Question

the television show 50 minutes has been successful for many years. that show recently had a share of 31, which means, that among the tv sets in use, 31% were tuned to 50 minutes. an advertiser wants to verify that 31% share value by conducting its own survey, and a pilot survey begins with 11 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) = find the probability that at least one household is tuned to 50 minutes. p(at least one) = find the probability that at most one household is tuned to 50 minutes. p(at most one) = if at most one household is tuned to 50 minutes, does it appear that the 31% share value is wrong? (hint: is the occurrence of at most one household tuned to 50 minutes unusual?) no, it is not wrong yes, it is wrong question help: message instructor

Explanation:

Step1: Calculate \(P(\text{none})\)

This is a binomial probability problem. The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 11\) (number of trials, i.e., households), \(k = 0\) (number of successes, i.e., households tuned to the show), and \(p=0.31\) (probability of a household being tuned to the show). \(C(n,k)=\frac{n!}{k!(n - k)!}\), and when \(k = 0\), \(C(11,0)=\frac{11!}{0!(11-0)!}=1\). Then \(P(X = 0)=(1)\times(0.31)^{0}\times(1 - 0.31)^{11-0}\). Since \(a^{0}=1\) for \(a
eq0\), \(P(X = 0)=(0.69)^{11}\approx0.0084\).

Step2: Calculate \(P(\text{at least one})\)

Use the complement rule \(P(X\geq1)=1 - P(X = 0)\). We know \(P(X = 0)\approx0.0084\), so \(P(X\geq1)=1-0.0084 = 0.9916\).

Step3: Calculate \(P(\text{at most one})\)

\(P(X\leq1)=P(X = 0)+P(X = 1)\). \(P(X = 0)=(0.69)^{11}\approx0.0084\). For \(P(X = 1)\), \(C(11,1)=\frac{11!}{1!(11 - 1)!}=11\), \(P(X = 1)=C(11,1)\times(0.31)^{1}\times(0.69)^{10}\). \(P(X = 1)=11\times0.31\times(0.69)^{10}\approx11\times0.31\times0.0122 = 0.0419\). Then \(P(X\leq1)=0.0084 + 0.0419=0.0503\).

Step4: Determine if \(31\%\) share is wrong

A probability is considered unusual if it is less than \(0.05\). Since \(P(X\leq1)=0.0503>0.05\) (very close to \(0.05\), but using a strict cut - off of \(0.05\)), it does not appear that the \(31\%\) share value is wrong.

Answer:

\(P(\text{none})\approx0.0084\)
\(P(\text{at least one})\approx0.9916\)
\(P(\text{at most one})\approx0.0503\)
no, it is not wrong