Question

sams spaghetti station is a popular weekend restaurant. on a busy saturday night, the waiting time for take - out orders is normally distributed with a mean of 20 minutes and a standard deviation of 5 minutes. what percent of customers are likely to wait more than 25 minutes? hint: this graph shows the percent of data values centered around a mean of μ with standard deviation σ in a normal distribution. graph of normal distribution with 68%, 95%, 99.7% labeled multiple - choice options: 16%, 34%, 50%, 68%

Explanation:

Step1: Identify mean and standard deviation

Mean ($\mu$) = 20 minutes, Standard Deviation ($\sigma$) = 5 minutes. We need to find the percentage of customers waiting more than 25 minutes. First, find how many standard deviations 25 is from the mean: $25 - 20 = 5$, so $25 = \mu + \sigma$.

Step2: Use the normal distribution graph

The normal distribution graph (empirical rule) shows that within $\mu - \sigma$ to $\mu + \sigma$ (68% of data), outside of that, the remaining is $100\% - 68\% = 32\%$. This 32% is split equally between the two tails (less than $\mu - \sigma$ and more than $\mu + \sigma$). So the percentage more than $\mu + \sigma$ (25 minutes here) is $\frac{32\%}{2} = 16\%$.

Answer:

16%