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

Answer 1

Let $X$ be the random variable for the value of goods sold in any given month. Also let $Y$ be the random variable representing the number of times that sales in a given month surpass a value of 100.

$X$ follows a normal distribution with mean $\mu=100$ and standard deviation $\sigma=5$, while $Y$ would follow a binomial distribution with some success probability $p$ on $n=6$ trials. To find $p$, we use the distribution of $X$:

$p=P(X>100)=P\left(\dfrac{X-100}5>\dfrac{100-100}5\right)=P(Z>0)=\dfrac12$

where $Z$ is a random variable following the standard normal distribution.

a) The probability that exactly 3 of the next 6 months have sales greater than 100 is then

$P(Y=3)=\dbinom63\left(\dfrac12\right)^3\left(1-\dfrac12\right)^{6-3}=\dfrac{6!}{3!(6-3)!}\cdot\dfrac1{2^6}=\dfrac5{16}\approx0.31$

b) Let $W$ be the random variable representing the total value of sales over 4 months, and denote by $X_i$ the random variable for the value of sales during month $i$. Then $W=X_1+X_2+X_3+X_4$, and we want to find $P(W>420)$.

We know the $X_i$ are mutually independent and identically distributed, and we (you probably should, anyway) also know that the sum of i.i.d. normally distributed random variables also follows a normal distribution, whose mean is the sum of the means, and whose standard deviation is the sum of the squares of the standard deviations, of the i.i.d. random variables:

$X_i\sim\mathcal N(100,5)\implies W\sim\mathcal N(400,100)$

Then the probability we want is

$P(W>420)=P\left(\dfrac{W-400}{100}>\dfrac{420-400}{100}\right)=P(Z>0.2)\approx0.42$