In the following problem, check that it is appropriate to use the normal approxi

Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Do you take the free samples offered in supermarkets? About 57% of all customers will take free samples. Furthermore, of those who take the free samples, about 34% 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, 327 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) = 0.34, while P(sample) = 0.57.


(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 p=0.57 and n=327 ;therefore np <10 and n(1-p) >10

hence we can use normal approximation of binomial distribution

here mean =np=0.57*327 =186.39

and std deviaiton =(np(1--p))1/2 =8.95

a) probability that more than 180 will take your free sample =P(X>180)=1-P(X<=179) =1-P(Z<(179.5-186.39)/8.95)

=1-P(Z<-0.7696)=1-0.2208 =0.7792

b) probability that fewer than 200 will take your free sample=P(X<=199)=P(Z<(199.5-186.39)/8.95)=P(Z<1.4644)

=0.9285

c)probability that a customer will take a free sample and buy the product = P(sample)*P(buy|sample)=0.57*0.34

=0.1938

d) mean =np=327*0.1938 =63.3726

std deviation =(np(1--p))1/2 =7.148

probability that between 60 and 80 customers will take the free sample and buy the product

=P((60<X<80))=P((59.5-63.3726)/7.148<Z<(79.5-63.3726)/7.148)=P(-0.5418<Z<2.2563)=0.9880-0.2940=0.6940

(here 60 and 80 values are taken exclusive of above interval)

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