At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 26 minutes and a standard deviation of 5 minutes. What is the probability that a randomly selected customer will have to wait longer than 32 minutes, to the nearest thousandth?
To solve this problem, we need to calculate the z-score for a wait time of 32 minutes and then find the probability associated with that z-score.
First, we calculate the z-score using the formula:
z = (X - μ) / σ
where X is the value we are interested in (32 minutes), μ is the mean (26 minutes), and σ is the standard deviation (5 minutes).
z = (32 - 26) / 5 = 1.2
Next, we look up the cumulative probability for a z-score of 1.2 in a standard normal distribution table. The probability associated with a z-score of 1.2 is approximately 0.8849.
Therefore, the probability that a randomly selected customer will have to wait longer than 32 minutes is 0.8849 (or 88.49%, when converted to a percentage) to the nearest thousandth.