2 5 of the customers with reservations at a restaruant dont

2) 5% of the customers with reservations at a restaruant don\'t show up. Use a normal distribution to approximate the probability that at least 20 customers won\'t show up from a random group of 200 reservations. (Use your calcuator) (Please show the calculator command used)

Solution

Normal Distribution
Mean ( np ) = 10
Standard Deviation ( npq )= 200*0.05*0.95 = 3.0822
Normal Distribution = Z= X- u / sd                   
P(X < 20) = (20-10)/3.0822
= 10/3.0822= 3.2444
= P ( Z <3.2444) From Standard Normal Table
= 0.9994                  
P(X > = 20) = (1 - P(X < 20)
= 1 - 0.9994 = 0.0006