The ?Centers for Disease Control and Prevention reported in 2012 that 1 in 88 American children had been diagnosed with an autism spectrum disorder (ASD). a. If a random sample of 200 American children is selected, what are the expected value and standard deviation of the number who have been diagnosed with ASD? b. Referring back to (a), calculate the approximate probability that at least 2 children in the sample have been diagnosed with ASD? c. If the sample size is 352, what is the approximate probability that fewer than 5 of the selected children have been diagnosed with ASD?

Problem 84E Answer: Step1: We have The Centers for Disease Control and Prevention reported in 2012 that 1 in 88 American children had been diagnosed with an autism spectrum disorder (ASD). 1 That is p = 88 = 0.0114 q = 1 - p = 1 - 0.0114 = 0.9886 We need to find, a. If a random sample of 200 American children is selected, what are the expected value and standard deviation of the number who have been diagnosed with ASD b. Referring back to (a), calculate the approximate probability that at least 2 children in the sample have been diagnosed with ASD c. If the sample size is 352, what is the approximate probability that fewer than 5 of the selected children have been diagnosed with ASD Step2: a). Let X be the number of children, in a random sample of n children that are diagnosed with ASD. We have n = 200 and p = 0.0114. Hence X follows poisson distribution with parameter “np” The probability mass function of poisson distribution is given by P(X) = e x , x = 0,1,2,3,...,n. x! Where, = np (mean of the poisson distribution) 1).Expected value X is given by E(X) = np = 200×0.0114 = 2.28 Therefore,expected value of the number who have been diagnosed with ASD is 2.28. 2).Standard deviation of X is given by x= nq = 200 × 0.0114 × 0.9886 = 2.254 = 1.5013. Step3: b). The approximate probability that at least 2 children in the sample have been diagnosed with ASD is given by P(at least 2 children in the sample have been diagnosed with ASD) = P(X 2) Consider, P(X 2) = 1 - P(X1) e2.(2.28) e2.2(2.28) = 1 - { + } 0! 1! (0.1022)1 (0.1022)(2.28) = 1 - { 1 + 1 } = 1 - {0.1022+0.2330 } = 1 - 0.3352 = 0.6647. Therefore, The approximate probability that at least 2 children in the sample have been diagnosed with ASD is 0.6647. Step4: c). We have n = 352 and p = 1/88 Hence, =np = 352× 1 88 = 4 Now, The approximate probability that fewer than 5 of the selected children have been diagnosed with ASD is given by P(fewer than 5 of the selected children have been diagnosed with ASD) = P(X < 5) Consider, 4 0 4 1 4 2 4 3 4 4 P(X < 5) = { e (4) + e (4) + e (4) + e (4) + e (4) } 0! 0 1!1 2! 3 3! 4 4! 4 (4) (4) (4) (4) (4) = e { 0! + 1! + 2! + 3! + 4! } 1 4 16 64 256 = (0.0184){ 1 + 1 + 2 + 6 + 24 } = (0.0184){1+4+8+10.67+10.57} = (0.0184){34.34} = 0.6319. Therefore, The approximate probability that fewer than 5 of the selected children have been diagnosed with ASD is 0.6319.