Question

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.Find the probability that none of the households are tuned to Found.P(none) =(round answer to 4 decimal places)Find the probability that at least one household is tuned to Found.P(at least one) =(round answer to 4 decimal places)Find the probability that at most one household is tuned to Found.P(at most one) =(round answer to 4 decimal places)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?)yes, it is wrongno, it is not wrong

Answer 1

a) 0.0523 = 5.23%

b) 0.9477 = 94.77%

c) 0.2242 = 22.42%

d) YES

Explanation:

This situation could be modeled with a binomial distribution where

The probability of getting exactly k “successes” in n trials is given by

`P(X=k)=\binom{n}{k}p^k(1-p)^(n-k)`

Where p is the probability of “success”.

In this case we can assume that “success” is the fact that the household interviewed is tuned to Found.

So, p=0.19 and n = 14 (the households interviewed).

a)

The probability that none of the households are tuned to Found is P(X=0)

`P(X=0)=\binom{14}{0}0.19^0(0.81)^(14)=0.0523=5.23\%`

b)

The probability that at least one household is tuned to Found is

1- P(X=0) = 1-0.0523 = 0.9477 = 94.77%

c)

The probability that at most one household is tuned to Found is P(X=0)+P(X=1)

`P(X=0)+P(X=1)=0.0523+\binom{14}{1}0.19^1(0.81)^(13)=0.2242=22.42\%`

d)

According to this sample, the probability that more than one household is tuned to Found would be 100%-22.42% = 77.58%, so it does appear that the 19% share value is wrong.