1. Normal Probability Distributions Chapter 5

2. § 5.2 Normal Distributions: Finding Probabilities

3. Larson & Farber, Elementary Statistics: Picturing the World, 3e 3 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

4. Larson & Farber, Elementary Statistics: Picturing the World, 3e 4 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)

5. Larson & Farber, Elementary Statistics: Picturing the World, 3e 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 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

6. Larson & Farber, Elementary Statistics: Picturing the World, 3e 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 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

7. Larson & Farber, Elementary Statistics: Picturing the World, 3e 7 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

8. Larson & Farber, Elementary Statistics: Picturing the World, 3e 8 Homework Page 227-228; 44, 53-62 P233; 13-17 odd

9. Larson & Farber, Elementary Statistics: Picturing the World, 3e 9 P 227-228; 44, 53-62, P233; 13-17 odd 44 A=24,750 B=30,000 C=33,000 D=35,150 -2.10 0 1.2 2.06 A and D are a little unusual… 54 0.6736 56 0.5987 58 0.481260 0.9500 62 0.0500