Question: Use the probability distribution to complete parts (a) through (e). The probability distribution…

Use the probability distribution to complete parts (a) through (e).

The probability distribution of the number of overtime hours worked in one week per employee

\[ \begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & 0.035 & 0.075 & 0.12 & 0.29 & 0.215 & 0.165 & 0.1 \end{array} \]

(Type an integer or a decimal. Do not round.)

(b) Find the probability of randomly selecting an employee whose overtime is two hours or less.

The probability is 0.23.

(Type an integer or a decimal. Do not round.)

(c) Find the probability of randomly selecting an employee whose overtime is from three to six hours, inclusive.

The probability is \boxed{}.

(Type an integer or a decimal. Do not round.)

Solution

To find the probabilities for parts (b) and (c), we’ll use the provided probability distribution. The distribution is: \[ \begin{array}{c|ccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & 0.035 & 0.075 & 0.12 & 0.29 & 0.215 & 0.165 & 0.1 \\ \end{array} \] (b) Find the probability of randomly selecting an employee whose overtime is two hours or less. Calculate the sum of probabilities for \(x = 0\), \(x = 1\), and \(x = 2\). \[ P(\text{overtime} \leq 2) = P(0) + P(1) + P(2) \] \[ P(\text{overtime} \leq 2) = 0.035 + 0.075 + 0.12 = 0.23 \] (c) Find the probability of randomly selecting an employee whose overtime is from three to six hours, inclusive. Calculate the sum of probabilities for \(x = 3\), \(x = 4\), \(x = 5\), and \(x = 6\). \[ P(3 \leq \text{overtime} \leq 6) = P(3) + P(4) + P(5) + P(6) \] \[ P(3 \leq \text{overtime} \leq 6) = 0.29 + 0.215 + 0.165 + 0.1 \] \[ P(3 \leq \text{overtime} \leq 6) = 0.77 \]