Question

A Wendy’s fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday, the demand for hamburgers is normally distributed with a mean of 450 and standard deviation of 80 and the demand for chicken sandwiches is normally distributed with a mean of 120 and standard deviation of 30.

How many chicken sandwiches must the restaurant stock to be 99% sure of not running out on a given day? Place you answer, rounded to the nearest whole number in the blank. For example, 345 would be a legitimate entry.

Answer 1

Answer on Question#50333 Economics – Accounting

190

Question

A Wendy's fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday, the demand for hamburgers is normally distributed with a mean of 450 and standard deviation of 80 and the demand for chicken sandwiches is normally distributed with a mean of 120 and standard deviation of 30.

How many chicken sandwiches must the restaurant stock to be 99% sure of not running out on a given day? Place you answer, rounded to the nearest whole number in the blank. For example, 345 would be a legitimate entry.

Solution

Note: we treat demand for chicken sandwiches like real number, not like natural number.

Normal distribution. General formula:

$$f(x, \mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}},$$

where $x$ – variable, $\mu$ – mean, $\sigma$ – standard deviation.

We know from the task that $\mu = 120$, $\sigma = 30$.

Thus,

$$f(x, 120, 30) = f(x) = \frac{1}{30\sqrt{2\pi}} e^{-\frac{(x - 120)^2}{2 \cdot 30^2}} = \frac{1}{30\sqrt{2\pi}} e^{-\frac{(x - 120)^2}{1800}}$$ $$99\% \text{ sure of not running out } \equiv \int_{-\infty}^{a} \frac{1}{30\sqrt{2\pi}} e^{-\frac{(x - 120)^2}{1800}} dx = 0.99$$

We able to solve it only numerically.

Result:

$$a = 189.7 \dots \approx 190$$

http://www.AssignmentExpert.com/