Use the partial probability distribution below to determine each of the following. a. Determine the missing probability, P(X = 2). b. Compute the probability, P(X > 3). c. Compu... Use the partial probability distribution below to determine each of the following. a. Determine the missing probability, P(X = 2). b. Compute the probability, P(X > 3). c. Compute E(X). Read more

Steps to Solve

1. Find $P(X=2)$ The sum of all probabilities in a probability distribution must equal 1. We can use this to find the missing probability $P(X=2)$. $$ P(X=-2) + P(X=0) + P(X=2) + P(X=4) + P(X=6) = 1 $$ $$ \frac{14}{80} + \frac{11}{80} + P(X=2) + \frac{23}{80} + \frac{17}{80} = 1 $$ $$ P(X=2) = 1 - \frac{14}{80} - \frac{11}{80} - \frac{23}{80} - \frac{17}{80} $$ $$ P(X=2) = 1 - \frac{14+11+23+17}{80} = 1 - \frac{65}{80} = \frac{80}{80} - \frac{65}{80} = \frac{15}{80} $$ Therefore, $P(X=2) = \frac{15}{80}$.

2. Find $P(X > 3)$ This is the probability that $X$ is greater than 3. From the table, the values of $X$ that are greater than 3 are 4 and 6. Thus we need to find the sum of the corresponding probabilities: $$ P(X > 3) = P(X=4) + P(X=6) $$ $$ P(X > 3) = \frac{23}{80} + \frac{17}{80} = \frac{23+17}{80} = \frac{40}{80} $$ Therefore, $P(X > 3) = \frac{40}{80}$.

3. Find $E(X)$ The expected value $E(X)$ is calculated as the sum of each value of $X$ multiplied by its corresponding probability: $$ E(X) = \sum X \cdot P(X) $$ $$ E(X) = (-2) \cdot \frac{14}{80} + (0) \cdot \frac{11}{80} + (2) \cdot \frac{15}{80} + (4) \cdot \frac{23}{80} + (6) \cdot \frac{17}{80} $$ $$ E(X) = \frac{-28}{80} + \frac{0}{80} + \frac{30}{80} + \frac{92}{80} + \frac{102}{80} $$ $$ E(X) = \frac{-28 + 0 + 30 + 92 + 102}{80} = \frac{196}{80} = \frac{49}{20} = 2.45 $$ Therefore, $E(X) = \frac{196}{80} = 2.45$.