The probability density function of a normal distribution with mean $\mu$ and variance $\sigma^{2}$ is of the form $$ f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-0.5\left(\frac{x-\mu}{\sigma}\right)^{2}}, \quad-\infty<x<\infty . $$ Let $X$ be a random variable with mean $\mathbb{E}(X)=\mu$ and variance $\mathbb{V a r}(X)=\sigma^{2}$.
Let $X_{1}, X_{2}, \ldots, X_{n}$ be $n$ independently sampled values of $X$, and let $\overline{X}=\frac{1}{n} \sum_{i} X_{i}$ be the sample mean. The central limit theorem states that if the sample size $n$ is large enough then $\overline{X}$ approximately follows the normal distribution with mean $\mu$ and variance $\sigma^{2} / n$. That is, $\overline{X} \approx N\left(\mu, \sigma^{2} / n\right)$. This in turn implies that $$ Y=\frac{\sqrt{n}(\overline{X}-\mu)}{\sigma} \approx N(0,1) .$$
Let $Z$ be a random variable that follows the normal distribution with mean $0$ and variance $1.$ For any real number $a$ let $\Phi(a)=\mathbb{P}(Z<a)$ be the probability that $Z$ takes values smaller than $a$. Then $$ \Phi(a)=\mathbb{P}(Z<a)=\int_{-\infty}^{a} \frac{e^{-u^{2} / 2}}{\sqrt{2 \pi}} d u $$
For solving the next two problems you may assume the following approximations: $\Phi(-2)=0.02, \Phi(-1)=$ $0.16, \Phi(0)=0.5, \Phi(1)=0.84, \Phi(2)=0.98$.
The weekly number of sales $S$ at a certain car dealership is known to follow a probability distribution with mean $\mathbb{E}(X)=10$ and variance $\operatorname{Var}(X)=144$. A performance audit picks a random sample of $36$ weekly sales figures $S_{1}, S_{2}, \ldots, S_{36}$ from the last two years. They find that the sample mean is $10$ and the sample variance is $144.$ Use this information to answer the next two questions. You may assume that $n=36$ is a large enough sample size.
- What is the probability that the average number of sales in a week will be more than $8?$
- What is the total number of sales over the $36$ weeks which were sampled?
