Answer to Question #97672 in Statistics and Probability for Juliet Beglaryan

Question #97672
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.
1
Expert's answer
2019-10-31T06:49:33-0400

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"



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS