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the time spent waiting in the line is approximately normally distributed. the mean waiting time is 7 minutes and the variance of the waiting time is 4 . find the probability that a person will wait for less than 11 minutes. round your answer to four decimal places.

User Racker
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We know that the waiting time is approximately normally distributed with a mean of 7 minutes and a variance of 4. Therefore, the standard deviation is the square root of the variance, which is 2.

To find the probability that a person will wait for less than 11 minutes, we need to standardize the value using the standard deviation and mean.

z = (x - μ) / σ

where x is the waiting time we want to find the probability for, μ is the mean waiting time, and σ is the standard deviation.

Plugging in the values, we get:

z = (11 - 7) / 2
z = 2

Using a standard normal distribution table or calculator, we can find the probability that a z-score is less than 2. This probability is approximately 0.9772.

Therefore, the probability that a person will wait for less than 11 minutes is 0.9772 (rounded to four decimal places).
User Chris Colbert
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