An industry representative claims that 44 percent of all satellite dish owners s
ID: 3230000 • Letter: A
Question
An industry representative claims that 44 percent of all satellite dish owners subscribe to at least one premium movie channel. In an attempt to justify this claim, the representative will poll a randomly selected sample of dish owners.
(a) Suppose that the representative's claim is true, and suppose that a sample of 4 dish owners is randomly selected. Assuming independence, use an appropriate formula to compute. (Do not round your intermediate calculation and round your answers to 4 decimal places.)
The probability that none of the dish owners in the sample subscribes to at least one premium movie channel.
.44^4=.0375
Probability
2 .The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.
Probability
(b) Suppose that the representative's claim is true, and suppose that a sample of 20 dish owners is randomly selected. Assuming independence, what is the probability that: (Do not round your intermediate calculation and round your answers to 4 decimal places.)
1. Nine or fewer dish owners in the sample subscribe to at least one premium movie channel?
Probability
2. More than 11 dish owners in the sample subscribe to at least one premium movie channel?
Probability
3. Fewer than five dish owners in the sample subscribe to at least one premium movie channel?
Probability
Explanation / Answer
(a) The probability that satellite dish owners subscribe to at least one premium movie channel, p = 0.44
The probability that satellite dish owners does not subscribe to at least one premium movie channel, q = 1 - 0.44 = 0.56
The probability that none of the dish owners in the sample subscribes to at least one premium movie channel.
= q^4 = 0.56^4 = 0.0983
The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.
= P(X>2) = P(X=3) + P(X=4) = C(4,3) p^3 q + C(4,4) p^4 (By binomial ditribution formula)
= 4 * 0.44^3 * 0.56 + 0.44^4 = 0.2283
(b) 1. Nine or fewer dish owners in the sample subscribe to at least one premium movie channel?
The probability of binomial distribution is
b(x; n, P) = nCx * Px * (1 - P)n - x
P(X < 9) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9)
= C(20,0) * 0.44^0 * 0.56^20 + C(20,1)* 0.44^1* 0.56^19 + C(20,2)* 0.44^2 *0.56^18 + C(20,3)* 0.44^3* 0.56^17 + C(20,4)* 0.44^4* 0.56^16 + C(20,5)* 0.44^5* 0.56^15 + C(20,6)* 0.44^6* 0.56^14 + C(20,7)* 0.44^7 *0.56^13 + C(20,8)* 0.44^8 *0.56^12 + C(20,9)* 0.44^9* 0.56^11
= 0.6264
2. P(X > 11) = P(X=12) + P(X=13) + P(X=14) + P(X=15) + P(X=16) + P(X=17) + P(X=18) + P(X=19) + P(X=20)
= C(20,12)* 0.44^12 *0.56^8 + C(20,13)* 0.44^13 *0.56^7 + C(20,14) *0.44^14 *0.56^6 + C(20,15)* 0.44^15* 0.56^5 + C(20,16)* 0.44^16* 0.56^4 + C(20,17)* 0.44^17* 0.56^3 + C(20,18)* 0.44^18* 0.56^2 + C(20,19)* 0.44^19* 0.56^1 + C(20,20)* 0.44^20* 0.56^0
= 0.1123
3. Fewer than five dish owners in the sample subscribe to at least one premium movie channel?
P(X < 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
= C(20,0) * 0.44^0 * 0.56^20 + C(20,1)* 0.44^1* 0.56^19 + C(20,2)* 0.44^2 *0.56^18 + C(20,3)* 0.44^3* 0.56^17 + C(20,4)* 0.44^4* 0.56^16
= 0.0660