normal Distributions Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score less than 90. The probability that a student receives a test score less than 90 is 0.9332. 1.5 P(x < 90) = P(z < 1.5) = 0.9332

  5. μ = 78 σ = 8 P(x > 85) x μ =78 85 z μ =0 ? Probability and Normal Distributions Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score greater than than 85. The probability that a student receives a test score greater than 85 is 0.1894. 0.88 P(x > 85) = P(z > 0.88) = 1 P(z < 0.88) = 1  0.8106 = 0.1894

  6. P(60 < x < 80) μ = 78 σ = 8 x 60 μ =78 80 z μ =0 ? ? Probability and Normal Distributions Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score between 60 and 80. The probability that a student receives a test score between 60 and 80 is 0.5865. 0.25 2.25 P(60 < x < 80) = P(2.25 < z < 0.25) = P(z < 0.25)  P(z <2.25) = 0.5987  0.0122 = 0.5865

  7. Complete 5.3 Practice

More Related