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At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 18 minutes and a standard deviation of 2 minutes. What is the probability that a randomly selected customer will have to wait less than 15 minutes, to the nearest thousandth?

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of $18$ minutes and a standard deviation of $2$ minutes. What is the probability that a randomly selected customer will have to wait less than $15$ minutes, to the nearest thousandth?

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Find probabilities using the binomial distribution

Full solution

Q. At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of

18

minutes and a standard deviation of

2

minutes. What is the probability that a randomly selected customer will have to wait less than

15

minutes, to the nearest thousandth?

Calculate z-score: To find the probability, we need to calculate the z-score for $15$ minutes. $\newline$ $Z = \frac{X - \mu}{\sigma}$ $\newline$ Where $X$ is $15$ minutes, $\mu$ is the mean ( $18$ minutes), and $\sigma$ is the standard deviation ( $2$ minutes). $\newline$ $Z = \frac{15 - 18}{2}$
Calculate z-score: Calculate the z-score. $\newline$ $Z = \frac{-3}{2}$ $\newline$ $Z = -1.5$
Look up in table: Now we look up the $z$ -score in the standard normal distribution table to find the probability. The table gives the probability that a value is less than a given $z$ -score.
Find probability: The probability for $Z = -1.5$ is approximately $0.0668$ .

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