Imum Coeli
Junior Member
- Joined
- Dec 3, 2012
- Messages
- 86
Hi I have a question and I'm afraid that i missed something.
Q: A retail grocery merchant figures that her daily gain from sales $\(\displaystyle X \) is a normally distributed random variable with \(\displaystyle \mu =50 \) and \(\displaystyle \sigma = 3 \). \(\displaystyle X \) can be negative if she has to dispose of enough perishable goods. Also, she figures daily overhead costs are $\(\displaystyle Y \), which have an Erlang distribution of order \(\displaystyle n=4 \) and common rate \(\displaystyle \lambda = 0.5\). if \(\displaystyle X\) and \(\displaystyle Y\) are independent, find the expected value and variance of her daily net gain.
A:
Let \(\displaystyle P \) be her daily net gain then
\(\displaystyle P = X - Y\)
So the expected valued of \(\displaystyle P\) is
\(\displaystyle E[P] = E[X]-E[Y] \)
and since \(\displaystyle X \sim \text{N}(50,3) \implies E[X] = 50 \) and \(\displaystyle Y \sim \text{Erl}(4,0.5) \implies E[Y]= \frac{4}{0.5} = 8 \) then
\(\displaystyle E[P]= 42 \) (coincidentally her expected profit is the meaning of life).
Now the variance of \(\displaystyle P \) is
\(\displaystyle \text{var}(P) = var(X) + var(Y) \) since independent.
and \(\displaystyle X \sim \text{N}(50,3) \implies \text{var}(X)= 3^2=9 \) and \(\displaystyle Y \sim \text{Erl}(4,0.5) \implies \text{var}(Y)= \frac{4}{(0.5)^2} = 16 \) then
\(\displaystyle \text{var}(P) = 25 \).
(Sorry if this seems like a stupid question but we have just had a baby two days ago and I need to talk about this problem. Hopefully I don't look too much like an idiot
)
Q: A retail grocery merchant figures that her daily gain from sales $\(\displaystyle X \) is a normally distributed random variable with \(\displaystyle \mu =50 \) and \(\displaystyle \sigma = 3 \). \(\displaystyle X \) can be negative if she has to dispose of enough perishable goods. Also, she figures daily overhead costs are $\(\displaystyle Y \), which have an Erlang distribution of order \(\displaystyle n=4 \) and common rate \(\displaystyle \lambda = 0.5\). if \(\displaystyle X\) and \(\displaystyle Y\) are independent, find the expected value and variance of her daily net gain.
A:
Let \(\displaystyle P \) be her daily net gain then
\(\displaystyle P = X - Y\)
So the expected valued of \(\displaystyle P\) is
\(\displaystyle E[P] = E[X]-E[Y] \)
and since \(\displaystyle X \sim \text{N}(50,3) \implies E[X] = 50 \) and \(\displaystyle Y \sim \text{Erl}(4,0.5) \implies E[Y]= \frac{4}{0.5} = 8 \) then
\(\displaystyle E[P]= 42 \) (coincidentally her expected profit is the meaning of life).
Now the variance of \(\displaystyle P \) is
\(\displaystyle \text{var}(P) = var(X) + var(Y) \) since independent.
and \(\displaystyle X \sim \text{N}(50,3) \implies \text{var}(X)= 3^2=9 \) and \(\displaystyle Y \sim \text{Erl}(4,0.5) \implies \text{var}(Y)= \frac{4}{(0.5)^2} = 16 \) then
\(\displaystyle \text{var}(P) = 25 \).
(Sorry if this seems like a stupid question but we have just had a baby two days ago and I need to talk about this problem. Hopefully I don't look too much like an idiot
)
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