archaic
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- Homework Statement
- The weight of ##25## individuals trying to squeeze themselves into an elevator is normally distributed with a mean of ##76\,kg## and a standard deviation of ##16\,kg##. The elevator is designed to hold ##1950\,kg##.
a) Determine the mean and standard deviation of the total weight inside the elevator.
b) What's the probability that that weight exceeds the limit?
c) Knowing that the electricity costs ##0.1\,$## per ##kg## for the first ##10## persons, and ##0.2\,$## per ##kg## for any extra person, find the mean and the variance of the cost for this scenario.
- Relevant Equations
- n/a
a) Total weight ##W=W_1+W_2+...+W_{25}##.$$E[W]=E[W_1]+E[W_2]+...+E[W_{25}]=25\times76=1900\,kg$$$$\sigma_W=\sqrt{V(W_1)+V(W_2)+...+V(W_{25})}=\sqrt{25\times(16)^2}=80\,kg$$
b) Since ##W## is a linear combination of normal distributions, the reproductive property tells us that ##W## is also normally distributed.$$P(W>1950)=P(Z>0.625)=1-P(Z<0.625)=0.26599$$
c) Cost ##C=0.1W_1+...+0.1W_{10}+0.2W_{11}+...+0.2W_{25}##.$$\begin{align*}E[C]&=0.1E[W_1]+...+0.1E[W_{10}]+0.2E[W_{11}]+...+0.2E[W_{25}]\\&=10\times0.1\times76+15\times0.2\times76\\&=304\,$\end{align*}$$$$\begin{align*}
V(C)&=(0.1)^2V(W_1)+...+(0.1)^2V(W_{10})+(0.2)^2V(W_{11})+...+(0.2)^2V(W_{25})\\
&=10\times(0.1)^2\times(16)^2+15\times(0.2)^2\times(16)^2\\
&=179.2\,$^2
\end{align*}$$
Correct, right? Thanks!
b) Since ##W## is a linear combination of normal distributions, the reproductive property tells us that ##W## is also normally distributed.$$P(W>1950)=P(Z>0.625)=1-P(Z<0.625)=0.26599$$
c) Cost ##C=0.1W_1+...+0.1W_{10}+0.2W_{11}+...+0.2W_{25}##.$$\begin{align*}E[C]&=0.1E[W_1]+...+0.1E[W_{10}]+0.2E[W_{11}]+...+0.2E[W_{25}]\\&=10\times0.1\times76+15\times0.2\times76\\&=304\,$\end{align*}$$$$\begin{align*}
V(C)&=(0.1)^2V(W_1)+...+(0.1)^2V(W_{10})+(0.2)^2V(W_{11})+...+(0.2)^2V(W_{25})\\
&=10\times(0.1)^2\times(16)^2+15\times(0.2)^2\times(16)^2\\
&=179.2\,$^2
\end{align*}$$
Correct, right? Thanks!