Do you take the free samples offered in supermarkets? About 65% of all customers

Question

Do you take the free samples offered in supermarkets? About 65% of all customers will take free samples. Furthermore, of those who take the free samples, about 33% will buy what they have sampled. Suppose you set up a counter in a supermarket offering free samples of a new product. The day you were offering free samples, 305 customers passed by your counter. (Round your answers to four decimal places.) (a) What is the probability that more than 180 will take your free sample? (b) What is the probability that fewer than 200 will take your free sample? (c) What is the probability that a customer will take a free sample and buy the product? Hint: Use the multiplication rule for dependent events. Notice that we are given the conditional probability P(buy|sample) = 0.33, while P(sample) = 0.65. (d) What is the probability that between 60 and 80 customers will take the free sample and buy the product? Hint: Use the probability of success calculated in part (c).

Explanation / Answer

a)

We first get the z score for the critical value:          
          
x = critical value =    180.5      
u = mean = np =    198.25      
          
s = standard deviation = sqrt(np(1-p)) =    8.329915966      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -2.130873837      
          
Thus, the left tailed area is          
          
P(z <   -2.130873837   ) =    0.983450231 [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    199.5      
u = mean = np =    198.25      
          
s = standard deviation = sqrt(np(1-p)) =    8.329915966      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    0.150061538      
          
Thus, the left tailed area is          
          
P(z <   0.150061538   ) =    0.559641968 [ANSWER]

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c)

P = P(S) P(B|S) = 0.65*0.33 = 0.2145 [ANSWER]

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d)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    59.5      
x2 = upper bound =    80.5      
u = mean = np =    65.4225      
          
s = standard deviation = sqrt(np(1-p)) =    7.168638207      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.826168071      
z2 = upper z score = (x2 - u) / s =    2.103258606      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.204354379      
P(z < z2) =    0.982278415      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.777924036   [ANSWER]  

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