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 64% of
all customers will take free samples. Furthermore, of those who
take the free samples, about 37% 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,
303customers 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) = 0.37, while P(sample)
= 0.64.
(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:
Given,
p = 0.64
n = 303
np = 303*0.64
= 193.92
standard deviation = sqrt(npq)
= sqrt(303*0.64*(1-0.64))
= 8.36
a)
To give P(X > 180)
= P((x-mu)/s > (180 - 193.92)/8.36)
= P(z > -1.67)
= 0.9525403 [since from z table]
b)
P(x < 200)
= P((x-mu)/s < (200 - 193.92)/8.36)
= P(z < 0.73)
= 0.7673049 [since from z table]
= 0.7673
c)
P(buy|sample) = 0.37
P(sample) = 0.64
Now the probability that a customer will take a free sample and buy the product
= P(buy|sample)
= P(buy & sample) / P(sample)
P(buy & sample) = P(buy|sample) * P(sample)
= 0.37*0.64
= 0.2368
d)
p = 0.2368
n = 303
mean = np
= 303*0.2368
= 71.75
standard deviation = sqrt(npq)
= sqrt(303*0.2368*(1-0.2368))
= 7.4
P(60 < x < 80) = P((60 - 71.75)/7.4 < z < (80 - 71.75)/7.4)
= P(-1.59 < z < 1.115)
= P(z < 1.115) - P(z < - 1.59)
= 0.8675748 - 0.0559174 [since from z table]
= 0.8117
Get Answers For Free
Most questions answered within 1 hours.