Normal Probability Distributions Chapter 5. § 5.2 Normal Distributions: Finding Probabilities.

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Presentation transcript:

Normal Probability Distributions Chapter 5

§ 5.2 Normal Distributions: Finding Probabilities

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

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 = μ =10 P(x < 15) μ = 10 σ = 5 Normal Distribution x 1μ =0 σ = 1 Standard Normal Distribution z P(z < 1)

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) = The probability that a student receives a test score less than 90 is μ =0 z ? μ =78 P(x < 90) μ = 78 σ = 8 x

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  = The probability that a student receives a test score greater than 85 is μ =0 z ? μ =78 P(x > 85) μ = 78 σ = 8 x

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 μ =0 z ?? 0.25  2.25 =  = μ =78 P(60 < x < 80) μ = 78 σ = 8 x

Larson & Farber, Elementary Statistics: Picturing the World, 3e 8 Homework Page ; 44, P233; odd

Larson & Farber, Elementary Statistics: Picturing the World, 3e 9 P ; 44, 53-62, P233; odd 44 A=24,750 B=30,000 C=33,000 D=35, A and D are a little unusual…