Express Courier Service has found that the delivery time for

Express Courier Service has found that the delivery time for packages is normally distributed, with mean 14 hours and standard deviation 2 hours.

(a) For a package selected at random, what is the probability that it will be delivered in 18 hours or less? (Round your answer to four decimal places.)
.9772  (this is the correct answer, how do I get it?)

(b) What should be the guaranteed delivery time on all packages in order to be 95% sure that the package will be delivered before this time? (Hint: Note that 5% of the packages will be delivered at a time beyond the guaranteed time period.) (Round your answer to one decimal place.)

17.3 (this is the correct answer, how do I get it?)

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    18      
u = mean =    14      
n = sample size =    1      
s = standard deviation =    2      
          
Thus,          
          
z =    2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   2   ) =    0.977249868 [ANSWER]

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    14      
z = the critical z score =    1.644853627      
s = standard deviation =    2      
n = sample size =    1      
Then          
          
x = critical value =    17.28970725   [ANSWER]