Starbucks uses a filling machine to fill aluminum cans with

Starbucks uses a filling machine to fill aluminum cans with coffee. The cans are independently filled. The cans are supposed to contain 300 milliliters (ml) of coffee, but in reality, the contents of the cans vary according to a normal distribution with a mean of 298 ml and a standard deviation of 4.5 ml. What is the probability that fewer than three cans in a 24-pack have contents less than 293 ml?

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    293      
u = mean =    298      
n = sample size =    1      
s = standard deviation =    4.5      
          
Thus,          
          
z =    -1.111111111      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -1.111111111   ) =    0.133260263

Thus, this is the probability that a certain can has less than 293 mL of contents.

Now, using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    24      
p = the probability of a success =    0.133260263      
x = the maximum number of successes =    2 (it says fewer than 3, so 3 is not included)      
          
Then the cumulative probability is          
          
P(x <=   2   ) =    0.362326067 [ANSWER]