Applied Statistics and Probability for Engineers, 6th Edition

Published by Wiley
ISBN 10: 1118539710
ISBN 13: 978-1-11853-971-2

Chapter 3 - Section 3-3 - Cumulative Distribution Functions - Exercises - Page 73: 3-54

Answer

$F(x)=\begin{aligned} &\left\{\begin{array}{ll}{0} & {x<1} \\ {0.038} & {1 \leq x<2} \\ {0.14} & {2 \leq x<3} \\ {0.312} & {3 \leq x<4} \\ {0.516} & {4 \leq x<5} \\ {0.69} & {5 \leq x<6} \\ {0.814} & {6 \leq x<7} \\ {0.894} & {7 \leq x<8} \\ {0.93} & {8 \leq x<9} \\ {0.958} & {9 \leq x<10} \\ {0.98} & {10\leq x<15} \\ {1} & {x \geq 15}\end{array}\right.\end{aligned}$

Work Step by Step

Given, $f(x)=\left\{\begin{array}{ll}{0.038} & {x=1} \\ {0.102} & {x=2} \\ {0.172} & {x=3} \\ {0.204} & {x=4} \\ {0.174} & {x=5} \\ {0.124} & {x=6} \\ {0.080} & {x=7} \\ {0.036} & {x=8} \\ {0.028} & {x=9} \\ {0.022} & {x=10} \\ {0.020} & {x=15} \end{array}\right.$ then, $F(x)=\begin{aligned} &\left\{\begin{array}{ll}{0} & {x<1} \\ {f(1)=0.038} & {1 \leq x<2} \\ {0.038+f(2)=0.14} & {2 \leq x<3} \\ {0.14+f(3)=0.312} & {3 \leq x<4} \\ {0.312+f(4)=0.516} & {4 \leq x<5} \\ {0.512+f(4)=0.69} & {5 \leq x<6} \\ {0.69+f(5)=0.814} & {6 \leq x<7} \\ {0.814+f(5)=0.894} & {7 \leq x<8} \\ {0.894+f(8)=0.93} & {8 \leq x<9} \\ {0.93+f(9)=0.958} & {9 \leq x<10} \\ {0.958+f(10)=0.98} & {10\leq x<15} \\ {1} & {x \geq 15}\end{array}\right.\end{aligned}$
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