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 4 minutes. What is the probability that a randomly selected customer will have to wait between 18 minutes and 28 minutes, to the nearest thousandth?
To find the probability that a randomly selected customer will have to wait between 18 minutes and 28 minutes, we need to find the z-scores for both 18 minutes and 28 minutes and then find the area under the normal curve between those two z-scores.
The z-score formula is given by:
z = (X - μ) / σ
For X = 18 minutes:
z1 = (18 - 28) / 4
z1 = -2.5
For X = 28 minutes:
z2 = (28 - 28) / 4
z2 = 0
Using a standard normal distribution table or calculator, we can find the probabilities for z1 and z2.
P(z < -2.5) ≈ 0.0062
P(z < 0) = 0.5
Therefore, the probability that a randomly selected customer will have to wait between 18 minutes and 28 minutes is:
P(-2.5 < z < 0) = P(z < 0) - P(z < -2.5)
P(-2.5 < z < 0) = 0.5 - 0.0062
P(-2.5 < z < 0) ≈ 0.4938
So, the probability that a randomly selected customer will have to wait between 18 minutes and 28 minutes is approximately 0.494 (rounded to the nearest thousandth).