Chapter 6.3, Problem 92E

a.

To determine

To explain: The random variable X being binomial random variable.

Explanation of Solution

Given:

Probability of success ( p ) = 12% = 0.12.

Number of trials ( n ) = 1500

The random variable X being binomial random variable:

1. There are two successes and failures, that is, black and non-black.
2. Americans are independent of each other.
3. Number of blacks are fixed.
4. Probability of selecting black individual is fixed.

Since, all the required condition for a binomial distribution are fixed. Thus, the variable X is a binomial random variable.

b.

To determine

To check: The condition for normal approximation.

Explanation of Solution

The condition for normal approximation are:

  $\begin{array}{l}np=1500\left(0.12\right)=180>10\\ n\left(1-p\right)=1500\left(1-0.12\right)=1320>10\end{array}$

Thus, normal approximation to binomial distribution could be used.

c.

To determine

To find: The probability that the sample would contain 165 and 195 blacks.

Answer to Problem 92E

The probability is 0.7660.

Explanation of Solution

Calculation:

The probability that the sample would contain 165 and 195 blacks is computed as:

  $\begin{array}{c}P\left(165\le X\le 195\right)=P\left(\frac{165-0.12\left(1500\right)}{\sqrt{1500\left(0.12\right)\left(1-0.12\right)}}\le Z\le \frac{195-0.12\left(1500\right)}{\sqrt{1500\left(0.12\right)\left(1-0.12\right)}}\right)\\ =P\left(-1.19\le Z\le 1.19\right)\\ =0.7660\end{array}$

Thus, the required probability is 0.7660.