Question

The gross weekly sales at a certain restaurant is a normal random variable with mean $2200 and standard deviation $230. What is the probability that (a) the total gross sales over the next 2 weeks exceeds $5000. (b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?

Binomial probability
ANSWERED
asked 2021-08-22
The gross weekly sales at a certain restaurant is a normal random variable with mean $2200 and standard deviation $230.
What is the probability that
(a) the total gross sales over the next 2 weeks exceeds $5000.
(b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?

Expert Answers (1)

2021-08-23

(a)
Define random variables X and Y that marks the total gross sales over the first and the second week, respectively.
We are given that \(\displaystyle{X},{Y}\sim{N}{\left({2200},{230}^{{2}}\right)}\) and when we can assume that they are independent (knowing total sales in the first week does not affect total sales in the second week).
Know, we are considering random variable Z=X+Y. From the theory, we know that
\(\displaystyle{Z}={X}+{Y}\sim{N}{\left({2}\cdot{2200},{2}\cdot{230}^{{2}}\right)}\)
i.e., the sum of two independent Normals is again Normal.
Now, we are required to find:
\(\displaystyle{P}{\left({Z}{>}{5000}\right)}={1}-{P}{\left({Z}\le{5000}\right)}={1}-{P}{\left({\frac{{{Z}-{4400}}}{{\sqrt{{2}}\cdot{230}}}}\le{\frac{{{5000}-{4400}}}{{\sqrt{{2}}\cdot{230}}}}\right)}={1}-\Phi{\left({1.8467}\right)}\approx{1}-{0.9675}={0.0325}\)
(b)
Define random variable N as the number of weeks in the next three weeks where the total gross sales exceed 2000.
We have that N~ Binom(3,p), where
\(\displaystyle{p}={P}{\left({X}>{2000}\right)}={1}-{P}{\left(\le{2000}\right)}={1}-{P}{\left({\frac{{{X}-{2200}}}{{{230}}}}\le{\frac{{-{200}}}{{{230}}}}\right)}={1}-\Phi{\left(-{0.86957}\right)}={1}-{0.1923}=\frac{0}{{8077}}\)
Finally
\(\displaystyle{P}{\left({N}\ge{2}\right)}={P}{\left({N}={2}\right)}+{P}{\left({N}={3}\right)}={\left(\begin{matrix}{3}\\{2}\end{matrix}\right)}\cdot{0.8077}^{2}\cdot{\left({1}-{0.8077}\right)}+{\left(\begin{matrix}{3}\\{2}\end{matrix}\right)}\cdot{0.8077}^{3}={0.9033}\)
Result:
(a)0.0325
(b)0.9033

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