Answer:
[tex]\mu = 80.51\\\sigma = 1.56[/tex]
Step-by-step 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
[tex]X\sim \text{ Binom}(p = 0.97, n = 83)[/tex]
[tex]\mu = np = (83)(0.97) = 80.51[/tex]
Around 81 people  can be expected to show up.
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{(83)(0.97)(1-0.97)} = 1.5541 \approx 1.56[/tex]
The standard deviation of this distribution is 1.56
