The length of time for one individual to be served at a cafe

The length of time for one individual to be served at a cafeteria is a random variable having a normal distribution with a mean of 10 minutes and standard deviation of 2 minutes. What is the probability that a person will be served in less than 9 minutes on at least 4 of the next 6 days?

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    9      
u = mean =    10      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    -0.5      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -0.5   ) =    0.308537539

Now, this is the probability of success in our binomial distribution.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    6      
p = the probability of a success =    0.308537539      
x = our critical value of successes =    4      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   3   ) =    0.92254518
          
Thus, the probability of at least   4   successes is  
          
P(at least   4   ) =    0.07745482 [ANSWER]