Answer:
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
