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The Orchard Cafe has found that about 3% of the diners who make reservations don't show up. If 83 reservations have been made, how many diners can be expected to show up? Find the standard deviation of this distribution. (Round your answers to two decimal places.) μ = σ =

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Answer:


\mu = 80.51\\\sigma = 1.56

Explanation:

We are given the following information in the question:

Percentage of of the diners who make reservations don't show up = 3%

Number of reservations = 83

Thus, we are given a binomial distribution with n = 83 and p = 0.97


X\sim \text{ Binom}(p = 0.97, n = 83)


\mu = np = (83)(0.97) = 80.51

Around 81 people can be expected to show up.


\sigma = √(np(1-p)) = √((83)(0.97)(1-0.97)) = 1.5541 \approx 1.56

The standard deviation of this distribution is 1.56

User Sayantan Ghosh
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