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 \\(\mu\\) with standard deviation \\(\sigma\\) in a normal distribution.
graph of normal distribution with labels \\(\mu - 3\sigma\\), \\(\mu - 2\sigma\\), \\(\mu - \sigma\\), \\(\mu\\), \\(\mu + \sigma\\), \\(\mu + 2\sigma\\), \\(\mu + 3\sigma\\) and percentages 68%, 95%, 99.7%
options: 16%, 34%, 50%, 68%

Explanation:

Step1: Identify the mean and standard deviation

The mean ($\mu$) is 20 minutes and the standard deviation ($\sigma$) is 5 minutes. We need to find the number of standard deviations 25 minutes is from the mean.
To find the z - score (number of standard deviations from the mean), we use the formula $z=\frac{x - \mu}{\sigma}$, where $x = 25$.
Substituting the values: $z=\frac{25 - 20}{5}=\frac{5}{5}=1$. So 25 minutes is $\mu+\sigma$ (since $\mu = 20,\sigma = 5,20 + 5=25$).

Step2: Use the empirical rule for normal distribution

The empirical rule (68 - 95 - 99.7 rule) for a normal distribution states that:

• Approximately 68% of the data lies within $\mu\pm\sigma$.
• Approximately 95% of the data lies within $\mu\pm2\sigma$.
• Approximately 99.7% of the data lies within $\mu\pm3\sigma$.

For the data within $\mu-\sigma$ to $\mu+\sigma$, the percentage is 68%. So the percentage of data outside of $\mu-\sigma$ to $\mu+\sigma$ is $100\% - 68\%=32\%$. Since the normal distribution is symmetric about the mean, the percentage of data greater than $\mu+\sigma$ (and less than $\mu - \sigma$) is equal. So the percentage of data greater than $\mu+\sigma$ is $\frac{32\%}{2}=16\%$.

Answer:

16%