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) Yes

b) The mean is 9/8

The standard deviation, σ, is 0.99

c) Very unlikely as P(X≥3) is  0.09189 << μ =0.125 .

Step-by-step explanation:

Yes, A binomial random variable is the amount of success obtained from a given number of independent trials where the  there are only two possible outcomes and the probability of success or failure for each trial remains the same.

For a binomial random variable, we have

the sample size given as n = number of trials = 9

p = The probability of success on a trial = 1/8

The mean is given as

n×p = 9 × 1/8 = 9/8

The standard deviation, σ, is given by

= = 0.99

c) It is unlikely that three of the nine customers receive a free month of service as

The probability of P(X≥3)

= 1-P(X<3)

P(X<3) = P(X=0) + P(X=1) + P(X=2)

= ₉C₀×(1/8)⁰×(7/8)⁹ + ₉C₁×(1/8)¹×(7/8)⁸ + ₉C₂×(1/8)²×(7/8)⁷

= 0.3006 + 0.38656 +0.221  = 0.908

Therefore, P(X≥3) = 1 - 0.908 = 0.09189  much lower than the mean.

what? ?

what type of question is thiss

step-by-step explanation:

explanation:

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