Question

The television show Found has been successful for many years. That show recently had a share of 29, meaning that among the TV sets in use, 29% were tuned to Found. Assume that an advertiser wants to verify that 29% share value by conducting its own survey, and a pilot survey begins with 12 households have TV sets in use at the time of a Found broadcast.

Find the probability that none of the households are tuned to Found.
P(none) =

Find the probability that at least one household is tuned to Found.
P(at least one) =

Find the probability that at most one household is tuned to Found.
P(at most one) =

If at most one household is tuned to Found, does it appear that the 29% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Found unusual?)

• yes, it is wrong
• no, it is not wrong

Answer 1

Here sample size = n = 12

probability of success = p = 0.29

Here x ~ BINOMIAL (n = 12, p = 0.29)

Find the probability that none of the households are tuned to Found.

P(x = 0) = 12C0 (0.29)0 (0.71)12  = 0.0164

Find the probability that at least one household is tuned to Found.

P(at least one) = 1 - P(None of the households) = 1 - 0.0164 = 0.9836

probability that at most one household is tuned to Found = P(x 1)

= P(x = 0) + P(x = 1)

= 12C0 (0.29)0 (0.71)12  + 12C1 (0.29)1 (0.71)11

= 0.0164 + 0.0804 = 0.0968

As p value is greater than 0.0968, so it is not unusual.so the 29% share value is not wrong