Chapter 3.1, Problem 4E

Observers in the Frogwatch Program The number of observers in the Frogwatch USA program (a wildlife conservation program dedicated to helping conserve frogs and toads) for the top 10 states with the most observers is 484, 483, 422, 396, 378, 352, 338, 331, 318, and 302. The top 10 states with the most active watchers list these numbers of visits: 634, 464, 406, 267, 219, 194, 191, 150, 130, and 114. Find the mean, median, mode, and midrange for the data. Compare the measures of central tendency for these two groups of data.

Source: www.nwf.org/frogwatch

To determine

The mean median mode and mid-range for observers and active watchers, and compare them.

Answer to Problem 4E

The mean for observers is 380.4 and for active watchers is 276.9. The median for observers is 365 and for active watchers is 206.5. The mid-range for the observers is 393 and for the active watchers is 374. For both observers and active watchers has no mode.

Explanation of Solution

Given Info:

The total number of observers and active watchers in 10 states is,

$484,483,422,396,378,352,338,331,318,302$ $634,464,406,267,219,194,191,150,130,114$

Calculation:

The formula to calculate the mean is,

$\begin{array}{c}\overline{\text{X}}=\frac{{\displaystyle \sum _1^{n}X}}{n}\\ =\frac{X_1+X_2+X_3+\mathrm{...}+X_n}{n}\end{array}$ (1)

Where, $X$ is the number of observers and active watchers and $n$ is the total number of states.

The sum of the 10 observers is give below:

$\begin{array}{c}{\displaystyle \sum _1^{10}X}=484+483+422+396+378+352+338+331+318+302\\ =3804\end{array}$

Substitute ${\displaystyle \sum _1^{10}X}=3804$ and $n=10$ in equation (1),

$\begin{array}{c}\text{Mean}=\frac{3804}{10}\\ =380.4\end{array}$

For active watchers-

The sum of the 10 active watchers is give below:

$\begin{array}{c}{\displaystyle \sum _1^{10}X}=634+464+406+267+219+194+191+150+130+114\\ =2769\end{array}$

Substitute ${\displaystyle \sum _1^{10}X}=2769$ and $n=10$ in equation (1),

$\begin{array}{c}\text{Mean}=\frac{2769}{10}\\ =276.9\end{array}$

Formula to calculate median when the number of terms is even,

$\text{Median}=\frac{{\left(\frac{n}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{n}{2}+1\right)}^{\text{th}}\text{value}}{2}$

For observers,

Substitute 10 for $n$ in the above formula.

$\begin{array}{c}\text{Median}=\frac{{\left(\frac{10}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{10}{2}+1\right)}^{\text{th}}\text{value}}{2}\\ =\frac{{\left(5\right)}^{\text{th}}\text{value}+{\left(6\right)}^{\text{th}}\text{value}}{2}\\ =\left(\frac{352+378}{2}\right)\\ =365\end{array}$

For active watchers,

Substitute 10 for $n$ in the above formula.

$\begin{array}{c}\text{Median}=\frac{{\left(\frac{10}{2}\right)}^{\text{th}}value+{\left(\frac{10}{2}+1\right)}^{\text{th}}value}{2}\\ =\frac{{\left(5\right)}^{\text{th}}value+{\left(6\right)}^{\text{th}}value}{2}\\ =\frac{194+219}{2}\\ =206.5\end{array}$

For observers,

The lowest value is 302 and highest value is 484.

For active watchers,

The lowest value is 114 and highest value is 634.

Formula for the mid-range

$\text{Mid-Range}=\frac{\text{Lowest}\text{value}+\text{Highest}\text{value}}{\text{2}}$

For observers,

Substitute 302 for lowest value and 484 for Highest value.

$\begin{array}{c}\text{Mid-Range}=\frac{302+484}{2}\\ =393\end{array}$

For active watchers,

Substitute 114 for lowest value and 634 for Highest value.

$\begin{array}{c}\text{Mid-Range}=\frac{114+634}{2}\\ =374\end{array}$

The value occurs most often in data is called mode.

Here, no value occurs more than once for both observers and active watchers. Hence, no mode for both data.