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 solve this problem, we need to first standardize the value 33 using the formula for z-scores:

z = (X - μ) / σ

where:
X = value we're interested in (33 minutes)
μ = mean (28 minutes)
σ = standard deviation (5 minutes)

Plugging in the values, we get:

z = (33 - 28) / 5
z = 1

Next, we need to find the probability of a z-score of 1 using a standard normal distribution table.

Looking up the z-score of 1 in the table, we find that the probability is approximately 0.8413.

Therefore, the probability that a randomly selected customer will have to wait less than 33 minutes is 0.8413 (or 84.13%) to the nearest thousandth.