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Section 07 x has equal probability over entire interval Parameters: a – beginning of interval b – end of interval 𝑓 𝑥 𝐸 𝑋 𝑉𝑎𝑟 1 = 𝑏−𝑎 𝑎+𝑏 = 2 (𝑏−𝑎)2 𝑋 = 12 Parameters: μ – mean σ2 – variance n – sample size 1 𝜎 2𝜋 ∗𝑒 −(𝑥−𝜇)2 2𝜎2 𝑓 𝑥 = 𝐸 𝑋 =𝜇 𝑉𝑎𝑟 𝑋 = 𝜎 2 Standard normal distribution is 𝑍~(0,1) Standardizing Given a normal random variable 𝑋~𝑁(𝜇, 𝜎 2 ), find 𝑃(𝑟 < 𝑋 < 𝑠) Define 𝑍 = 𝑋−𝜇 𝜎 𝑟−𝜇 𝑥−𝜇 𝑠−𝜇 Then 𝑃 𝑟 < 𝑋 < 𝑠 = 𝑃( < < ) 𝜎 𝜎 𝜎 𝑟−𝜇 𝑠−𝜇 𝑠−𝜇 𝑟−𝜇 =𝑃 <𝑍< =φ − 𝜑( ) 𝜎 𝜎 𝜎 𝜎 Approximating another distribution A random variable X with mean μ and variance σ2 is sometimes approximated by assuming the distribution of X is approximately 𝑁(𝜇, 𝜎 2 ) Often the question will ask for an approximate probability of some interval – assume normal distribution if not specifically mentioned This is also often used for sums of random variables Integer correction for the normal approximation When estimating a discrete distribution using the normal distribution If 𝑛 and 𝑚 are integers, to find discrete probability 𝑃(𝑛 ≤ 𝑋 ≤ 𝑚) We assign normal r. v. 𝑌 the same mean and 1 1 variance as X and find 𝑃(𝑛 − ≤ 𝑌 ≤ 𝑚 + ) 2 2 We do this because endpoints matter in discrete distributions, but not continuous x is often time between events Parameters: λ 𝑓 𝑥 = λ𝑒 −λ𝑥 𝐹 𝑋 = 1 − 𝑒 −λ𝑥 1 𝐸 𝑋 = λ 𝑉𝑎𝑟 𝑋 = 1 λ2 Exponential and Poisson are often used in tandem; eg, exponential represents time until an event occurs, Poisson represents number of events occurring in a period of time. This is because they both use the variable λ. Can also be described with 𝜃 = 1 λ so that 𝑓 𝑥 = 1 −𝑥 𝑒𝜃 𝜃 Questions can be ambiguous – look for description of mean The minimum of a collection of independent exponential random variables Suppose independent r.v. 𝑌1 , 𝑌2 , … 𝑌𝑛 each have 1 1 1 exponential distributions with means , , … , λ1 λ2 λ𝑛 If 𝑌 = min{𝑌1 , 𝑌2 , … 𝑌𝑛 } then 𝑌 has an exponential distribution with mean 1 λ1 +λ2 +⋯+λ𝑛 These types of word problems are pretty common with other distributions as well The minimum of a collection of independent exponential random variables - alternative ▪ Suppose independent r.v. 𝑌1 , 𝑌2 , … 𝑌𝑛 each have 1 1 1 exponential distributions with means , , … , and λ1 λ2 λ𝑛 𝑌 = min{𝑌1 , 𝑌2 , … 𝑌𝑛 } Convert 𝑌1 , 𝑌2 , … , 𝑌𝑛 to Poisson variables with means λ1 , λ2 , … , λ𝑛 Then add the means to get the mean of 𝑌 ▪ 𝑌 now has a Poisson distribution with mean λ1 + λ2 + ⋯ + λ𝑛 ▪ This is the same as an exponential distribution with 1 mean λ1 +λ2 +⋯+λ𝑛 Parameters: α and β 𝑓 𝑥 = βα ∗𝑥 α−1 ∗𝑒 −β𝑥 Γ(α) for x>0 only Γ 𝑛 = 𝑛 − 1 ! for positive integers n 𝐸 𝑋 = α β 𝑉𝑎𝑟 𝑋 = α β2 𝑀𝑋 𝑡 = 𝛽 𝛼 ( ) 𝛽−𝑡 for t<β DEFINITELY know Uniform Normal Exponential Try to know Gamma A student received a grade of 80 on a math final where the mean grade was 72 and the standard deviation was s. In the statistics final, he received a 90, where the mean grade was 80 and the standard deviation was 15. If the standardized scores (i.e., the scores adjusted to a mean of 0 and standard deviation of 1) were the same in each case, find s. The lifetime of a printer costing 200 is exponentially distributed with mean 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, and a one-half refund if it fails during the second year. If the manufacturer sells 100 printers, how much should it expect to pay in refunds? An actuary determines that the claim size for a certain class of accidents is a random variable, X, with a moment generating function M(t) = 1/(1-2500t)^4 Calculate the standard deviation of the claim size for this class of accidents. The waiting time for the first claim from a good driver and the waiting time for the first claim from a bad driver are independent and follow exponential distributions with means 6 years and 3 years, respectively. What is the probability that the first claim from a good driver will be filed within 3 years and the first claim from a bad driver will be filed within 2 years? Write as an expression, not a value In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true age and the rounded age is assumed to be uniformly distributed on the interval from -2.5 years to 2.5 years. The healthcare data are based on a random sample of 48 people. Calculate the approximate probability that the mean of the rounded ages is within 0.25 years of the mean of the true ages The minimum force required to break a particular type of cable is normally distributed with mean 12,432 and standard deviation 25. A random sample of 400 cables of this type is selected. Calculate the probability that at least 349 of the selected cables will not break under a force of 12,400. A machine has two components and fails when both components fail. The number of years from now until the first component fails, X, and the number of years from now until the machine fails, Y, are random variables with joint density function f(x,y) = 1/18 * exp[-(x+y)/6], for 0<x<y, and 0 otherwise. Calculate Var(Y|X=2).