Question

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 28 minutes and a standard deviation of 5 minutes. What is the probability that a randomly selected customer will have to wait less than 33 minutes, to the nearest thousandth?

Answer 1

To find the probability that a randomly selected customer will have to wait less than 33 minutes, you can use the z-score formula and the standard normal distribution (z-distribution).

First, calculate the z-score for 33 minutes using the given mean (μ = 28 minutes) and standard deviation (σ = 5 minutes):

Z
=
X
−
μ
σ
=
33
−
28
5
=
1
Z=
σ
X−μ
​
=
5
33−28
​
=1

Now, you want to find the probability that a z-score is less than 1. You can look up this value in a standard normal distribution table or use a calculator. The probability that Z is less than 1 is approximately 0.8413 (rounded to four decimal places).

So, the probability that a randomly selected customer will have to wait less than 33 minutes is approximately 0.8413 to the nearest thousandth.