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Understanding Normal Distributions: Probability Calculations and Examples

This article explores how to find probabilities using normal distributions, particularly focusing on examples with specified means (μ) and standard deviations (σ). With a mean of 10 and standard deviation of 5, the probability that a random variable falls below a certain value, such as 15, is calculated. Further examples illustrate cases with a mean of 78 and standard deviation of 8, analyzing test scores to determine the likelihood of scoring below or above specific thresholds, such as 90 and 85. The concepts of z-scores and area under the curve are emphasized for clarity.

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Understanding Normal Distributions: Probability Calculations and Examples

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  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

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