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?
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.