Q. 6.29 - A First Course in Probability

Question

The gross weekly sales at a certain restaurant are a normal random variable with mean$\backslash (2200$ and standard deviation $\backslash )230$ . What is the probability that

(a) the total gross sales over the next $2$ weeks exceeds $\backslash (5000$;

(b) weekly sales exceed $\backslash )2000$ in at least $2$ of the next $3$ weeks? What independence assumptions have you made?

Step-by-Step Solution

Answer

(a) Probability that the total gross sales over the next two weeks exceed $\$5000 is 0.03252$  

(b) Probability that daily sales exceed $\$2000$in at least $2$of the $3$ days is $0.90342$

Step 1: Given information (part a)

Mean of the gross sales in a day$=\$2200$

The standard deviation of the gross sales in a day $=\$230$

Formula used:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ \text{where μ is the mean} \\ \sigma \text{is the standard deviation} \end{gathered}$$

Step 2: Explanation (part a)

If the two events and identically distributed then

$X+Y~N\left({\mu }_x+{\mu }_y,{{\sigma }^2}_x+{{\sigma }^2}_y\right)$

So, Let X be the gross daily sales in a day and Y be the gross sales in two days

So, $X~N\left(2200,{230}^2\right)$

Now, using the above concept the gross sales for two days will follow

$$\begin{gathered} Y=X_1+X_2~N\left(2200+2200,{230}^2+{230}^2\right) \\ Y~N\left(4400,2\times {230}^2\right) \end{gathered}$$

the required probability can be calculated as follows

$$\begin{gathered} P\left(Y>5000\right) \\ P\left(\frac{y-\mu }{\sigma }>\frac{5000-4400}{\sqrt{2\times {230}^2}}\right) \\ P\left(z>1.845\right) \\ 1-P\left(z<1.845\right) \end{gathered}$$

form  z tables

$$\begin{gathered} P\left(z<1.84\right)=0.96712 \\ P\left(z<1.85\right)=0.96784 \end{gathered}$$

using linear interpolation

$$\begin{gathered} P\left(z<1.845\right)=0.96712+\left(0.96784-0.96712\right)\times 0.5 \\ =0.96748 \end{gathered}$$

hence the required probability is  $$\begin{gathered} P\left(Y>5000\right)=1-0.96748 \\ =0.03252 \end{gathered}$$

Step 1: Given information (part b)

The pdf of the binomial distribution is

$P(X=k)={}^n{C}_k\times p^k\times {(1-p)}^{n-k}$

Step 2: Explanation (part b)

Independence assumption

If the probability of occurrence of events is the same and are Independent, then it can be modeled by the binomial distribution

Probability that the daily scale exceeds $\$2000$

$$\begin{gathered} P\left(X>2000\right)=P\left(\frac{x-\mu }{\sigma }>\frac{2000-2200}{230}\right) \\ P\left(z>-0.87\right) \\ P\left(z<0.87\right) \end{gathered}$$

From z tables

$P\left(z<0.087\right)=0.80785$

using the above independence property

number of trials (n) $=3$

probability of success (p) $=0.80785$

let the daily sales exceeding $\$2000$ in at least $2$ of the $3$ days be denoted as Z

So, $Z~Bin\left(3,0.80785\right)$

The required probability will be

$$\begin{gathered} P\left(Z\ge 2\right)=P\left(Z=2\right)+P\left(Z=3\right) \\ P\left(Z\ge 2\right)={}^3{C}_2\times {(0.080785)}^2\times {\left(1-0.80785\right)}^{3-2}{+}^3{C}_3\times {\left(0.80785\right)}^3 \\ P\left(Z\ge 2\right)=0.90342 \end{gathered}$$