Chapter 6.3, Problem 89E

(a)

To determine

Probability for 14 or more students would prefer the last kiss tasted.

Answer to Problem 89E

Probability,

  $P\left(X\ge 14\right)=0.0000$

Explanation of Solution

Given information:

Number of trials, $n=22$

Probability of success, $p=0.20$

Calculations:

According to the binomial probability,

  $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)\cdot p^k\cdot {\left(1-p\right)}^{n-k}$

Addition rule for mutually exclusive event:

  $P\left(A\cup B\right)=P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$

At $k=14$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=14\right)=\left(\begin{array}{c}22\\ 14\end{array}\right)\cdot {\left(0.20\right)}^{14}\cdot {\left(1-0.20\right)}^{22-14}\\ =\frac{22!}{14!\left(22-14\right)!}\cdot {\left(0.20\right)}^{14}\cdot {\left(0.80\right)}^8\\ \approx 0.0000\end{array}$

At $k=15$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=15\right)=\left(\begin{array}{c}22\\ 15\end{array}\right)\cdot {\left(0.20\right)}^{15}\cdot {\left(1-0.20\right)}^{22-15}\\ =\frac{22!}{15!\left(22-15\right)!}\cdot {\left(0.20\right)}^{15}\cdot {\left(0.80\right)}^7\\ \approx 0.0000\end{array}$

At $k=16$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=16\right)=\left(\begin{array}{c}22\\ 16\end{array}\right)\cdot {\left(0.20\right)}^{16}\cdot {\left(1-0.20\right)}^{22-16}\\ =\frac{22!}{16!\left(22-16\right)!}\cdot {\left(0.20\right)}^{16}\cdot {\left(0.80\right)}^6\\ \approx 0.0000\end{array}$

At $k=17$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=17\right)=\left(\begin{array}{c}22\\ 17\end{array}\right)\cdot {\left(0.20\right)}^{17}\cdot {\left(1-0.20\right)}^{22-17}\\ =\frac{22!}{17!\left(22-17\right)!}\cdot {\left(0.20\right)}^{17}\cdot {\left(0.80\right)}^5\\ \approx 0.0000\end{array}$

At $k=18$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=18\right)=\left(\begin{array}{c}22\\ 18\end{array}\right)\cdot {\left(0.20\right)}^{18}\cdot {\left(1-0.20\right)}^{22-18}\\ =\frac{22!}{18!\left(22-18\right)!}\cdot {\left(0.20\right)}^{18}\cdot {\left(0.80\right)}^4\\ \approx 0.0000\end{array}$

At $k=19$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=19\right)=\left(\begin{array}{c}22\\ 19\end{array}\right)\cdot {\left(0.20\right)}^{19}\cdot {\left(1-0.20\right)}^{22-19}\\ =\frac{22!}{19!\left(22-19\right)!}\cdot {\left(0.20\right)}^{19}\cdot {\left(0.80\right)}^3\\ \approx 0.0000\end{array}$

At $k=20$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=20\right)=\left(\begin{array}{c}22\\ 20\end{array}\right)\cdot {\left(0.20\right)}^{20}\cdot {\left(1-0.20\right)}^{22-20}\\ =\frac{22!}{20!\left(22-20\right)!}\cdot {\left(0.20\right)}^{20}\cdot {\left(0.80\right)}^2\\ \approx 0.0000\end{array}$

At $k=21$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=21\right)=\left(\begin{array}{c}22\\ 21\end{array}\right)\cdot {\left(0.20\right)}^{21}\cdot {\left(1-0.20\right)}^{22-21}\\ =\frac{22!}{21!\left(22-21\right)!}\cdot {\left(0.20\right)}^{21}\cdot {\left(0.80\right)}^1\\ \approx 0.0000\end{array}$

At $k=22$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=22\right)=\left(\begin{array}{c}22\\ 22\end{array}\right)\cdot {\left(0.20\right)}^{22}\cdot {\left(1-0.20\right)}^{22-22}\\ =\frac{22!}{14!\left(22-14\right)!}\cdot {\left(0.20\right)}^{22}\cdot {\left(0.80\right)}^0\\ \approx 0.0000\end{array}$

Since two different numbers of successes are impossible on same simulation.

Apply addition rule for mutually exclusive events:

  $\begin{array}{l}P\left(X\ge 14\right)=P\left(X=14\right)+P\left(X=15\right)+P\left(X=16\right)+P\left(X=17\right)+P\left(X=18\right)\\ +P\left(X=19\right)+P\left(X=20\right)+P\left(X=21\right)+P\left(X=22\right)\\ =0.0000+0.0000+0.0000+0.0000+0.0000+0.0000\\ +0.0000+0.0000+0.0000\\ =0.0000\\ =0.00\%\end{array}$

(b)

To determine

Whether the statement gives convincing evidence for the participants have a preference for the last thing they taste.

Answer to Problem 89E

Yes, there is convincing evidence that the participants have a preference for the last thing they taste.

Explanation of Solution

Given information:

Number of trials, $n=22$

Probability of success, $p=0.20$

Calculations:

When the probability is less than 0.05, it is considered to be small.

In this case,

Note that

The probability is small enough.

Thus,

It is unlikely to obtain 14 students who gave the final kiss the highest rating among the 22 students.

This implies

There is convincing evidence that the participants have a preference for the last thing they taste.