Question

Do you take the free samples offered in supermarkets? About 58% 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, 309 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) = 0.58.

(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).

Answer 1

We are given P( sample ) = 0.58 and n = 309

Let x be the number of customers take free samples , x follows binomial with n = 309 and p = 0.58 , q = 1-0.58 = 0.42

As n*p and n*q both are greater than 5, we have to use normal approximation.

So mean = n.*p = 309* 0.58 = 179.22 and standard deviation = = 8.676

We are asked to find P( x > 180 ) , we have to use continuity correction by adding 0.5 to 180

a)So we have to find P( x>180.5 ) using normal distribution

P( x > 180.5 ) =

P( z > 0.15) = 1 - P( z < 0.15 ) = 1 - 0.5596 ---- ( from z score table )

= 0.4404

b) We are asked to find P ( x < 200 ) so we have to use continuity correction by subtracting the 0.5 from 200

P( x < 199.5 ) =

=P( z < 2.34)

= 0.9903

c) We have P(buy|sample) = 0.32, while P(sample) = 0.58

We are asked to find P( buy and sample ) = P(buy|sample) * P(sample) = 0.32*0.58 = 0.1856

d) We have P( buy and sample ) = 0.1856

Let y be the number of customers will take the free sample and buy the product.

Y follows binomial with n = 309 and p = 0.1856

As n*p and n*q both are greater than 5, we have to use normal approximation.

So mean = n.*p = 309* 0.1856 = 57.3504 and standard deviation = = 6.8342

We are asked to find P(60<y<80) , we have to use continuity correction by adding 0.5 to 80 and subtracting 0.5 from 60

P( 59.5 < y < 80.5 ) = P( y < 80.5 ) - P( y <59.5 )

=

=P( z < 3.39) - P( z < 0.31)

= 0.9997 - 0.6217

= 0.3780