To increase sales, a cell phone store began giving a free month of service on some cell plan purchases. Customers were unaware whether they would receive the free month until after purchase. The store manager claimed that one in eight customers received the free month. Nine customers each buy a cell phone plan. Let X = the number of customers that receive a free month of cell phone service.

Part A: Is X a binomial random variable? Explain. (3 points)

Part B: What is the mean and standard deviation of X? Provide an interpretation for each value in context. (4 points)

Part C: Three of the nine customers receive a free month of service. Is the store's claim accurate? Compute P(X ≥ 3) and use the result to justify your answer. (3 points)

A) For each customer, there are only two possible outcomes, either they receive a free month of cell phone service, or they do not. The probability of a customer receiving a free month of cell phone service is independent of any other customer, hence, X is a binomial variable.

B) The mean of X is of 1.125, while the standard deviation is of 0.99. It means that in samples of nine customers, around 1.125 are expected to receive a free month of cell phone service, with a spread of 0.99 plus/minus customers.

C) [tex]P(X \geq 3) = 0.0918[/tex], which is greater than 0.05, hence the store's claim is accurate.

What is the binomial probability distribution formula?

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

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.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Item a:

For each customer, there are only two possible outcomes, either they receive a free month of cell phone service, or they do not. The probability of a customer receiving a free month of cell phone service is independent of any other customer, hence, X is a binomial variable.

Item b:

The store manager claimed that one in eight customers received the free month, hence [tex]p = \frac{1}{8} = 0.125[/tex].

Nine customers each buy a cell phone plan, hence [tex]n = 9[/tex].

Then:

[tex]E(X) = np = 9(0.125) = 1.125[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{9(0.125)(0.875)} = 0.99[/tex]

The mean of X is of 1.125, while the standard deviation is of 0.99. It means that in samples of nine customers, around 1.125 are expected to receive a free month of cell phone service, with a spread of 0.99 plus/minus customers.

Item c:

The probability is:

[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]

In which:

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{9,0}.(0.125)^{0}.(0.875)^{9} = 0.3007[/tex]

[tex]P(X = 1) = C_{9,1}.(0.125)^{1}.(0.875)^{8} = 0.3866[/tex]

[tex]P(X = 2) = C_{9,2}.(0.125)^{2}.(0.875)^{7} = 0.2209[/tex]

Then:

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3007 + 0.3866 + 0.2209 = 0.9082[/tex]

[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.9082 = 0.0918[/tex]

[tex]P(X \geq 3) = 0.0918[/tex], which is greater than 0.05, hence the store's claim is accurate.

You can learn more about the binomial distribution at https://brainly.com/question/24863377