Continuous Distributions. Math 10. Continuous Random Variable. Discrete vs Continuous. Examples of Exponential Distributiuon. Exponential distribution

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1 Continuous Distributions Math 1 Part 4 Slides Continuous Random Variables and the Central Limit Theorem Maurice Geraghty, 15 Uncountable Number of possibilities Probability of a point makes no sense Probability is measured over intervals Comparable to Relative Frequency Histogram Find Area under curve. 1 Discrete vs Continuous Continuous Random Variable Countable Discrete Points p() is probability distribution function p() Σp() =1 Uncountable Continuous Intervals f() is probability density function f() Total Area under curve =1 f() is a density function P(<) is a distribution function. P(a<<b) = area under function between a and b 3 4 Eponential distribution Waiting time Memoryless f() = (1/μ)e (1/μ) P(>a) = e (a/μ) μ=μ =μ P(>a+b >b) = e (a/μ) Eamples of Eponential Distributiuon Time until a circuit will fail the net RM 7 Earthquake the net customer calls An oil refinery accident you buy a winning lotto ticket 5 6 Maurice Geraghty, 15 1

2 Relationship between Poisson and Eponential Distributions If occurrences follow a Poisson Process with mean = μ, then the waiting time for the net occurrence has Eponential distribution with mean = 1/μ. Eample: If accidents occur at a plant at a constant rate of 3 per month, then the epected waiting time for the net accident is 1/3 month. Eponential Eample The life of a digital display of a calculator has eponential distribution with μ=5hours. (a) Find the chance the display will last at least 6 hours. P(>6) = e -6/5 = e -1. =.31 (b) Assuming it has already lasted 5 hours, find the chance the display will last an additional 6 hours. P(>11 >5) = P(>6) = Eponential Eample The life of a digital display of a calculator has eponential distribution with μ=5 hours. (a) Find the median of the distribution ib ti P(>med) = e -(med)/5 =.5 med = -ln(.5)5= p th Percentile = -ln(1-p)μ 9 Uniform Distribution Rectangular distribution Eample: Random number generator 1 f ( ) = a b b a b + a μ = E( ) = = Var( ) = ( b ) a 1 1 Uniform Distribution - Probability Uniform Distribution - Percentile Area = p a c d b d c P( c < < d) = b a a p b Formula to find the pth percentile p : p = a + p ( b a) 11 1 Maurice Geraghty, 15

3 Uniform Eample 1 Find mean, variance, P(<3) and 7 th percentile for a uniform distribution from 1 to 11. ( ) μ = = 6 = = P( < 3) = = = = 8 7 ( ) Uniform Eample A tea lover orders 1 grams of Tie Guan Yin loose leaf when his supply gets to 5 grams. The amount of tea currently in stock follows a uniform random variable. Determine this model Find the mean and variance Find the probability of at least 7 grams in stock. Find the 8 th percentile Uniform Eample 3 Normal Distribution A bus arrives at a stop every minutes. Find the probability of waiting more than 15 minutes for the bus after arriving randomly at the bus stop. If you have already waited 5 minutes, find the probability of waiting an additional 1 minutes or more. (Hint: recalculate parameters a and b) The normal curve is bell-shaped The mean, median, and mode of the distribution are equal and located at the peak. The normal distribution is symmetrical about its mean. Half the area under the curve is above the peak, and the other half is below it. The normal probability distribution is asymptotic -the curve gets closer and closer to the -ais but never actually touches it. μ 1 ( ) e f ( ) = π The Standard Normal Probability Distribution A normal distribution with a mean of and a standard deviation of 1 is called the standard normal distribution. Z value: The distance between a selected value, designated, and the population mean μ, divided by the population standard deviation, Z μ = 7-9 Areas Under the Normal Curve Empirical Rule About 68 percent of the area under the normal curve is within one standard deviation of the mean. μ ± 1 About 95 percent is within two standard deviations of the mean μ ± 99.7 percent is within three standard deviations of the mean. μ ± Maurice Geraghty, 15 3

