According to the Center for Disease Control (CDC), about 1 in 88 children in the
ID: 3054106 • Letter: A
Question
According to the Center for Disease Control (CDC), about 1 in 88 children in the U.S. have been diagnosed with autism (“CDC-data and statistics„” 2013). Suppose you consider a group of 10 children and the number of them that have autism. CDC-data and statistics, autism spectrum disorders - ncbdd. (2013, October
a) State the random variable.
b) Argue that this is a binomial experiment.
c) What is the probability that none have autism?
d) What is the probability seven have autism? e) What is the probability at least ?ve have autism? f) What is the probability at most two have autism?
g) Suppose ?ve children out of ten have autism. Is this unusual? What does that tell you
please show all work and thanks!
Explanation / Answer
a) The random variable X = the number of children that have autism
b) This a binomial experiment because the n trials that is the 10 children are independent of each other and the probability of a children have been diagonised with the autism is same for every children.
c) P(have autism) = 1/88 = 0.011
n = 10
P(X = 0) = 10C0 * (0.011)^0 * (1 - 0.011)^10 = 0.8953
d) P(X = 7) = 10C7 * (0.011)^7 * (1 - 0.011)^3 = 0.000001
e) P(X > 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
= 10C5 * (0.011)^5 * (1 - 0.011) 2)^5 + 10C6 * (0.011)^6 * (1 - 0.011)^4 + 10C7 * (0.011)^7 * (1 - 0.011)^3 + 10C8 * (0.011)^8 * (1 - 0.011)^2 + 10C9 * (0.011)^9 * (1 - 0.011)^1 + 10C10 * (0.011)^10 * (1 - 0.011)^0 = 0.000001
f) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 10C0 * (0.011)^0 * (1 - 0.011)^10 + 10C1 * (0.011)^1 * (1 - 0.011)^9 + 10C2 * (0.011)^2 * (1 - 0.011)^8
= 0.9998
g) P(X = 5) = 10C5 * (0.011)^5 * (1 - 0.011) 2)^5 = 0.000001
As the probability is less than 0.05, so it is unusual.
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