Number of Channels Probabilty of Subscriptions 2. [Hint: Let pi = 0.06 as shown

Question

Number of Channels                                                Probabilty of Subscriptions

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

Please provide explanations for all inputs

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.

Explanation / Answer

a)

Note that P(fewer than x) = P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    50      
p = the probability of a success =    0.06      
x = our critical value of successes =    3      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   2   ) =    0.416246472
          
Which is also          
          
P(fewer than   3   ) =    0.416246472 [ANSWER]

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b)

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    50      
p = the probability of a success =    0.06      
x = the maximum number of successes =    1      
          
Then the cumulative probability is          
          
P(at most   1   ) =    0.190003258 [ANSWER]

***********

c)

Note that P(more than x) = 1 - P(at most x).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    50      
p = the probability of a success =    0.06      
x = our critical value of successes =    4      
          
Then the cumulative probability of P(at most x) from a table/technology is          
          
P(at most   4   ) =    0.820596047
          
Thus, the probability of at least   5   successes is  
          
P(more than   4   ) =    0.179403953 [ANSWER]

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d)

We see that the mean of this distrbution is E(x) = n p = 50*0.06 = 3. Hence, we expect the probabilities for values near x = 3 to be quite high.

That is why part a) probability is greater than both part b and c, because part a includes values nearer to the mean that b and c. Also, as parts b and c include values that are about as far from the mean, they are of similar probabilities.

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