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after collecting the data, shawn finds that the monthly number of take-out orders at a restaurant is normally distributed with mean 132 and standard deviation 6. what is the probability that a randomly selected month's number of orders is more than 150?

Sagot :

varshamittal029

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%.

Learn more about Normal distribution here:

brainly.com/question/4079902

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