The probability that a randomly selected month's number of order is more than 150 is 0.13%
Given, shawn finds that the monthly number of take-out orders at a restaurant is normally distributed with mean 132 and standard deviation 6.
⇒ mean = 132
⇒ standard deviation = 6
Analysis:
Set the monthly number of take out order as x.
From the question, we know:
P(x > 150) = P(x-132/ > 150-132/6)
= P(x=132/6 > 3)
= 1 - P(x-132/6 ≤ 3)
≈ 1 - 0.9987 {standard normal distribution table}
≈ 0.0013
= 0.13%
Hence we get the probability as 0.13%.
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