Question

the television show degenerate housewives has been successful for many years. that show recently had a share of 21, meaning that among the tv sets in use, 21% were tuned to degenerate housewives. assume that an advertiser wants to verify that 21% 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 degenerate housewives broadcast. (round answers to four decimal places.) find the probability that none of the households are tuned to degenerate housewives. ( p(\text{none}) = ) find the probability that at least one household is tuned to degenerate housewives. ( p(\text{at least one}) = ) find the probability that at most one household is tuned to degenerate housewives. ( p(\text{at most one}) = ) if at most one household is tuned to degenerate housewives, does it appear that the 21% share value is wrong? (hint: is the occurrence of at most one household tuned to degenerate housewives unusual?) ( \bigcirc ) no, it is not wrong ( \bigcirc ) yes, it is wrong question help: ( \bigcirc ) post to forum

Explanation:

Step1: Identify the probability of a household not being tuned

The probability that a household is tuned to the show \(p = 0.21\), so the probability that a household is not tuned \(q=1 - p=1 - 0.21 = 0.79\). The number of households \(n = 14\).
The probability mass function of a binomial distribution is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate \(P(\text{none})\)

For \(k = 0\), \(C(14,0)=\frac{14!}{0!(14-0)!}=1\)
\(P(X = 0)=C(14,0)\times(0.21)^{0}\times(0.79)^{14}\)
\(P(X = 0)=1\times1\times0.79^{14}\)
\(0.79^{14}\approx0.0341\)

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

Using the complement rule \(P(X\geq1)=1 - P(X = 0)\)
Since \(P(X = 0)\approx0.0341\), then \(P(X\geq1)=1 - 0.0341=0.9659\)

Step4: Calculate \(P(\text{at most one})\)

\(P(X\leq1)=P(X = 0)+P(X = 1)\)
For \(k = 1\), \(C(14,1)=\frac{14!}{1!(14 - 1)!}=\frac{14!}{1!13!}=14\)
\(P(X = 1)=C(14,1)\times(0.21)^{1}\times(0.79)^{13}\)
\(P(X = 1)=14\times0.21\times0.79^{13}\)
\(0.79^{13}\approx0.0432\), \(14\times0.21\times0.0432 = 14\times0.009072=0.1270\)
\(P(X\leq1)=0.0341+0.1270 = 0.1611\)

Step5: Determine if \(21\%\) share is wrong

A probability less than \(0.05\) is considered unusual. Since \(P(X\leq1)=0.1611>0.05\), it is not unusual.

Answer:

\(P(\text{none}) = 0.0341\)
\(P(\text{at least one})=0.9659\)
\(P(\text{at most one})=0.1611\)
\(P(\text{at most one})\) is not unusual, so \(\text{no, it is not wrong}\)