Answer to Question #74304 in Statistics and Probability for sam leung

Question

A plumber loads his truck each morning with faucets that will be needed for the service calls and other emergency calls that come in that day. Based on past experience, the number of faucets required each day (N) has the following distribution: p(0) =0.05; p(1) = 0.25;
p(2) = 0.5; p(3) = 0.15; p(4) = 0.05.

The plumber loads 5 faucets in his truck each day. Over all days, what is the
average number of faucets he can expect to have left at the end of the day?

Answer 1

p(0)+p(1)+p(2)+p(3)+p(4)=0.05+0.25+0.5+ 0.15+0.05=1
than p(i)=0,where i>4,
than, number of faucets he can expect to have left at the end of the day is:
(5-0)∙p(0)+(5-1)∙p(1)+(5-2)∙p(2)+(5-3)*p(3)+(5-4)∙p(4)=
=5 ∙1-( 0 ∙0.05+1∙0.25+2∙0.5+3∙0.15+4 ∙0.05)=5- 1.9=3.1
Answer:
Number of faucets he can expect to have left at the end of the day is: 3.1