A restaurant charges $8.95$ pp. Management finds it's expenses per person has a distribution that is skewed to the right with a mean of $8.20$ and a standard deviation of $3.00$.
Q: If $100$ customers have the characteristics of a random sample from their customer base, find the mean and standard deviation of the sampling distribution of the restaurants sample mean expense per customer.
Q: Find the probability that the restaurant makes a profit, with the sample mean expense being less than $8.95$.
I'm not just wanting an answer to cheat, but want to know how to solve it. I have a link for a Z-table, but not sure where to put what numbers to calculate. This is my first time here-please be nice and helpful. :) Thanks so much for any direction.
Let $X_i$ be the expense for the $i$th customer and $\overline{X}$ the sample mean. Then, $$\overline{X} = \frac{1}{100} \sum\limits_{i = 1}^{100} X_i.$$
Use the linearity of the expectation operator to evaluate $$\mathop{\mathrm{E}} \left[\overline{X}\right] = \mathop{\mathrm{E}} \left[\frac{1}{100} \sum\limits_{i = 1}^{100} X_i\right].$$ (Hint: $\mathop{\mathrm{E}} [aX] = a \mathop{\mathrm{E}} [X]$ and $\mathop{\mathrm{E}} [X + Y] = \mathop{\mathrm{E}} [X] + \mathop{\mathrm{E}}[Y]$.)
Similar properties of the variance operator can be used to evaluate $$\mathop{\mathrm{Var}} \left[\overline{X}\right] = \mathop{\mathrm{Var}} \left[\frac{1}{100} \sum\limits_{i = 1}^{100} X_i\right].$$ (Hint: $\mathop{\mathrm{Var}} [aX] = a^2 \mathop{\mathrm{Var}} [X]$ and $\mathop{\mathrm{Var}} [X + Y] = \mathop{\mathrm{Var}} [X] + \mathop{\mathrm{Var}}[Y]$. The latter relationship assumes $X$ and $Y$ are independent of each other; we assume the expense of one customer is independent of the expense of any other customer and, therefore, the relationship applies.)
We want to calculate $\Pr(\overline{X} < 8.95)$. We lack a knowledge of the underlying probability distribution of $\overline{X}$, without which we cannot calculate exact probabilities. We content ourselves by calculating approximate probabilities, and we do so by first assuming $\overline{X}$ follows a normal distribution and then standardizing $\overline{X}$.
Assume $\overline{X}$ follows a normal distribution. Then, $Z = \frac{\overline{X} - \mu_{\overline{X}}}{\sigma_{\overline{X}}}$, the standardization of $\overline{X}$, follows the standard normal distribution and $$\Pr(\overline{X} < 8.95) = \Pr\left(\frac{\overline{X} - \mu_{\overline{X}}}{\sigma_{\overline{X}}} < \frac{8.95 - \mu_{\overline{X}}}{\sigma_{\overline{X}}} \right) = \Pr\left(Z < 2.5\right).$$ The last probability, $\Pr(Z < 2.5)$, can be calculated with a z-table.