Chapter 6.1, Problem 13P

Fishing: Trout The following data are based on information taken from Daily Creel Summary, published by the Paiute Indian Nation. Pyramid Lake. Nevada. Movie stars and U.S. presidents have fished Pyramid Lake. It is one of the best places in the lower 48 states to catch trophy cutthroat trout. In this table, x = number of fish caught in a 6-hour period. The percentage data are the percentages of fishermen who catch x fish in a 6-hour period while fishing from shore.

X	0	1	2	3	4 or more
%	44%	36%	15%	4%	1%

(a) Convert the percentages to probabilities and make a histogram of the probability distribution.

(b) Find the probability that a fisherman selected at random fishing from shore catches one or more fish in a 6-hour period.

(c) Find the probability that a fisherman selected at random fishing from shore catches two or more fish in a 6-hour period.

(d) Compute $\mu$, the expected value of the number of fish caught per fisherman in a 6-hour period (round 4 or more to 4).

(c) Compute $\sigma$, the standard deviation of the number of fish caught per fisherman in a 6-hour period (round 4 or more to 4).

To determine

The provided percentage in terms of probability and histogram.

Answer to Problem 13P

Solution: The table that shows the percentage in terms of probability is shown below:

x	0	1	2	3	4 or more
P(x)	0.44	0.36	0.15	0.04	0.01

Explanation of Solution

Given: The following table is given as:

x	0	1	2	3	4 or more
Percentage	44%	36%	15%	4%	1%

Calculation: The percentage can be converted into probability by dividing the percentage by 100.

For $x=0$, the probability is:

$\begin{array}{c}\text{Probability = }\frac{\text{44}}{100}\\ =0.44\end{array}$

Similarly, the probabilities for the values of $x=1,2,3,4$ can be obtained. The obtained probabilities are shown below in the form of a table:

x	0	1	2	3	4 or more
P(x)	0.44	0.36	0.15	0.04	0.01

To graph the data, follow the steps given below in MS Excel:

Step 1: Enter the data into an MS Excel sheet. The screenshot is given below:

Step 2: Select the data and click on ‘Insert’. Go to ‘charts’ and select the ‘Clustered Column’ option as the chart type.

Step 3: Select the first plot and click the ‘add chart element’ option that is provided in the left-hand corner of the menu bar. Insert the ‘Chart title’. The histogram for the provided data is shown below.

Interpretation: It can be seen that the histogram has a long right tail. So, it is skewed to the right.

To determine

The probability that a fisherman who selected fishes at random from the shore, catches one or more fish in a 6-hour period.

Answer to Problem 13P

Solution: The probability is 0.56.

Explanation of Solution

Given: The probability distribution table is given below:

x	0	1	2	3	4 or more
P(x)	0.44	0.36	0.15	0.04	0.01

Calculation: The probability that a fisherman who selected fishes at random from the shore, catches one or more fish in a 6-hour period can be obtained by:

$P(1\text{or more})=1-P(0)$

According to the probability distribution table, the value of the probability $P\left(0\right)$ is 0.44. Thus,

$\begin{array}{c}P(1\text{or more})=1-P(0)\\ =1-0.44\\ =0.56\end{array}$

Interpretation: There is a 56% chance of selecting the fishes at random from the shore, catches one or more fish in a 6-hour period.

To determine

The probability that a fisherman who selected fishes from the shore, catches two or more fish in a 6-hour period.

Answer to Problem 13P

Solution: The probability is 0.20.

Explanation of Solution

Given: The probability distribution table is given below:

x	0	1	2	3	4 or more
P(x)	0.44	0.36	0.15	0.04	0.01

Calculation: The probability that that a fisherman who selected fishes at random from the shore catches two or more fish in a 6-hour period can be obtained by:

$P\left(2\text{or more}\right)=P\left(2\right)+P\left(3\right)+P\left(4\text{or more}\right)$

According to the probability distribution table, the probabilities- $P\left(2\right)$, $P\left(3\right)$, and $P\left(4\text{ or more}\right)$ are 0.15, 0.04, and 0.01 respectively. Thus,

$\begin{array}{c}P\left(2\text{or more}\right)=P\left(2\right)+P\left(3\right)+P\left(4\text{or more}\right)\\ =0.15+0.04+0.01\\ =0.20\end{array}$

Interpretation: There is a 20% chance of selecting the fishes at random from the shore, catches two or more fish in a 6-hour period.

To determine

The expected value ‘$\mu$’.

Answer to Problem 13P

Solution: The expected value ‘$\mu$’ is 0.82.

Explanation of Solution

Given: The probability distribution table is given below:

x	0	1	2	3	4 or more
P(x)	0.44	0.36	0.15	0.04	0.01

Calculation: The formula to calculate the expected value is:

$\mu ={\displaystyle \sum _i{x}_{i}p(x_i)}$

Here, $x_i$ is the random variable and $p(x_i)$ is its corresponding probability. Substitute the values in the above formula. Thus:

$\begin{array}{c}\mu ={\displaystyle \sum xP\left(x\right)}\\ =0\times \left(0.44\right)+1\times \left(0.36\right)+2\times \left(0.15\right)+3\times \left(0.04\right)+4\times \left(0.01\right)\\ =0.82\end{array}$

Hence, the expected value is 0.82.

To determine

The standard deviation ‘$\sigma$’ of the number of fishes caught per fisherman in a 6-hour period.

Answer to Problem 13P

Solution: The standard deviation ‘$\sigma$’ is 0.8986.

Explanation of Solution

Given: The probability distribution table is given below:

x	0	1	2	3	4 or more
P(x)	0.44	0.36	0.15	0.04	0.01

Calculation: The formula to calculate the standard deviation is:

$\sigma =\sqrt{{\displaystyle \sum {(x-\mu )}^{2}P(x)}}$

Here, $\mu$ is the expected value and $P(x)$ is the probability. Substitute the values in the above formula:

$\begin{array}{c}\sigma =\sqrt{{\left(0-0.82\right)}^2\left(0.44\right)+{\left(1-0.82\right)}^2\left(0.36\right)+{\left(2-0.82\right)}^2\left(0.15\right)+{\left(3-0.82\right)}^2\left(0.04\right)+{\left(3.18\right)}^2\left(0.01\right)}\\ =\sqrt{0.295856+0.011664+0.20886+0.19009+0.10112}\\ =\sqrt{0.8076}\\ =0.8986\end{array}$

Hence, the standard deviation is 0.899.