The PDF and CDF are used to describe the behavior of a population Variables are described and quantified through the mean and variance If X is a continuous random variable with a PDF of f x then the mean of x is x x f x dx Gives the center of mass of the PDF Often called expected value The variance of X is given by x2 x f x dx 2 2 x x 2 x f x dx This describes the dispersion or scatter on the possible values of X The standard deviation is given by the square root of the variance It is also sometimes useful to find the median of X Given f x How can we find the median xm P X xm 0 5 xm f x dx 0 5 f x 0 05 0 x 20 What is the mean x x f x dx xm What is the median f x dx 0 5 What is the variance 2 x x f x dx 2 What is the standard deviation 2 x x 2 x f x dx Normal Distribution By far the most commonly used distribution in statistics Good model for many continuous populations Looks like this The PDF for a Normal Distribution 1 f x e 2 Mean expected value x 2 2 2 Variance The proportion of a normal population that is within a given number of standard deviations from the mean is the same for any normal population Convert to Standard Units z score z x Z score is an item sampled from a population with a mean of zero and a standard deviation of one Standard Normal Population b Want to find P a x b f x dx a 1 Can t integrate f x 2 e x 2 2 2 1 Convert to standard units 2 Look up area under the standard normal curve on standard normal table or z table Ok let s walk through this Lifetimes of batteries for cell phones are normally distributed with a mean of 50 hours and a standard deviation of 5 hours What is the probability that the lifetime of a randomly selected battery will be less than 51 hours Find z z x 0 2 P x 51 P z 0 2 We found P 0 z 0 2 0 0793 P z 0 2 P z 0 P 0 z 0 2 0 5 0 0793 0 5793 Lifetimes of batteries for cell phones are normally distributed with a mean of 50 hours and a standard deviation of 5 hours What is the probability that the lifetime of a randomly selected battery will be less than 51 hours What is the probability that the battery will last more than 51 hours Battery life Mean 50 Standard Deviation 5 P 45 X 51 hours Need 2 z values z1 z2 51 45 Going to the table we find P z1 0 0793 P z2 0 3413 0 2 1 P 45 X 51 Battery life Mean 50 Standard Deviation 5 At what lifetime do we expect 95 of the phones to have died z x A bit of a backwards problem We need to find z for what probability 0 45 From the table z 1 65 x 50 5 x 58 3 hours 1 65 X follows a normal distribution with a mean of 6 and a standard deviation of 2 Find P 5 x 8 Variables in an experiment sometimes have relation x exp w If w itself follows a Gaussian Distribution with a mean q and variance w2 then x is a log normal random variable with a PDF ln x q 2 f x exp 2 2w xw 2 1 mean variance w2 q 2 2q w 2 e e e 2 w2 1 x exp w w itself follows a Gaussian Distribution with a mean q and variance w2 How can we find P x 1000 P x 1000 P ln x ln 1000 P w 6 907 w has a Normal Distribution and given the mean and variance we can figure this out using the z table
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