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Normal Probability Distributions

Chapter 5. Normal Probability Distributions. Normal Distributions: Finding Probabilities. § 5.2. μ = 10 σ = 5. P ( x < 15). x. μ = 10. 15. Probability and Normal Distributions.

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Normal Probability Distributions

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  1. Chapter 5 Normal Probability Distributions

  2. Normal Distributions: Finding Probabilities § 5.2

  3. 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.

  4. 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

  5. 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

  6. 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

  7. 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

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