Expert: Sherry Wallin - 4/26/2011

Question
QUESTION: In one popular new restaurant, data was collected to determine the mean waiting time to be seated at a table. Assume the data was normally distributed with a mean of 45 min and standard deviation of 12 min.
---Determine the probability that a randomly selected person has to wait less than 20 min. I am supposed to express my answer to the nearest tenth of a percent.

ANSWER: Hi Andrea~

If the mean is 45 and the std dev is 12, then you want to know how many std dev 20 min is from 45. So 45-20 = 25. Now 25/12 = 2.083. This 2.083 is the number of std dev away from the mean. You want the area under the curve to the left of the mean since 20 < 45 so realistically we should use the difference 20-45 = -25 and then -25/12 = -2.083. Using the normal table find this z-score and that will be your answer.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Determine the probability that a randomly selected person has to wait more than 1 h.

Answer
Andrea~

More than one hour means the wait is greater than 60 min. The difference between 60 and 45 is 15 min and how many std dev of 12 are there in 15? This means 15/12 = 1.2. You want to know how much area is under the curve to the right of 1.2 std dev or a z score greater than 1.2. Use a table of a normal distribution to find this value.

Math Prof