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Normal Probability Distributions

Learn how to find probabilities in normal distributions based on interval calculations under the curve. Discover the relationship between standard and normal distributions with examples of probability calculations.

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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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