Question

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?
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Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value we are interested in, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Given that $\mu = 26$, $\sigma=5$, and $x = 32$.
Substitute these values into the formula: $z=\frac{32 - 26}{5}=\frac{6}{5}=1.2$.

Step2: Find the probability that $Z>1.2$

We know that the total area under the standard normal curve is 1. The probability that $Z\leq1.2$ can be found using a standard normal table or a calculator.
Looking up the value for $z = 1.2$ in the standard normal table, we find that $P(Z\leq1.2)=0.8849$.
To find $P(Z > 1.2)$, we use the formula $P(Z>1.2)=1 - P(Z\leq1.2)$.
So, $P(Z > 1.2)=1 - 0.8849 = 0.1151$.
Rounding to the nearest thousandth, we get $0.115$.

Answer:

$0.115$