Question

question
at a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 44 minutes and a standard deviation of 3 minutes. what percentage of customers have to wait less than 44 minutes, to the nearest tenth?
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Explanation:

Step1: Recall Normal Distribution Property

In a normal distribution, the mean ($\mu$) is the center of the distribution, and the distribution is symmetric about the mean. So, the probability that a value is less than the mean in a normal distribution is 0.5 (or 50%), because half of the data lies to the left of the mean and half to the right. Here, the mean $\mu = 44$ minutes. We want to find $P(X < 44)$, where $X$ is the waiting time. Since the distribution is symmetric about $\mu = 44$, the area to the left of 44 (which represents the percentage of customers waiting less than 44 minutes) is 0.5.

Step2: Convert to Percentage

To convert the probability 0.5 to a percentage, we multiply by 100: $0.5\times100 = 50.0\%$ (to the nearest tenth).

Answer:

50.0%