The television show CSI: Shoboygan has been successful for many years. That show recently had a share of 18, meaning that among the TV sets in use, 18% were tuned to CSI: Shoboygan. Assume that an advertiser wants to verify that 18% share value by conducting its own survey, and a pilot survey begins with 14 households have TV sets in use at the time of a CSI: Shoboygan broadcast. Find the probability that none of the households are tuned to CSI: Shoboygan. P(none) = Find the probability that at least one household is tuned to CSI: Shoboygan. P(at least one) = Find the probability that at most one household is tuned to CSI: Shoboygan. P(at most one) = If at most one household is tuned to CSI: Shoboygan, does it appear that the 18% share value is wrong? (Hint: Is the occurrence of at most one household tuned to CSI: Shoboygan unusual?)
X ~ Binomial (n,p)
Where n = 14, p = 0.18
Binomial probability distribution is
P(X) = nCx px ( 1 - p)n-x
a)
P( X = 0) = ( 1 - 0.18)14
= 0.0621
b)
P( X >= 1) = 1 - P( X = 0)
= 1 - 0.0621
= 0.9379
c)
P( X <= 1) = P(X = 0) + P( X = 1)
= ( 1 - 0.18)14 + 0.18 * ( 1 - 0.18)13
= 0.2531
Since 0.2531 is greater than 0.05, the event is not unusual.
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