Chapter 10.4, Problem 50E

Solution

a.

To determine: the probability that exactly 17 athletes should have graduated.

The probability that exactly 17 athletes should have graduated is $0.4\%$ .

Given information:

A university widely known for its track and field program claims that 75% of its track athletes get degrees. A journalist investigates what happened to the 32 athletes who began the program over a 6-year period that ended 7 years ago. Of these athletes, 17 have graduated and the remaining 15 are no longer attending any college. If the university’s claim is true, the number of athletes who graduate among the 32 examined should have been governed by binomial probability with $p=0.75$ .

Concept Used:

Binomial probability Distribution:

Consider a simple event with these properties:

Each trial has two possible outcomes, called success and failure.

The probability of success on each trial is the same. (We denote the probability of success as $p$ and the probability of failure as $q$ . Note that $q=1-p$ .)

The trials are independent.

If the binomial random variable $X$ counts the number of successes in $n$ trials with probability of success $p$ and the probability of getting $k$ successes in $n$ trials, then $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)p^k{q}^{n-k}$ .

Calculation:

From the situation, the following information could be identified.

Here, $n=32,k=17\text{and}p=0.75$ .

The probability of failure $q$ is $1-0.75=0.25$ .

  $\begin{array}{}$ P\left(X=17\right)=\left(\begin{array}{c}32\\ 17\end{array}\right){(0.75)}^{17}{(0.25)}^{32-17}$$ =\frac{32!}{17!15!}{(0.75)}^{17}{(0.25)}^{15}$$ =0.4\%$\end{array}$

b.

To determine: the probability that 17 or fewer athletes should have graduated.

The probability that 17 or fewer athletes should have graduated is $0.59\%$ .

Given information:

A university widely known for its track and field program claims that 75% of its track athletes get degrees. A journalist investigates what happened to the 32 athletes who began the program over a 6-year period that ended 7 years ago. Of these athletes, 17 have graduated and the remaining 15 are no longer attending any college. If the university’s claim is true, the number of athletes who graduate among the 32 examined should have been governed by binomial probability with $p=0.75$ .

Concept Used:

Binomial probability Distribution:

Consider a simple event with these properties:

Each trial has two possible outcomes, called success and failure.

The probability of success on each trial is the same. (We denote the probability of success as $p$ and the probability of failure as $q$ . Note that $q=1-p$ .)

The trials are independent.

If the binomial random variable $X$ counts the number of successes in $n$ trials with probability of success $p$ and the probability of getting $k$ successes in $n$ trials, then $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)p^k{q}^{n-k}$ .

Calculation:

The probability that exactly 17 athletes should have graduated is $0.4\%$ .

Here, $n=32,k=16\text{and}p=0.75$ .

The probability of failure $q$ is $1-0.75=0.25$ .

  $\begin{array}{c}P\left(17\le X\le 1\right)=P\left(X=1\right)+P\left(X=2\right)+\mathrm{.......}+P\left(X=17\right)\\ ={\displaystyle \sum _{k=1}^{17}\left(\begin{array}{c}32\\ k\end{array}\right){(0.75)}^k{(0.25)}^{32-k}}\\ =0.0059\\ =0.59\%\end{array}$

c.

To explain: the story on the investigation if you were journalist.

Given information:

A university widely known for its track and field program claims that 75% of its track athletes get degrees. A journalist investigates what happened to the 32 athletes who began the program over a 6-year period that ended 7 years ago. Of these athletes, 17 have graduated and the remaining 15 are no longer attending any college. If the university’s claim is true, the number of athletes who graduate among the 32 examined should have been governed by binomial probability with $p=0.75$ .

Concept Used:

Binomial probability Distribution:

Consider a simple event with these properties:

Each trial has two possible outcomes, called success and failure.

The probability of success on each trial is the same. (We denote the probability of success as $p$ and the probability of failure as $q$ . Note that $q=1-p$ .)

The trials are independent.

If the binomial random variable $X$ counts the number of successes in $n$ trials with probability of success $p$ and the probability of getting $k$ successes in $n$ trials, then $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)p^k{q}^{n-k}$ .

Calculation:

The probability that exactly 17 athletes should have graduated is $0.4\%$ .

The probability that 17 or fewer athletes should have graduated is $0.59\%$ .

Since the probability is very small, it is very unusual that 17 or fewer athletes graduated and thus we would say that the universities claim is incorrect.