Example 5.5 from Statistics, 5.2 The Uniform Distribution

(part 2)

b. Find the probability that a different nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the doughnut for more than 1.5 minutes.

The second question has a conditional probability. You are asked to find the probability that a nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example 5.3). You must reduce the sample space. First way: Since you know the child has already been eating the doughnut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes.

Write a new f(x):

f (x) = \frac{1}{4 - 1.5} = \frac{2}{5} for 1 .5 \leq x \leq 4.

Find P(x > 2|x > 1.5). Draw a graph.

f(X)=2/5 graph displaying a boxed region consisting of a horizontal line extending to the right from point 2/5 on the y-axis, a vertical upward line from points 1.5 and 4 on the x-axis, and the x-axis. A shaded region from points 2-4 occurs within this area.

Figure 5.17

P (x > 2 | x > 1.5) = (base)(new height) = (4 – 2) (\frac{2}{5}) = \frac{4}{5}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
b. $\frac{4}{5}$
Discussion
The probability that a nine-year-old child eats a donut in more than two minutes given that the child has already been eating the doughnut for more than 1.5 minutes is $\frac{4}{5}$ .
Second way: Draw the original graph for X ~ U(0.5, 4). Use the conditional formula
$P (x > 2 | x > 1.5) = \frac{P (x > 2 A N D x > 1.5)}{P (x > 1.5)} = \frac{P (x > 2)}{P (x > 1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0.8 = \frac{4}{5}$

0 of 2 steps open

How did it go?

Solution

Discussion

Solution

Discussion