Chapter 7.2, Problem 33E

To determine

Tocalculate:The probability of 3 or fewer Hispanics if the executive committee were chosen at random from all the workers.

Answer to Problem 33E

The required probability is 0.2969.

Explanation of Solution

Given information:

Number of unionized workers = 3000

Proportion of workers that are Hispanic $=30\%$

Number of members which are Hispanic in the 15-member union executive committee = 3

Formula used:

The probability mass function of $x$is,

  $p\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){\left(p\right)}^x{\left(1-p\right)}^{n-x};x=0,1,2,\cdots ,n$

Consider that number of Hispanics in the union executive committee and the proportion be $x$and $\widehat{p}$respectively.

So, we can write $p=0.30$and $n=15$

Now, find the product of n and p as per the requirement.

  $\begin{array}{c}np=15\times 0.30\\ =4.5\\ <10\end{array}$

Also, find the product of n and $1-p$ .

  $\begin{array}{c}n\left(1-p\right)=15\times \left(1-0.30\right)\\ =15\times 0.70\\ =10.5\\ \ge 10\end{array}$

The given condition does not satisfy because the value of $np<10$ .

This means, we cannot find the required probability by using normal approximation.

To find the required probability, use the formula probability of mass function $p\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){\left(p\right)}^x{\left(1-p\right)}^{n-x};x=0,1,2,\cdots ,n$ .

Substitute 15 for n, 0.3 for p and 0.7 for $\left(1-p\right)$in the above formula.

  $p\left(x\right)=\left(\begin{array}{c}15\\ x\end{array}\right){\left(0.3\right)}^x{\left(0.7\right)}^{15-x}$

Write the required probability as,

  $\begin{array}{c}P\left(x\le 3\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)\\ =\left[\begin{array}{l}\left(\begin{array}{c}15\\ 0\end{array}\right){\left(0.3\right)}^0{\left(0.7\right)}^{15}+\left(\begin{array}{c}15\\ 1\end{array}\right){\left(0.3\right)}^1{\left(0.7\right)}^{14}+\left(\begin{array}{c}15\\ 2\end{array}\right){\left(0.3\right)}^2{\left(0.7\right)}^{15}\\ +\left(\begin{array}{c}15\\ 3\end{array}\right){\left(0.3\right)}^3{\left(0.7\right)}^{15}\end{array}\right]\\ =0.0047+0.0305+0.0916+0.1700\\ =0.2969\end{array}$

Hence, the probability of 3 or fewer Hispanics if the executive committee were chosen at random from all the workers is 0.2969.