4 7-11 EAMPLE normally distributed with a mean of gallons and a standard deviation of 5 gallons. About 68% of the daily water usage per person in New Providence lies between what two values? μ± 1 = ± 1( 5). That is, about 68% of the daily water usage will lie between 15 and 5 gallons. Normal Distribution probability problem procedure Given: Interval in terms of Convert to Z by μ Z = Look up probability in table EAMPLE EAMPLE continued normally distributed with a mean of gallons and a standard deviation of 5 gallons. What is the probability that a person from the town selected at random will use less than 18 gallons per day? The associated Z value is Z=(18-)/5=. Thus, P(<18)=P(Z<-.4)=.3446 normally distributed with a mean of gallons and a standard deviation of 5 gallons. What proportion of the people uses between 18 and 4 gallons? The Z value associated with =18 is Z=-.4 and with =4, Z=(4-)/5=.8. Thus, P(18<<4)=P(-.4<Z<.8) = = EAMPLE continued normally distributed with a mean of gallons and a standard deviation of 5 gallons. What percentage of the population uses more than 6. gallons? The Z value associated with =6., Z=(6.-)/5=1.4. Thus P(>6.)=P(Z>1.4) =1-.895=.175 Normal Distribution percentile problem procedure Given: probability or percentile desired. Look up Z value in table that corresponds to probability. Convert to by the formula: = μ + Z 3 4 Maurice Geraghty, 15 4

5 EAMPLE normally distributed with a mean of gallons and a standard deviation of 5 gallons. A special ta is going to be charged on the top 5% of water users. Find the value of daily water usage that generates the special ta The Z value associated with 95 th percentile =1.645 = + 5(1.645) = 8. gallons per day EAMPLE Professor Kurv has determined that the final averages in his statistics course is normally distributed with a mean of 77.1 and a standard deviation of 11.. He decides des to assign his grades for his current course such that the top 15% of the students receive an A. What is the lowest average a student can receive to earn an A? The top 15% would be the finding the 85 th percentile. Find k such that P(<k)=.85. The corresponding Z value is 1.4. Thus we have =77.1+(1.4)(11.), or = EAMPLE The amount of tip the servers in an eclusive restaurant receive per shift is normally distributed with a mean of $8 and a standard deviation of $1. Shelli feels she has provided poor service if her total tip for the shift is less than $65. What percentage of the time will she feel like she provided poor service? Let y be the amount of tip. The Z value associated with =65 is Z= (65-8)/1= Thus P(<65)=P(Z<-1.5)=.668. Distribution of Sample Mean Random Sample: 1,, 3,, n Each i is a Random Variable from the same population All i s are Mutually Independent is a function of Random Variables, so is itself Random Variable. In other words, the Sample Mean can change if the values of the Random Sample change. What is the Probability Distribution of? 7 8 Eample Roll 1 Die Eample Roll Dice Probabilit y Dist ribut ion of Sample Mean - 1 Die Roll Probabilit y Dist ribut ion of Sample Mean - Die Rolls Maurice Geraghty, 15 5

6 Eample Roll 1 Dice Probability Distribution of Sample Mean - 1 Die Rolls Eample Roll 3 Dice Probability Distribution of Sample Mean - 3 Die Rolls Eample - Poisson Central Limit Theorem Part 1 p() μ=1.5 μ= IF a Random Sample of any size is taken from a population with a Normal Distribution with mean= μ and standard deviation = μ THEN the distribution of the sample mean has a Normal Distribution with: μ = μ = n 34 Central Limit Theorem Part IF a random sample of sufficiently large size is taken from a population with any Distribution with mean= μ and standard deviation = THEN the distribution of the sample mean has approimately a Normal Distribution with: μ = μ = n μ 35 Central Limit Theorem 3 important results for the distribution of Mean Stays the same μ = μ Standard Deviation Gets Smaller = n If n is sufficiently large, has a Normal Distribution 36 Maurice Geraghty, 15 6

7 Eample The mean height of American men (ages -9) is μ = 69. inches. If a random sample of 6 men in this age group is selected, what is the probability the mean height for the sample is greater than 7 inches? Assume =.9. (7 69.) P( > 7) = P Z >.9 6 = P( Z >.14) =.16 Eample (cont) μ = 69. = μ = = = Eample Central Limit Theorem The waiting time until receiving a tet message follows an eponential distribution with an epected waiting time of 1.5 minutes. Find the probability that the mean waiting time for the 5 tet messages eceeds 1.6 minutes. μ = 1.5 = 1.5 n = 4 Use Normal Distribution (n>3) ( ) P( > 1.6) = P Z > = ( >.47) = P Z 39 Maurice Geraghty, 15 7