Question

Satistics: Suppose the hourly dollar amount of food sold by a Burger King franchise follows an approximately mound shaped distribution with a mean sales level of $400/hour and a standard deviation of $60/hour. What percentage of the working hours does the Burger King franchise sell between $340 and $460 worth of food/hour.

Note: Use the empirical rule.

Answer 1

\[ P(340<X<460)=P\left( \frac{340-400}{60}<\frac{X-400}{60}<\frac{460-400}{60}\right)\\
=P(-1<Z<1)\]
Thus, you have values between -1 and 1 of a standard normal distribution. Do you recall what percentage of data values fall within 1 standard deviation of the mean?

Answer 2

If you don't remember, you could always look at a normal distribution table and see what is the cumulative probability at Z=1, and then establish using symmetry arguments, what is the cumulative probability at Z=-1, and then subtract both of these probabilities.

Answer 3

\[(\mu \pm \sigma)\] 68%

Answer 4

\[(\mu \pm 2\sigma)\] = 95%

Answer 5

\[(\mu \pm 3\sigma) =99.7 %\]

Answer 6

I have the use one of the three, just not sure how

Answer 7

The mean of the standard normal distribution is 0 (\(\mu=0\)), and the st. deviation is 1 (\(\sigma=1\)). So, you have \((0 \pm 1)=(-1,1)\)

Answer 8

Without standardizing, you can see that the mean is \(\mu=400\), and the standard deviation is \(\sigma=60\), so if you calculate \((400 \pm 1\cdot60) = (340, 460)\), so you are again just 1 standard deviation from the mean.

Answer 9

Ohhh okay that makes sense, so it would jusst be 68%, right?

Answer 10

Yep :)

Answer 11

Thank you: )