Customer checkout time at Osco Drug is uniformly distributed

Customer checkout time at Osco Drug is uniformly distributed between 0.6 and 2.4 minutes. The check out time at Walgreens is normally distributed with mean 0f 102 seconds and standard deviation 30 seconds.

Which store checks customers out more quickly on average?

            a)         Osco Drug                   b)         Walgreens

What is the probability that the customers in Osco Drug wait more than 100 seconds?

a).41

b).59

c).4761

d).5239

e)None of the above

What is the probability that the customer in Walgreens wait more than 100 seconds?

a).41

b)0.59

c).4761

d).5239

e)None of the above

Only 10% of all customers at Walgreens require more than t minutes to check out. Find t.

a).78

b)1.06

c)2.22

d)2.34

e)None of the above

Only 10% of all customers at Osco Drug require more than t minutes to check out. Find t.

a).78

b)1.06

c)2.22

d)2.34

e)None of the above

Solution

1.

The mean of Osco drug is (0.6 + 2.4)/2 = 1.5 minutes = 90 s.

The mean of Walgreens is 102 s.

Thus, ANSWER IS OSCO DRUG. [ANSWER]

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2.

Note that here,          
          
a = lower fence of the distribution =    0.6      
b = upper fence of the distribution =    2.4      
          
Thus, the mean, variance, and standard deviations are          
          
u = mean = (b - a)/2 =    0.9      
s^2 = variance = (b -a)^2 / 12 =    0.27      
s = standard deviation = sqrt(s^2) =    0.519615242      
          
Also, the area between the said numbers is          
          
c = lower number =    1.666666667      
d = higher number =    2.4      
          
Thus,          
          
A = (d - c)/(b - a) =    0.407407407   [NONE OF THE ABOVE]

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3.
We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    100      
u = mean =    102      
          
s = standard deviation =    30      
          
Thus,          
          
z = (x - u) / s =    -0.06      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.06   ) =    0.523922183 [OPTION D]

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4.

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    102      
z = the critical z score =    1.281551566      
s = standard deviation =    30      
          
Then          
          
x = critical value =    140.446547 seconds or 2.340775783 min [ANSWER, OPTION D]

5.

(2.4 - t) / (2.4 - 0.6) = 0.9

Thus,

t = 0.78 [answer]