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

Question

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.

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? Please show the step by step explanations.

PLEASE SHOW THE STEP BY STEP EXPLANATIONS. THANKS! Their is no table for this assignments. It is asking for explanations or a PHStat 4 printout. Here is the chart I think you guys were talking about. I tried copying the chart my prof made in excel but this was the best I could do. Could I possible email everything to you, including the chart. This is due @ 5:30 pm tomorrow.

Sp Assign 1 Data Sample size 50 Probability of an event of interest 0.01 Statistics Mean 0.5 Variance 0.4950 Standard deviation 0.7036 Binomial Probabilities Table X P(X) P(<=X) P(<X) P(>X) P(>=X) 0 0.6050 0.6050 0.0000 0.3950 1.0000 1 0.3056 0.9106 0.6050 0.0894 0.3950 2 0.0756 0.9862 0.9106 0.0138 0.0894 3 0.0122 0.9984 0.9862 0.0016 0.0138 4 0.0015 0.9999 0.9984 0.0001 0.0016 5 0.0001 1.0000 0.9999 0.0000 0.0001 6 0.0000 1.0000 1.0000 0.0000 0.0000 7 0.0000 1.0000 1.0000 0.0000 0.0000 8 0.0000 1.0000 1.0000 0.0000 0.0000 9 0.0000 1.0000 1.0000 0.0000 0.0000 10 0.0000 1.0000 1.0000 0.0000 0.0000 11 0.0000 1.0000 1.0000 0.0000 0.0000 12 0.0000 1.0000 1.0000 0.0000 0.0000 13 0.0000 1.0000 1.0000 0.0000 0.0000 14 0.0000 1.0000 1.0000 0.0000 0.0000 15 0.0000 1.0000 1.0000 0.0000 0.0000 16 0.0000 1.0000 1.0000 0.0000 0.0000 17 0.0000 1.0000 1.0000 0.0000 0.0000 18 0.0000 1.0000 1.0000 0.0000 0.0000 19 0.0000 1.0000 1.0000 0.0000 0.0000 20 0.0000 1.0000 1.0000 0.0000 0.0000 21 0.0000 1.0000 1.0000 0.0000 0.0000 22 0.0000 1.0000 1.0000 0.0000 0.0000 23 0.0000 1.0000 1.0000 0.0000 0.0000 24 0.0000 1.0000 1.0000 0.0000 0.0000 25 0.0000 1.0000 1.0000 0.0000 0.0000 26 0.0000 1.0000 1.0000 0.0000 0.0000 27 0.0000 1.0000 1.0000 0.0000 0.0000 28 0.0000 1.0000 1.0000 0.0000 0.0000 29 0.0000 1.0000 1.0000 0.0000 0.0000 30 0.0000 1.0000 1.0000 0.0000 0.0000 31 0.0000 1.0000 1.0000 0.0000 0.0000 32 0.0000 1.0000 1.0000 0.0000 0.0000 33 0.0000 1.0000 1.0000 0.0000 0.0000 34 0.0000 1.0000 1.0000 0.0000 0.0000 35 0.0000 1.0000 1.0000 0.0000 0.0000 36 0.0000 1.0000 1.0000 0.0000 0.0000 37 0.0000 1.0000 1.0000 0.0000 0.0000 38 0.0000 1.0000 1.0000 0.0000 0.0000 39 0.0000 1.0000 1.0000 0.0000 0.0000 40 0.0000 1.0000 1.0000 0.0000 0.0000 41 0.0000 1.0000 1.0000 0.0000 0.0000 42 0.0000 1.0000 1.0000 0.0000 0.0000 43 0.0000 1.0000 1.0000 0.0000 0.0000 44 0.0000 1.0000 1.0000 0.0000 0.0000 45 0.0000 1.0000 1.0000 0.0000 0.0000 46 0.0000 1.0000 1.0000 0.0000 0.0000 47 0.0000 1.0000 1.0000 0.0000 0.0000 48 0.0000 1.0000 1.0000 0.0000 0.0000 49 0.0000 1.0000 1.0000 0.0000 0.0000 50 0.0000 1.0000 1.0000 0.0000 0.0000

Explanation / Answer

Let p = 0.06, and X has Binomail distribution with n= 50 and p = 0.06

a. P(X < 3) = 0.4162

b. P(X = 0) + P(X = 1) = 0.19

c. P(X > 4) = 1 – P(X 4) = 0.1794

Now for different values of p and fixed n = 50

a.) 1 free premium channel, 5 subscriptions, p = 0.04, P(X 5) = 0.0489

b. 3 free premium channels, 6 subscriptions, p = 0.07, P(X 6) = 0.1351

c. 4 free premium channels, 6 subscriptions,p = 0.08, P(X 6) = 0.2081

d. 5 free premium channels, 7 subscriptions, p = 0.085, P(X 7) = 0.1291

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