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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 46 minutes and a standard deviation of 4 minutes. what is the probability that a randomly selected customer will have to wait less than 41 minutes, to the nearest thousandth?

User HAXM
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Final answer:

The probability that a randomly selected customer will have to wait less than 41 minutes at a restaurant, where wait times are normally distributed with a mean of 46 minutes and a standard deviation of 4 minutes, is approximately 0.106, to the nearest thousandth.

Step-by-step explanation:

To calculate the probability that a randomly selected customer will have to wait less than 41 minutes for their food at a local restaurant, where the wait times are normally distributed with a mean of 46 minutes and a standard deviation of 4 minutes, we need to find the z-score and then use the standard normal distribution to find the probability.

First, we calculate the z-score using the following formula:

Z = (X - μ) / σ

Where:

  • X is the value we are comparing to the mean (41 minutes),
  • μ is the mean (46 minutes),
  • σ is the standard deviation (4 minutes).

Plugging in the values we get:

Z = (41 - 46) / 4 = -5 / 4 = -1.25

To find the probability of a z-score being less than -1.25, we refer to the standard normal distribution table (z-table), which gives us the probability that a standard normally distributed variable is less than -1.25. This is equivalent to the cumulative probability up to that point.

Looking up the z-score of -1.25 on the z-table gives us a probability of approximately 0.1056. Therefore, the probability that a randomly selected customer will have to wait less than 41 minutes is 0.106, to the nearest thousandth.

User MaxThom
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