Question

1. determine if the following are probability distributions (if no, state why).

a. x 3 6 9 12 15
p(x) 4/9 2/9 1/9 1/9 1/9
b. x 1 2 3 4 5
p(x) 3/10 1/10 1/10 2/10 3/10
c. x 20 30 40 50
p(x) 1.1 0.2 0.9 0.3

Explanation:

Step1: Check probability distribution conditions

A probability distribution must satisfy two conditions:

1. \(0\leq P(X)\leq1\) for all \(X\).
2. \(\sum P(X) = 1\).

Step2: Analyze part a

• Calculate \(\sum P(X)\):

\(\frac{4}{9}+\frac{2}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{4 + 2+1+1+1}{9}=\frac{9}{9}=1\).

• Also, \(0\leq\frac{4}{9},\frac{2}{9},\frac{1}{9},\frac{1}{9},\frac{1}{9}\leq1\).

Step3: Analyze part b

• Calculate \(\sum P(X)\):

\(\frac{3}{10}+\frac{1}{10}+\frac{1}{10}+\frac{2}{10}+\frac{3}{10}=\frac{3 + 1+1+2+3}{10}=\frac{10}{10}=1\).

• And \(0\leq\frac{3}{10},\frac{1}{10},\frac{1}{10},\frac{2}{10},\frac{3}{10}\leq1\).

Step4: Analyze part c

• Check the first condition: \(P(X = 20)=1.1>1\).

Answer:

a. Yes, it is a probability distribution.
b. Yes, it is a probability distribution.
c. No, because \(P(X = 20)=1.1>1\) (violates \(0\leq P(X)\leq1\) condition).