Q. 33 - The Practice of Statistics for AP

Question

Hispanic workers A factory employs 3000 unionized workers, of whom $30\%$are Hispanic. The 15 -member union executive committee contains 3 Hispanics. What would be the probability of 3 or fewer Hispanics if the executive committee were chosen at random from all the workers?

CHALLENGE: See if you can compute the probability using another method.

Step-by-Step Solution

Answer

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

Step 1: 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 $=3$

Step 2: Explanation

Given:

The mass function of $x$ probability   is,

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

We can write that the proportion $x$ and $p$ correspond to the number of Hispanics on the union executive committee.

So, $p=0.30$ and $n=15$.

Now, we can solve for the product of $n$ and $p$.

$$\begin{gathered} \begin{array}{r}np=15\times 0.30 \end{array} \\ =10.5\ge 10 \end{gathered}$$

Because $n p<10$ does not satisfy the specified criterion, a normal approximation cannot be used to obtain the requisite probability.

The probability of mass function can be used to determine the required probability.

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

Substitute the values of $n=15$

$$\begin{gathered} p=0.3 \\ (1-D)=0.7 \\ p\left(x\right)=\left(\begin{array}{c}15\\ x\end{array}\right)\left(0.3{)}^x\right(0.7{)}^{15-x} \end{gathered}$$

Needed probability is:

$P(x\le 3)=P(x=0)+P(x=1)+P(x=2)+P(x=3)$

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