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7.4 M2 Normal Distributions EQ:

7.4 M2 Normal Distributions EQ: How do you find the probability of an event given that the data is normally distributed and its mean and standard deviation are known?. The Normal Curve - Characteristics.

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7.4 M2 Normal Distributions EQ:

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  1. 7.4 M2 Normal Distributions • EQ: • How do you find the probability of an event given that the data is normally distributed and its mean and standard deviation are known?

  2. The Normal Curve - Characteristics • You can use the characteristics of a normal distribution to find probabilities of many real-world events, such as the number of deaths caused by lightning each month. • This curve is called a normal curve. • A normal curve is defined by the mean and the standard deviation. • The total area of the curve is 1. • The total area to left and right = .5

  3. A little more… • The curve is symmetric about the mean, x-bar. • The mean, median and mode are about equal.

  4. And finally… • About 68% of the area falls within 1 standard deviation of the mean. • About 95% of the area falls within 2 standard deviations of the mean. • About 99.7% of the area falls within 3 standard deviations of the mean. • If what the graph shows is symmetrical, you can say that the data represents a normal distribution.

  5. Example 1: A city’s annual rainfall is approximately normally distributed with a mean of 40” and a standard deviation of 6”. Find the probability for each example. a)less than 34” P(x<34)= 13.5% + 2.35% + .15% = 16% b) greater than 46” P(x>46) = 13.5% +2.35% + .15% = 16% c) between 34” and 46” P(34<x<46) = 34% + 34% = 68%

  6. Example 2: Scores on a professional exam are normally distributed with a mean of 500 and a standard deviation of 50. Out of 28,000 randomly selected exams, find the number of exams that could be expected to have each score. a) greater than 500 P(x>500) = 34% + 13.5% + 2.35% + .15% = 50%* 28,000 = 14,000 people b) less than 600 P(x<600) =50% + 34% + 13.5% = 97.5% * 28,000 = 27, 300 people c) between 400 and 600 P(400<x600)= 13.5% + 34% + 13.5% + 34% = 95% * 28,000 = 26,600 people

  7. Homework p. 267 #1-17 odds; 18-20 all

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