Chapter 6.3, Problem 85E

To determine

Whether it is surprising for more than 4 elk in the sample to survive to adulthood and calculate an appropriate probability.

Answer to Problem 85E

The event of more than 4 elk surviving to adulthood is not surprising and the appropriate probability is 0.1402.

Explanation of Solution

Given information:

Number of trials, $n=7$

Probability of success, $p=44\%=0.44$

Calculations:

According to the binomial probability,

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

Addition rule for mutually exclusive event:

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

At $k=5$ ,

The binomial probability to be evaluated as:

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

At $k=6$ ,

The binomial probability to be evaluated as:

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

At $k=7$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=7\right)=\left(\begin{array}{c}7\\ 7\end{array}\right)\cdot {\left(0.44\right)}^7\cdot {\left(1-0.44\right)}^{7-7}\\ =\frac{7!}{7!\left(7-7\right)!}\cdot {\left(0.44\right)}^7\cdot {\left(0.56\right)}^0\\ =1\cdot {\left(0.44\right)}^7\cdot {\left(0.56\right)}^0\\ \approx 0.0032\end{array}$

Since two different numbers of successes are impossible on same simulation.

Apply addition rule for mutually exclusive events:

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

The probability less than 0.05 is considered to be small.

But in this case,

The probability is not small.

This implies

The event is likely to occur by chance.

Thus,

The event of more than 4 elk surviving to adulthood is not surprising.