Question

The television show 50 Minutes 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 50 Minutes. Assume that an advertiser wants to verify that 29% share value by conducting its own survey, and a pilot survey begins with 15 households have TV sets in use at the time of a 50 Minutes broadcast.

Find the probability that none of the households are tuned to 50 Minutes. P(none) = (round answer to 4 decimal places)
Find the probability that at least one household is tuned to 50 Minutes. P(at least one) = (round answer to 4 decimal places)
Find the probability that at most one household is tuned to 50 Minutes. P(at most one) = (round answer to 4 decimal places)
If at most one household is tuned to 50 Minutes, does it appear that the 19% share value is wrong? (Hint: Is the occurrence of at most one household tuned to 50 Minutes unusual?)
A. yes, it is wrongB. no, it is not wrong

Answer 1

(a) P (none) = 0.005873

(b) P ( at least one household ) = 0.9941

(c) P ( at most one household ) = 0.9960

(d) not wrong

Explanation:

Given data

probability P = 29% = 0.29

no of sample n = 15

tv show X = 50

to find out

the probability that none of the households

the probability that at least one household

the probability that at most one household

If at most one household is tuned to 50 Minutes, does it appear that the 19% share value is wrong

solution

the probability that none of the households is tuned to 50 Minutes is

P (none) = 10C0 ×
`P^(0)` ×
`(1 - 0.29)^(15)`

P (none) =
`(0.71)^(15)`

P (none) = 0.005873

and the probability that at least one household is tuned to 50 Minutes is

P ( at least one household ) = 1 - P(none)

P ( at least one household ) = 1 - 0.005873

P ( at least one household ) = 1 - 0.005873

P ( at least one household ) = 0.9941

the probability that at most one household is tuned to 50 Minutes is

P ( at most one household ) = P (none) + P (x=1)

P ( at most one household ) =0.9941 + 10C1 ×
`(0.23)^(1)` ×
`(0.71)^(14)`

P ( at most one household ) = 0.9960

and in last part If at most one household is tuned to 50 Minutes it appear that the 19% share value is not wrong

because P( at most one household ) is not zero