Chapter PV, Problem 23RE

(a)

To determine

To explain why a skewed model makes sense here.

Explanation of Solution

It is given in the question that organizers of a fishing tournament believe that the lake holds a sizable population of largemouth bass. They assume that the weights of these fish have a modal that is skewed to the right with a mean of $3.5$ pounds and a standard deviation of $2.2$ pounds. Thus, a skewed model makes sense here because most small largemouth bass will be released back into the lake but large ones will be kept.

(b)

To determine

To explain why you cannot determine the probability that a largemouth bass randomly selected from the lake weighs over three pounds.

Explanation of Solution

It is given in the question that organizers of a fishing tournament believe that the lake holds a sizable population of largemouth bass. They assume that the weights of these fish have a modal that is skewed to the right with a mean of $3.5$ pounds and a standard deviation of $2.2$ pounds. Thus, we cannot determine the probability that a largemouth bass randomly selected from the lake weighs over three pounds because the distribution of this modal is not normal but it is skewed to the right as given.

(c)

To determine

To explain can you find out the probability that someone’s catch averages over three pounds.

Explanation of Solution

It is given in the question that organizers of a fishing tournament believe that the lake holds a sizable population of largemouth bass. They assume that the weights of these fish have a modal that is skewed to the right with a mean of $3.5$ pounds and a standard deviation of $2.2$ pounds. Thus, no, we cannot find out the probability that someone’s catch averages over three pounds because a sample size of five is not large enough sample to say the sampling modal for the mean is approximately normal, with a skewed distribution.

(d)

To determine

To find out what is the probability that the total catch of $60$ fish averaged more than three pounds.

Answer to Problem 23RE

  $0.9608$ .

Explanation of Solution

It is given in the question that organizers of a fishing tournament believe that the lake holds a sizable population of largemouth bass. They assume that the weights of these fish have a modal that is skewed to the right with a mean of $3.5$ pounds and a standard deviation of $2.2$ pounds. Thus, the probability that the total catch of $60$ fish averaged more than three pounds is calculated as:

  $\begin{array}{l}\text{Mean}=np\\ =3.5\times 60\\ =210\\ \text{Standard deviation}=\sqrt{npq}\\ =\sqrt{60\times {2.2}^2}\\ =17.04\end{array}$

The average number of fish is as:

  $\begin{array}{l}=3\times 60\\ =180\end{array}$

Then the z -score will be as:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{180-210}{17.04}\\ =-1.76\end{array}$

And then the probability is calculated by the table and the calculator as:

  $\begin{array}{l}Normalcdf(-1.76,99.9)\\ =0.9608\end{array}$