Chapter 10.3, Problem 18E

To determine

To calculate:

The group should be awarded the prize

Answer to Problem 18E

The group $1$ has the best probability of getting awarded the prize.

Explanation of Solution

Given information:

  

Calculation:

This question deals with study of two way table with provided the values of experiment is being considered.

	Exceed Expectations	Meet Expectations	Total
Group $1$	$12$	$4$	$16$
Group $2$	$8$	$5$	$13$
Group $3$	$9$	$6$	$15$
Total	$29$	$15$	$44$

Use findings to make the two ways table shows the marginal relative frequencies and the joint. There are total $41$ observations in this table.

		Result
		Exceed Expectations	Meet Expectations	Total
Group	$1$	$0.273$	$0.091$	$0.364$
	$2$	$0.182$	$0.114$	$0.295$
	$3$	$0.205$	$0.136$	$0.341$
	Total	$0.659$	$0.341$	$1$

The conditional probabilities by dividing the joint relative frequency in the column “on the time” by the marginal relative frequency in its row:

Probability of on time for route $1$

  $P\left(C_1\right)=\frac{P\left(A_1\right)}{P\left(B_1\right)}$

Where,

  $P\left(A_1\right)$ : Probability of route $1$ and on time.

  $P\left(B_1\right)$ : Probability of Route $1$

  $P\left(C_1\right)$ : Probability of route $1$ for on time.

  $\begin{array}{l}P\left(C_1\right)=\frac{0.171}{0.268}\\ P\left(C_1\right)=0.638\end{array}$

Probability of on time for route $2$

  $P\left(C_2\right)=\frac{P\left(A_2\right)}{P\left(B_2\right)}$

Where,

  $P\left(A_2\right)$ : Probability of route $2$ and on time.

  $P\left(B_2\right)$ : Probability of Route $2$

  $P\left(C_2\right)$ : Probability of route $2$ for on time.

  $\begin{array}{l}P\left(C_2\right)=\frac{0.268}{0.342}\\ P\left(C_2\right)=0.784\end{array}$

Probability of on time for route $3$

  $P\left(C_3\right)=\frac{P\left(A_3\right)}{P\left(B_3\right)}$

Where,

  $P\left(A_3\right)$ : Probability of route $3$ and on time.

  $P\left(B_3\right)$ : Probability of Route $3$

  $P\left(C_3\right)$ : Probability of route $3$ for on time.

  $\begin{array}{l}P\left(C_3\right)=\frac{0.205}{0.341}\\ P\left(C_3\right)=0.601\end{array}$

As per the observation, the group $1$ has the best probability of getting awarded the prize.