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? About57%of all customers will take free samples. Furthermore, of those who take the free samples, about34%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,317customers passed by your counter.(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 fordependentevents. Notice that we are given the conditional probabilityP(buy|sample) =0.34,whileP(sample) =0.57. (d) What is the probability that between 60 and 80 customers will take the free sampleandbuy the product?Hint:Use the probability of success calculated in part (c).

Step 1

(a) What is the probability that more than 180 will take your free sample?

We are asked to find the probability that more than 180 people of the317that walk by will take a free sample in a supermarket. We are told that about57%of all customers take free samples. First we must test to see if we can use the normal approximation to the binomial distribution. Here we will define success as "someone takes a free sample." In this scenario, the number of trials is

n=317

317

.

The probability of success is

p=0.57

0.57

,

which means that the probability of a failure,q, is

q= 1p=0.43

0.43

.

Since

np=180.69

180.69

andnq=136.31, wecan

can

use the normal approximation to the binomial distribution because these values are both greater than 5.

Step 2

We wish to find the probability that more than 180 take samples, so

P(r> 180).

Because the normal approximation to the binomial will be used, the original probability statement

P(r> 180)

must be rewritten using the continuity correction. The continuity correction to the normal approximation is the process of converting the discrete random variabler(number of successes) to the continuous normal random variablexby doing the following.Ifris aleft pointof an interval, subtract 0.5 to obtain the corresponding normal variablex; that is,

x=r0.5.

Ifris aright pointof an interval, add 0.5 to obtain the corresponding normal variablex; that is,

x=r+ 0.5.

We know that

P(r> 180)

can be rewritten as

P(r181).

Applying the continuity correction, the desired probability statement becomes

P

x180.5

180.5

.

Step 3

We must now convert

P(x180.5)

to standard units. To convert to standard units, we need the values ofandrounded to four decimal places.

=

np

=

180.69

=

npq

=

12.9967

Now use the formula

z=

x

to find the value ofzrounded to two decimal places that corresponds to

x= 180.5.

z=

The probability is found as follows. Recall for areas to the right of a specifiedzvalue, we must subtract thetableentry from 1. (Round your answers to four decimal places.)

P(x180.5)	=	P z
	=	1P z
	=	1
	=

Therefore, the probability that more than 180 people will take a free sample is .