1. § 5.3 Normal Distributions: Finding Values

2. 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. P(x < 15) μ = 10 σ = 5 15μ =10 x

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

4. 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. Probability and Normal Distributions P(x < 90) = P(z < 1.5) = 0.9332 The probability that a student receives a test score less than 90 is 0.9332. μ =0 z ? 1.5 90μ =78 P(x < 90) μ = 78 σ = 8 x

5. 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. Probability and Normal Distributions P(x > 85) = P(z > 0.88) = 1  P(z < 0.88) = 1  0.8106 = 0.1894 The probability that a student receives a test score greater than 85 is 0.1894. μ =0 z ? 0.88 85μ =78 P(x > 85) μ = 78 σ = 8 x

6. 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. Probability and Normal Distributions P(60 < x < 80) = P(  2.25 < z < 0.25) = P(z < 0.25)  P(z <  2.25) The probability that a student receives a test score between 60 and 80 is 0.5865. μ =0 z ? ? 0.25  2.25 = 0.5987  0.0122 = 0.5865 6080μ =78 P(60 < x < 80) μ = 78 σ = 8 x

7. Complete 5.3 Practice