Question

The owner of a small deli is trying to decide whether to discontinue selling magazines. He suspects that only 9.8% of his customers buy a magazine and he thinks that he might be able to use the display space to sell something more profitable. Before making a final decision, he decides that for one day he will keep track of the number of customers that buy a magazine.(b) Assuming his suspicion that 9.8% of his customers buy a magazine is correct, what is the probability that exactly 2 out of the first 11 customers buy a magazine? Give your answer as a decimal number rounded to two digits.(c) What is the expected number of customers from this sample that will buy a magazine?

Answer 1

Answer:

Step-by-step explanation:

Let x be a random variable representing the number of his customers that buy a magazine. This is a binomial distribution since the outcomes are two ways. It is either a selected customer buys a magazine or he doesn't. The probability of success, p = 9.8/100 = 0.098

The probability of failure, q would be 1 - p = 1 - 0.098 = 0.902

b We want to determine P(x = 2)

n = 11

From the binomial distribution calculator,

P(x = 2) = 0.21

c) the expected number of customers from this sample that will buy a magazine is same as the mean.

mean = np

mean = 11 × 0.098 = 1.08

Answer 2

Using the binomial distribution, it is found that:

b) There is a 0.2088 = 20.88% probability that exactly 2 out of the first 11 customers buy a magazine.

c) The expected number of customers from this sample that will buy a magazine is of 1.078.

For each customer, there are only two possible outcomes, either they buy a magazine, or they do not. The probability of a customer buying a magazine is independent of any other customer, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• 9.8% of his customers buy a magazine, hence $p = 0.098$

• A sample of 11 customers is taken, hence $n = 11$.

Item b:

The probability is P(X = 2), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{11,2}.(0.098)^{2}.(0.902)^{9} = 0.2088$

There is a 0.2088 = 20.88% probability that exactly 2 out of the first 11 customers buy a magazine.

Item c:

The expected value of the binomial distribution is given by:

$E(X) = np$

Hence:

$E(X) = 11(0.098) = 1.078$

The expected number of customers from this sample that will buy a magazine is of 1.078.

To learn more about the binomial distribution, you can check https://brainly.com/question/24863377