Q. 6.32 - A First Course in Probability

Question

Monthly sales are independent normal random variables with mean $100$ and standard deviation $5$.

(a) Find the probability that exactly $3$ of the next $6$ months have sales greater than $100$.

(b) Find the probability that the total of the sales in the next $4$ months is greater than $420$. 

Step-by-Step Solution

Answer

The probability that exactly 3 of the next 6 months have sales greater than 100 is $0.3125$

The probability that the total of the sales in the next 4 months is greater than 420 is $0.0228$

Step 1: Information (part a)

Mean of monthly sales =100

Standard deviation of monthly sales =5

Monthly sales are independent and follow the normal distribution

The pdf of the binomial distribution is$P(X=k)={}^{n}C_k\times p^k\times (1-p{)}^{n-k}$

Where n is the number of trials and p is the probability of success.

Step 2: Explanation (part a)

Let M be the monthly sale

So, $M~N\left(100,5^2\right)$

Probability of monthly sales greater than $100$

$$\begin{gathered} P(M>100)=P\left(\frac{m-{\mu }_m}{{\sigma }_m}>\frac{100-100}{5}\right) \\ =P(z>0) \\ =1-P(z\le 0) \end{gathered}$$

From z table

$$\begin{gathered} P(z<0)=0.5 \\ P(z<0)=1-0.5 \\ =0.5 \end{gathered}$$

As each month sale is independent has the same probability of being greater than 100 , so the number of sales greater than 100 (let it be denoted as X) can be modeled by a binomial distribution with parameter,

$n=6\text{ and }p=0.5$

so, $X~Bin(6,0.5)$

$$\begin{gathered} P(X=3)={}^{6}C_3\times 0.5^3\times (1-0.5{)}^{6-3} \\ P(X=3)=0.3125 \end{gathered}$$

Step 1: Given information (part b)

if the events are independent and follow normal distribution with the same parameters then 

$\sum _{i=1}^n{X}_i~N\left(n\times \mu ,n\times {\sigma }^2\right)$

Step 2: Explanation (part b)

Let Y denote the total sales in 4 months

so, the distribution of Y will be

$$\begin{gathered} Y~N(4\times 100, 4\times 5^2) \\ Y~N(400, 100) \end{gathered}$$

The required probability can be calculated as follows

$$\begin{gathered} P(Y>420) \\ =P\left(\frac{Y-{\mu }_y}{{\sigma }_y}>\frac{420-400}{\sqrt{100}}\right) \\ =P\left(z>2\right) \\ =1-P\left(z\le 2\right) \end{gathered}$$

From z tables

$$\begin{gathered} P\left(z<2\right)=0.97725 \\ P(z>2)=1-0.97725 \\ P(z<2)=0.02275 \end{gathered}$$