Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Ch 8 Some Useful Continuous Probability Distributions 8.1 Properties of Continuous Probability Distributions A smooth curve known as the density function, f x is used to represent the probability distribution of a continuous random variable. The curve must never fall below the x-axis … f x 0 for all x The total area under the curve must be 1 … f x 1 For continuous random variables we assign probability to intervals. (Not Points) P(a x b) the area under the curve between a and b. With continuous variables, each point has probability zero P( x a) 0 P ( x b) 0 Thus for continuous variables P ( a x b) P ( a x b) For continuous distributions Population Mean = xf x Population Variance = 2 x f x x 2 f x 2 2 Population Standard Deviation 2 8.2 The Uniform Distribution The density function for the uniform distribution is as follows 1 f x for a x b ba Calculating descriptive statistics ab Population Mean = 2 b xf x a b 1 x ba a b 1 x2 2b a a 1 1 b2 a2 2b a 2b a b2 a2 2b a b a b a 2b a ab 2 Population Median = Population variance = 2 b a 2 12 b x 2 f x 2 2 a b 1 x2 2 ba a b 1 x3 2 3b a a 1 1 b3 a3 2 3b a 3b a b3 a3 a b 3b a 2 2 b a b 2 ab a 2 a 2 2ab b 2 3b a 4 b 2 ab a 2 a 2 2ab b 2 3 4 2 2 2 4b ab a 3a 2ab b 2 12 2 2 b 2ab a 12 b a 2 12 Population standard deviation = 2 The probability that a value is between c and d is d c Pc x d where a c d b ba d f x c d 1 ba c d 1 x ba c d c ba ba d c ba Example The travel time from Lexington KY to Columbus Ohio is uniformly distributed between 200 and 240 minutes. Find the mean. a b 200 240 440 220 2 2 2 Find the median. 220 Find the variance. 2 2 2 b a 240 200 40 1600 2 133.333 12 12 12 12 Find the standard deviation. 2 133.333 11.547 Find the probability of arriving in less than 225 minutes. d c 225 200 25 P200 x 225 .625 b a 240 200 40 8.3 The Normal Distribution The density function for the normal distribution is as follows: f x x 2 1 2 2 e 2 2 The normal distribution is a very common type of continuous distribution. It is a bell shaped curve. The bell is symmetric about the mean of the random variable . The standard deviation of the random variable affects the spread of the bell. The larger is the more spread out the bell. The mean, median, and mode are all equal for the normal distribution. The value of and characterize which normal distribution we are talking about. The normal distribution with 0 and 1 is called the standard normal distribution. (This is used to calculate normal probabilities) Fact: If X is normal with mean and standard deviation , then Z x is standard normal. Explain why. Areas under the Standard Normal Curve Look at Standard Normal Distribution table in the back cover of the book; table gives P(0 Z a ) for the standard normal distribution Draw graph and show area Examples Draw Pictures of desired areas when doing problems!!! P(0 Z 1.55) .4394 P(0 Z 1.96) .4750 Facts: Total area under the curve is 1 Curve is symmetric about 0 P( Z 0) P( Z 0) 1 2 Combining these facts with the table allows us to compute all probability statements for Z Example PZ 1.64 .5 P0 Z 1.64 .5 .4945 .9495 P( Z 1.64) .5 P(0 Z 1.64) .5 .4495 .0505 P( Z 1.64) P( Z 1.64) .0505 by symmetry P(2.32 Z 0) P(0 Z 2.32) .4898 P(2 Z 2) 2 P(0 Z 2) 2 .4772 .9544 Does this make sense by the Empirical Rule? P(1.41 Z 2.18) P(0 Z 2.18) P(0 Z 1.41) .4854 .4207 .0647 Notice that Probabilities in the table stop at 3.9 are .5000. Beyond this Z value you will always have close to .5 the area. 8.4 Calculating Areas Under Any Normal Curve We have learned how to compute probabilities for the standard normal. We will now compute probabilities for any normal. Fact: If x is normal with mean and standard deviation , then Z x is standard normal. Write probability statement for X Rewrite in terms of Z Example The distribution of IQ scores for the general population is approximately normal with 100 and 10 . x = IQ score of randomly selected person Find P(100 X 120) Draw Picture P (100 X 120) 100 100 120 100 P( X ) 10 10 P (0 Z 2) .4772 Find P ( X 130) Draw Picture P( X 130) 130 100 P( Z ) 10 P( Z 3) .5 P(0 Z 3) .5 .4987 .0013 very unlikely Example Suppose the amount of Pepsi in a “12 oz” can has a Normal distribution with 12 oz. and .1 oz. x amount of Pepsi in a Randomly selected can Find P( X 11.90) Draw Picture P ( X 11.90) 11.9 12 P( Z ) .1 P ( Z 1) .5 P0 Z 1 .5 .3413 .1587