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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. - PowerPoint PPT Presentation
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7.4 M2 Normal DistributionsEQ: •How do you find the probability of an event given that the data is normally distributed and its mean and standard deviation are known?
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
The Normal Curve - Characteristics
A little more…The curve is symmetric about the mean, x-bar.
The mean, median and mode are about equal.
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.
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%
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 500P(x>500) = 34% + 13.5% + 2.35% + .15% = 50%* 28,000 = 14,000 people
b) less than 600P(x<600) =50% + 34% + 13.5% = 97.5% * 28,000 = 27, 300 people
c) between 400 and 600P(400<x600)= 13.5% + 34% + 13.5% + 34% = 95% * 28,000 = 26,600 people
p. 267 #1-17 odds; 18-20 all