Answer:
a) It is a valid distribution because the sum of all possible probabilities is 1. The sample space is discrete, because the number of diseases is a countable number.
b) P(X <2) = 0.924, P(1≤X≤3) = 0.212
c - i) 0.225
c - ii) 0.19
Step-by-step explanation:
(a) Explain why this is a valid probability distribution. Is the sample space discrete or continuous?
0.775 + 0.149 + 0.041 + 0.022 + 0.013 = 1
Since the sum of all possible probabilities is 1, this is a valid probability distribution.
Since the possible outcomes are 0, 1, 2, 3 or 4 diseases, that is, countable numbers, the sample space is discrete.
(b) Find P(X <2) and P(1≤X≤3).
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.775 + 0.149 = 0.924[/tex]
[tex]P(1 \geq X \geq 3) = P(X = 1) + P(X = 2) + P(X = 3) = 0.149 + 0.041 + 0.022 = 0.212[/tex]
(c) Obtain the probability that a randomly selected American adult suffers from
(i) at least one mental disorder
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.775 = 0.225[/tex]
(ii) either one or two mental disorders
[tex]P(X = 1) + P(X = 2) = 0.149 + 0.041 = 0.19[/tex]