Question

Baby elk Refer to Exercise 77 . How surprising would it be for more than 4 elk in the sample to survive to adulthood? Calculate an appropriate probability to support your answer.

Short answer

It's not surprising that more than four elk live to adulthood, because the acceptable likelihood is$0.1402$

Step-by-step solution

Given Information

Total number of trials,$\mathrm{n}=7$

The likelihood of success, $\mathrm{p}=44\%=0.44$

Simplifications

The binomial probability states that

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}\mathrm{n}\\ \mathrm{k}\end{array}\right)·{\mathrm{p}}^{\mathrm{k}}·(1-\mathrm{p}{)}^{\mathrm{n}-\mathrm{k}}$

Mutually exclusive event addition rule:

$\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\text{or}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)$

For$\mathrm{k}=5$,

The binomial probability is calculated as follows:

$\begin{array}{r}\mathrm{P}(\mathrm{X}=5)=\left(\begin{array}{c}7\\ 5\end{array}\right)\cdot (0.44{)}^5\cdot (1-0.44{)}^{7-5}\\ =\frac{7!}{5!(7-5)!}\cdot (0.44{)}^5\cdot (0.56{)}^2\\ =21\cdot (0.44{)}^5\cdot (0.56{)}^2\\ \approx 0.1086\end{array}$

For $k=6$,

The binomial probability is calculated as follows:

$\begin{array}{r}\mathrm{P}(\mathrm{X}=6)=\left(\begin{array}{c}7\\ 6\end{array}\right)\cdot (0.44{)}^6\cdot (1-0.44{)}^{7-6}\\ =\frac{7!}{6!(7-6)!}\cdot (0.44{)}^6\cdot (0.56{)}^1\\ =7\cdot (0.44{)}^6\cdot (0.56{)}^1\\ \approx 0.0284\end{array}$

For K=7,

The binomial probability is calculated as follows:
$\begin{array}{r}\mathrm{P}(\mathrm{X}=7)=\left(\begin{array}{c}7\\ 7\end{array}\right)\cdot (0.44{)}^7\cdot (1-0.44{)}^{7-7}\\ =\frac{7!}{7!(7-7)!}\cdot (0.44{)}^7\cdot (0.56{)}^0\\ =1\cdot (0.44{)}^7\cdot (0.56{)}^0\\ \approx 0.0032\end{array}$

Because it's impossible to have two distinct counts of successes in the same simulation.

For mutually exclusive events, use the addition rule:

$\begin{array}{r}\mathrm{P}(\mathrm{X}>4)=\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7)\\ =0.1086+0.0284+0.0032\\ =0.1402\end{array}$

Probabilities of less than 0.05 are deemed small.

However, in this scenario, the likelihood is not insignificant.

This indicates that the occurrence is most likely to happen by chance.

As a result, the fact that more than four elk have survived to adulthood is not remarkable.