Answer:
a) 0.0523 = 5.23%
b) 0.9477 = 94.77%
c) 0.2242 = 22.42%
d) YES
Step-by-step explanation:
This situation could be modeled with a binomial distribution where
The probability of getting exactly k “successes” in n trials is given by
[tex]P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}[/tex]
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)
[tex]P(X=0)=\binom{14}{0}0.19^0(0.81)^{14}=0.0523=5.23\%[/tex]
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)
[tex]P(X=0)+P(X=1)=0.0523+\binom{14}{1}0.19^1(0.81)^{13}=0.2242=22.42\%[/tex]
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.
