Questions and Answers #8 Random Variables

Questions and Answers Sheet 8 Random Variables Question #1 A population is normally distributed with 𝜇 = 100 and 𝜎 = 25 . Find the probability that a value randomly selected from this population will have a value greater than 135. Answer: From the provided information, Population mean (𝜇) = 100 Population standard deviation (𝜎) = 25 𝑋 ∼ 𝑁(100,25) The required probability that a value randomly selected from this population will have a value greater than 135 can be obtained as: 𝑃(𝑋 > 135) = 𝑃 ( 𝑥−𝜇 𝜎 > 135−100 25 ) = 𝑃(𝑍 > 1.4) = 1 − 𝑃(𝑍 < 1.4) =1-0.9192 (Using standard normal table) =0.0808 Thus, the required probability is 0.0808. Question #2 A population is normally distributed with 𝜇 = 100 and 𝜎 = 25 . Find the probability that a value randomly selected from this population will have a value less than 95. Answer: From the provided information, Population mean (𝜇) = 100 Population standard deviation (𝜎) = 25 𝑋 ∼ 𝑁(100,25) The required probability that a value randomly selected from this population will have a value less than 95 can be obtained as: 𝑃(𝑋 < 95) = 𝑃 ( 𝑥−𝜇 𝜎 < 95−100 25 ) = 𝑃(𝑍 < −0.2) =0.4207 (Using standard normal table) Thus, the required probability is 0.4207. Question #3 A population is normally distributed with 𝜇 = 100 and 𝜎 = 25 . Find the probability that a value randomly selected from this population will have a value between 95 and 135. Answer: From the provided information, Population mean (𝜇) = 100 Population standard deviation (𝜎) = 25 𝑋 ∼ 𝑁(100,25) The required probability that a value randomly selected from this population will have a value between 95 and 135 can be obtained as: 𝑃(95 < 𝑋 < 135) = 𝑃 ( 95−100 25 < 𝑥−𝜇 𝜎 < 135−100 25 ) = 𝑃(−0.2 < 𝑍 < 1.4) = 𝑃(𝑍 < 1.4) − 𝑃(𝑍 < −0.2) =0.9192-0.4207 (Using standard normal table) =0.4985 Thus, the required probability is 0.4985. Question #4 The number of cases of tetanus reported in the US in a single month has a Poisson distribution with a parameter of 𝜆 = 4.0. What is the probability that exactly one case of tetanus will be reported? Answer: Explanation: 𝑃(𝑋 = 1), 𝜆 = 4 𝑃(𝑋 = 1) = 𝑒 −𝜆𝜆𝑘 𝑘! = 𝑒 −4 (4)1 1! = (0.018316)×(4) 1 = 0.073263 Use excel function POISSON.DIST (X, mean, cumulative) to calculate 𝑃𝑋 (1): POISSON.DIST (1,4, FALSE) = 0.0732625555549 P(X=1)=0.0733 Question #5 Consider a Poisson distribution with 𝜇 = 3 . What is 𝑃(𝑥 ≥ 2) ? Answer: Explanation: The Poisson distribution is 𝑃(𝑥) = 𝑒 −𝜇 𝜇𝑥 𝑥! 𝜇= 3 3! = 3 ∗ 2 ∗ 1 𝑃(𝑥 ≥ 2) = 1 − 𝑃(1) − 𝑃(0) 𝑃(1) = 𝑒 −3 ∗31 𝑃(0) = 𝑒 1! −3 = = 1 3 𝑒3 𝑒3 Therefore, 𝑃(𝑥 ≥ 2) = 1 − 𝑃(1) − 𝑃(0) = 1 − 0.0498 − 0.1494 = 0.8009 Explanation: In a Poisson probability distribution, if mean value of success is 𝜇, the probability of getting x successes is given by 𝑃(𝑥) = 𝑒 −𝜇 𝜇𝑥 𝑥! Now 𝑃(𝑥 ≥ 2) means 1 − 𝑃(𝑥 = 0) − 𝑃(𝑥 = 1) Here 𝜇 = 3 and 𝑒 −𝜇 = 𝑒 −3 = 0.049787 and hence, desired probability is 1− 𝑒 −3 30 0! − 𝑒 −3 31 1! = 1 − 𝑒 −3 × (1 + 3) = 1 − 0.049787 × 4 =1-0.199148 =0.800852 Question #6 Suppose X has a Poisson distribution with a mean of .4. How would i find the probability when X=0, X=4, 𝑋 ≤ 2 , and when X=8? Answer: 𝑃(𝑋 = 0), 𝜆 = 0.4 𝑃(𝑋 = 0) = 𝑒 −𝜆𝜆𝑘 𝑘! = 𝑒 −0.4 (0.4)0 0! = (0.670320)×(1) 1 = 0.670320 Use excel function POISSON.DIST (X, mean, cumulative) to calculate 𝑃𝑋 (0): POISSON.DIST (0, 0.4, FALSE) =0.670 320 046 036 P(X=0)=0.6703 𝑃(𝑋 = 4), 𝜆 = 0.4 𝑃(𝑋 = 4) = 𝑒 −𝜆𝜆𝑘 𝑘! = 𝑒 −0.4 (0.4)4 4! = (0.670320)×(0.0256) 24 = 0.000715 Use excel function POISSON.DIST (X, mean, cumulative) to calculate 𝑃𝑋 (4): POISSON.DIST (4, 0.4, FALSE) =0.000 715 008 049 105 P(X=4)=0.0007 𝑃(𝑋 ≤ 2), 𝜆 = 0.4 𝑃(𝑋 ≤ 2) = 𝑃(0 ≤ 𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) 𝑃(𝑋 = 0) = 𝑃(𝑋 = 1) = 𝑃(𝑋 = 2) = 𝑒 −𝜆𝜆0 0! 