Revision as of 13:22, 14 May 2023 by Admin (Created page with "'''Key: B''' The number of repairs for each boat type has a binomial distribution. For power boats: <math display = "block"> \begin{aligned} &\operatorname{E}[ S ) = 100(0.3...")
Exercise
ABy Admin
May 14'23
Answer
Key: B
The number of repairs for each boat type has a binomial distribution. For power boats:
[[math]]
\begin{aligned}
&\operatorname{E}[ S ) = 100(0.3)(300) = 9, 000, \\
&\operatorname{E}[ S ) = 100(0.3)10, 000) + 100(0.3)(0.7)(300 2 ) = 2,190, 000
\end{aligned}
[[/math]]
For sail boats:
[[math]]
\begin{aligned}
&\operatorname{E}[ S ) = 300(0.1)(1, 000) = 30, 000, \\
&\operatorname{E}[ S ) = 300(0.1)(400, 000) + 300(0.1)(0.9)(1, 0002 ) = 39, 000, 000
\end{aligned}
[[/math]]
For luxury yachts:
[[math]]
\begin{aligned}
&\operatorname{E}[ S ) = 50(0.6)(5, 000) = 150, 000, \\
&\operatorname{E}[ S ) = 50(0.6)(0.4)(2, 000, 000) + 50(0.6)(0.4)(5, 0002 ) = 360, 000, 000
\end{aligned}
[[/math]]
The sums are 189,000 expected and a variance of 401,190,000 for a standard deviation of 20,030. The mean plus standard deviation is 209,030.