5 of the customers with reservations at a restaurant dont sh

5% of the customers with reservations at a restaurant don’t show up. Us 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 calculator.

Solution

Here,

mean = n p = 200*0.05 = 10

standard deviation = sqrt(n p (1-p)) = sqrt(200*0.05*(1-0.05)) = 3.082207001

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    19.5      
u = mean =    10      
          
s = standard deviation =    3.082207001      
          
Thus,          
          
z = (x - u) / s =    3.082207002      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.082207002   ) =    0.001027359 [ANSWER]