Do you take the free samples offered in supermarkets? About 62% of all customers

Question

Do you take the free samples offered in supermarkets? About 62% of all customers will take free samples. Furthermore, of those who take the free samples, about 32% 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, 303 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) = 0.32, while P(sample) = 0.62

(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

here n=303, p=P(sample)=0.62

mean=np=303*0.62=187.86

variance(p)=np(1-p)=71.3868

SE(p)=sqrt(variance(p))=sqrt(71.3868)=8.4491

we use standard normal variate z=(x-mean)/SE(p)

answer (a)for x=180 ,z=(180-187.86)/8.4491=0.9303

p(x>180)=p(z>0.9303)=1-p(z<0.9303)=1-0.8239=0.1761

answer b) similary as in a) for 200, z=1.5445

p(x<200)=p(z<1.5445)=0.9388

answer c)probability that a customer will take a free sample and buy the product=P(sample and buy)

=P(sample)*P(buy|sample)=0.62*0.32=0.1984

answer d)

here p=0.1984 ,mean=303*0.1984=60.1152, SE(p)=sqrt(np(1-p))=6.9418

for 60, z1=-0.0166

for 80, z2=2.8811

P(60<x<80)=P(-0.0166<z<2.8811)=P(z<2.8811)-P(z<-0.1166)=0.5046

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