Question

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m. Based on past experience the company has determined that the employee is equally likely to arrive at any time between 8:00 a.m. and 8:30 a.m. Find the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m. Round your answer to four decimal places, if necessary.

Answer 1

Answer:

0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m.

We can consider 8 am = 0, and 8:30 am  = 30, so $a = 0, b = 30$

Find the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

Between 15 and 25, so:

$P(15 \leq X \leq 25) = \frac{25 - 15}{30 - 0} = 0.3333$

0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.