The CDC reported in 2012 that 1 in 88 American children had been diagnosed with an autism spectrum disorder (ASD).

a. If a random sample of 300 children is selected, let X be the number of these children who have been diagnosed with ASD. What distribution does X follow? What is the expected value and standard deviation of X?
b. Approximate the probability that fewer than 2 children in the sample have been diagnosed with ASD, using a Poisson distribution.

Poisson Distribution , P(x<2) = 0.1446

Step-by-step explanation:

This is a binomial distribution question with

sample space, n = 300

probability of diagnosed with ASD, p = 1/88 = 0.0114

probability of not diagnosed with ASD, q = 1 - p = 0.9886

Mean, m is given as np = 300 * 0.0114 = 3.42

variance, v = npq = 300 * 0.0114 * 0.9886 = 3.38101

standard deviation, s = square root of variance = 3.38101^(0.5) = 1.83875

This binomial distribution can be approximated as Poisson Distribution since

n > 20 and p < 0.05

For a Poisson Distribution

P(X = x) = [e^(-m) * m^(x)]/ x!

b) x = 2

P(X < 2) = P(X = 0) + P(X = 1)

using the z score table and evaluating, we obtain

P(x<2) = 0.1446

add an image

step-by-step explanation:

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.