Chapter 17, Problem 30E

(a)

To determine

To find: that it is required to find the mean and standard deviation of the numbers of frogs with the trait researcher in his sample.

Answer to Problem 30E

Mean = 18.75

Standard Deviation = 4.05

Explanation of Solution

Given:

  $\begin{array}{c}n=150\\ p=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\end{array}$

Formula used:

  $\begin{array}{c}E\left(X\right)=np\\ SD\left(X\right)=\sqrt{npq}\end{array}$

Calculation:

Computing the mean and standard deviation.

Mean number is

  $\begin{array}{c}E\left(x\right)=np\\ =150\times \frac{1}{8}\\ =18.75\end{array}$

Standard deviation is

  $\begin{array}{c}SD\left(x\right)=\sqrt{npq}\\ =\sqrt{150\times \frac{1}{8}\times \frac{7}{8}}\\ =4.05\end{array}$

(b)

To determine

To Verify: A normal model may be used by that researcher to estimate the distribution of the number of frogs with the trait.

Explanation of Solution

Given:

  $\begin{array}{c}n=150\\ p=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\end{array}$

Calculation:

Verifying that, could use a normal model is about to appropriate or not.

Following conditions

  $\begin{array}{c}np=150\times \frac{1}{8}\\ =18.75\\ nq=150\times \frac{7}{8}\\ =131.25\end{array}$

Clearly both $np\ge 10\text{ and }nq\ge 10$

Hence, it could be said that, it is about to use the normal model in this situation.

(c)

To determine

To explain: this proves that this characteristic has become more prevalent.

Explanation of Solution

Given:

  $\begin{array}{c}n=150\\ p=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\end{array}$

researcher find that the trait in 22 frogs.

Calculation:

For 22 frogs calculate the corresponding Z value as follows.

  $\begin{array}{c}Z=\frac{x-np}{\sqrt{npq}}\\ =\frac{22-18.75}{4.05}\\ =0.80\end{array}$

Since the value of Z is very small, it can be said that in frogs, the trait has become more general.