1. [Hint: Let = 0.02 as shown in the table for no free premium channels.] (5 poi

Question

1. [Hint: Let = 0.02 as shown in the table for no free premium channels.] (5 points/part)

a.) P(X < 3) =

b.) P(X = 0) + P(X = 1) =

c.) P(X > 4) = 1 – P(X 4) =

d.) [Hint: Calculate P(X 4) and compare it to your answer in c. Basically, the question is asking whether it is less likely or more likely that you will get four or more subscriptions in a sample of 50 prospective customers, given that the probability of a subscription is estimated as 0.02.] Explain your reasoning.

2. [Hint: Let = 0.06 as shown in the table for two free premium channels.] (5 points/part)

a. P(X < 3) =

b. P(X = 0) + P(X = 1)

c. P(X > 4) = 1 – P(X 4).

d. [Hint: This question is asking you to compare the likelihood of your getting 4 or more subscribers in a sample of 50, given that the probability of a subscription is estimated as 0.06. Address sample proportions not pi values.] Talk about the comparison of probabilities in your explanation.

3. Study the table given and answer the question after you calculate four possibilities. (5 points)

a.) 1 free premium channel, 5 subscriptions, pi = 0.04,

P(X 5) =

b.) 3 free premium channels, 6 subscriptions, pi = 0.07,

P(X 6) =

3. [continued]

c. (5 points) 4 free premium channels, 6 subscriptions, = 0.08,

P(X 6) =

d. 5 free premium channels, 7 subscriptions, = 0.085,

P(X 7) =

e.) Given all the evidence that you have collected from all your surveys of 50 prospective customers, answer the question with full explanation. Use this entire assignment not only answers to #3. [Hint: Consider AMS’s profits in your answer.] How many free premium channels should the research director recommend for inclusion in the 3-For-All service?

Explanation / Answer

Solution:-
1) a.    P(X< 3) = 0.921572
Probability of less than three subscriptions is 0.921572.

b.   P(X = 0) + P(X = 1) =0.735771

Probability of zero or one subscription is 0.7357

c.   P(X > 4) = 1 – P(X 4) =0.00321

The probability of getting more than 4 subscriptions = 0.00321.

d. P(X 4) =0.017758
The probability of getting 4 or more subscriptions = 0.017758

Comparing with c, p(x>4) = p(x>=5)

P(x>4) is less than P(X 4)
Therefore, the probability of 5 or more subscriptions is less as compared to 4 or more subscriptions. It is more likely that you will get 4 or more subscriptions as compared to 5 or more subscriptions.

2) a. P(X < 3) = 0.416246
Probability of less than three subscriptions is 0.416246.

b. P(X = 0) + P(X = 1) = 0.190003
Probability of zero or one subscription is 0.190003
  
c. P(X > 4) = 1 – P(X 4).= 0.179404  
The probability of getting more than 4 subscriptions =0.179404.

d. P (x>=4) = 0.352697
Comparing to C. We are more likely to have 4 or more subscribers as compared to 5 or more (greater than 4) subscribers.

3) a. P(X 5) = 0.048971

b. P(X 6) = 0.135046

c. P(X 6) = 0.208126

d. P(X 7) =0.129051

e.   
service probability of subscription
0 free   0.02
1 free   0.04
2 free   0.06
3 free   0.07
4 free   0.08
5 free   0.085

As free channels increase, the subscription also increases.

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