Question

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: x 0 1 2 3 4 5 6 p(x) 0.078 0.052 0.154 0.301 0.198 0.164 0.053 (a) find the probability of randomly selecting an employee whose overtime is one or two hours. the probability is 0.206. (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 0.229. (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 0.716. (type an integer or a decimal. do not round.) (d) find the probability of randomly selecting an employee whose overtime is from one to three hours, inclusive. the probability is 0.507. (type an integer or a decimal. do not round.) (e) find the probability of randomly selecting an employee whose overtime is at most four hours. the probability is . (type an integer or a decimal. do not round.)

Explanation:

Step1: Identify relevant probabilities

We are given a probability - distribution table for the number of overtime hours worked in one week per employee. To find the probability that an employee has at most four hours of overtime, we need to sum the probabilities for \(x = 0\), \(x = 1\), \(x = 2\), \(x = 3\), and \(x = 4\).
Let \(P(0)=0.078\), \(P(1)=0.052\), \(P(2)=0.154\), \(P(3)=0.301\), \(P(4)=0.186\).

Step2: Calculate the sum

We use the formula \(P(X\leq4)=P(0)+P(1)+P(2)+P(3)+P(4)\).
\[

$$\begin{align*} P(X\leq4)&=0.078 + 0.052+0.154 + 0.301+0.186\\ &=(0.078+0.052)+0.154 + 0.301+0.186\\ &=0.13+0.154 + 0.301+0.186\\ &=(0.13 + 0.154)+0.301+0.186\\ &=0.284+0.301+0.186\\ &=(0.284 + 0.301)+0.186\\ &=0.585+0.186\\ &=0.771 \end{align*}$$

\]

Answer:

0.771