Question 1107808
 The television show Found has been successful for many years. That show recently had a share of 19, meaning that among the TV sets in use, 19% were tuned to Found. Assume that an advertiser wants to verify that 19% 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 Found broadcast.
P(watch) = 0.19 ; p(don't watch) = 0.81
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Find the probability that none of the households are tuned to Found. 
P(none) = (0.81)^14 = 0.0523
(round answer to 4 decimal places)
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Find the probability that at least one household is tuned to Found. 
P(at least one) = 1-P(none) = 0.9477
(round answer to 4 decimal places)
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Find the probability that at most one household is tuned to Found. 
P(at most one) = binomcdf(14,0.19,1) = 0.2242
(round answer to 4 decimal places)
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If at most one household is tuned to Found, does it appear that the 19% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Found unusual?)
Not unusual.
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Cheers,
Stan H.
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yes, it is wrong
no, it is not wrong