𝑒 −𝜆𝜆1 1! 𝑒 −𝜆𝜆2 2! = = = 𝑒 −0.4 (0.4)0 0! 𝑒 −0.4 (0.4)1 1! 𝑒 −0.4 (0.4)2 2! = 0.670320 = 0.268128 = 0.053626 Answer = 0.670320 + 0.268128 + 0.053626 = 0.992074 Use excel function POISSON.DIST (X, mean, cumulative) to calculate 𝑃(𝑋 ≤ 2) = 𝐹𝑋 (2): POISSON.DIST (2, 0.4, TRUE)=0.992073668133 𝑃(𝑋 ≤ 2) = 0.9921 𝑃(𝑋 = 8), 𝜆 = 0.4 𝑃(𝑋 = 8) = 𝑒 −𝜆𝜆𝑘 𝑘! = 𝑒 −0.4 (0.4)8 8! = (0.670320)×(0.000655) 40320 =0 Use excel function POISSON.DIST (X, mean, cumulative) to calculate 𝑃𝑋 (8): POISSON.DIST (8, 0.4, FALSE) = 1.08953607483 × 10−8 𝑃(𝑋 = 8) = 0.0000 Question #7 The number of cases of tetanus reported in the US in a single month has a Poisson distribution with a parameter of 𝜆 = 4.0 . What is the probability that five or more cases will be reported? Answer: Let X be the number of cases in 1 month X follows Poisson distribution with parameter (rate) 𝜆 = 4 X\sim Poisson (𝜆 = 4) 𝑃(𝑋 ≥ 5), 𝜆 = 4 𝑃(𝑋 ≥ 5) = 1 − 𝑃(0 ≤ 𝑋 ≤ 4) where 𝑃(0 ≤ 𝑋 ≤ 4) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) +𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) + 𝑃(𝑋 = 4) 𝑃(𝑋 = 0) = 𝑃(𝑋 = 1) = 𝑃(𝑋 = 2) = 𝑃(𝑋 = 3) = 𝑃(𝑋 = 4) = 𝑒 −𝜆𝜆0 0! 𝑒 −𝜆𝜆1 1! 𝑒 −𝜆𝜆2 2! 𝑒 −𝜆𝜆3 3! 𝑒 −𝜆𝜆4 4! = = = = = 𝑒 −4 (4)0 0! 𝑒 −4 (4)1 1! 𝑒 −4 (4)2 2! 𝑒 −4 (4)3 3! 𝑒 −4 (4)4 4! = 0.018316 = 0.073263 = 0.146525 = 0.195367 = 0.195367 Answer=1 − (0.018316 + 0.073263 + 0.146525 + 0.195367 + 0.195367) = 0.371163 Use excel function POISSON.DIST (X, mean, cumulative) to calculate 𝑃(𝑋 ≥ 5) = 1 − 𝑃(𝑋 ≤ 4) = 1 − 𝐹𝑋 (4): 1-POISSON.DIST (4,4 TRUE)=0.37116306482 𝑃(𝑋 ≥ 5) = 0.3712 Question #8 3 What are the mean and standard deviation of a binomial probability distribution with n=15 and 𝑝 = ? 7 Answer: Mean =6.43 SD =1.92 Explanation: Given n=15 3 𝑝= 7 3 4 7 7 Calculate 𝑞 = 1 − 𝑝 = 1 − = 3 Mean = 𝑛𝑝 = 15 × = 6.43 7 3 4 𝑆𝐷 = √𝑛𝑝𝑞 = √15 × × = √36.7 = 1.92 7 7 Question #9 What are the mean and standard deviation of a binomial probability distribution with n=114 and 𝑝 = 3 49 Answer: Mean = 6 48 49 SD=2.56 Explanation: Given n=114 3 𝑝= 49 𝑞 =1−𝑝 =1− 3 49 46 = 49 3 Mean 𝑥 = 𝑛𝑝 = 114 × Variance 𝜎 2 = 𝑛𝑝𝑞 = 114 × 3 49 49 × 46 49 =6 48 49 = 6.55 Standard Deviation 𝜎 = √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = √6.55 = 2.56 Question #10 What are the mean and standard deviation of a binomial probability distribution with n=13 and 𝑝 = 5 24 Answer: Explanation: Binomial distribution is a frequency distribution of possible number of successful outcomes in given number of trials n, in each of which there is the probability of success is p (and q=1-p is probability of failure) The mean and standard deviation of a binomial distribution are given by np and √𝑛𝑝𝑞. Here 𝑛 = 13, 𝑝 = 5 24 Hence mean is 13 × and 𝑞 = 1 − 5 24 = 2.708 Standard deviation is √13 × = 1 24 5 24 5 24 × × √1235 = √4.8696 = 1.464 19 24 = 19 24 . ? ?