Question

A study found that the mean amount of time cars spent in drive thru of a certain fast food restaurant was 130.7 seconds. Assuming drive thru times are normally distributed with a standard deviation of 28 seconds complete a through dA what is the probability that a randomly selected car will get through the restaurants drive try in less than 98 seconds?

Answer 1

In order to find the probability, first let's calculate the z value with the formula below:

`z=(x-\mu)/(\sigma)`

Using x = 98, mean = 130.7 and standard deviation = 28, we have:-

`\begin{gathered} z=(98-130.7)/(28) \\ z=-(32.7)/(28) \\ z=-1.17 \end{gathered}`

Looking at the z-table for the corresponding probability for z = -1.17, we have 0.1210, so the probability is equal to 12.1%.

This probability corresponds to the following area in the normal distribution:

The value of 0.379 in the z-table would correspond to the following area:

That is, in practical terms, it means the probability for a selected car taking between 98 seconds and 130.7 seconds (instead of the probability for lesser than 98 seconds)