Question

Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let x = the time needed to fix a furnace. Then x ~ U (1.5, 4).
a) Find the 30th percentile of furnace repair times.

Answer 1

Let $X$ be the time needed to fix a furnace and $X \sim U(1.5,4)$ with probability density ($b=4$ and $a=1.5$ )

$${p_X}(x) = \begin{cases} \frac{1}{{b - a}}, & a < x < b \\ 0, & \text{otherwise}\end{cases}$$

Then let's find the distribution function as ${F_X}(x) = \int\limits_{ - \infty }^x {{p_X}(\xi )d\xi }$

At the interval $( - \infty ,a)$ we have ${F_X}(x) = \int\limits_{ - \infty }^x {0 \cdot d\xi } = 0$

At the interval $(a,b)$ we have ${F_X}(x) = \int\limits_a^x {\frac{1}{{b - a}} \cdot d\xi } = \left. {\frac{\xi }{{b - a}}} \right|_a^x = \frac{{x - a}}{{b - a}}$

At the interval $(b, + \infty )$ we have ${F_X}(x) = {\left. {\frac{{x - a}}{{b - a}}} \right|_{x = b}} + \int\limits_x^{ + \infty } {0 \cdot d\xi } = 1$

Thus, it has the form

$${F_X}(x) = \begin{cases} 0, & x<a \\ {\frac{{x - a}}{{b - a}}}, & a<x<b \\1 & x>b \end{cases}$$

(We can use strict or non strict inequalities, it does not matter because $X$ is continuous random variable).

We were asked to find such value ${q_{0.3}}$ that $F({q_{0.3}}) = 0.3$. We have

$$\frac{{{q_{0.3}} - 1.5}}{{4 - 1.5}} = 0.3$$

Thus

$${q_{0.3}} - 1.5 = 0.75$$

and the final answer is

$${q_{0.3}} = 2.25$$