Chapter 6, Problem T6.14SPT

(a)

To determine

To find: the mean number of Hispanic in such samples and standard deviation.

Answer to Problem T6.14SPT

  $\begin{array}{c}\mu =156\\ \sigma =11.6499\end{array}$

Explanation of Solution

Given:

  $\begin{array}{c}p=13\%=0.13\\ n=1200\end{array}$

Formula used:

  $\mu =np$

  $\sigma =\sqrt{np\left(1-p\right)}$

Calculation:

The mean is

  $\mu =np=1200\left(0.13\right)=156$

The standard deviation of a binomial random variable is

  $\begin{array}{c}\sigma =\sqrt{np\left(1-p\right)}\\ =\sqrt{1200\left(0.13\right)\left(1-0.13\right)}\\ =11.6499\end{array}$

(b)

To determine

To Calculate: an appropriate probability to help the answer.

Answer to Problem T6.14SPT

Yes

Explanation of Solution

Given:

15% of the sample contains Hispanic.

  $p=13\%=0.13$

Formula used:

  $P\left(X=k\right)=\left({}^n{C}_k\right)\times p^k\times {\left(1-p\right)}^{n-k}$

Calculation:

The distribution is binomial, because of want to know the probability of the number of successes in the sample.

  $P\left(X=k\right)=\left({}^n{C}_k\right)\times p^k\times {\left(1-p\right)}^{n-k}$

To find when should be suspicious or not, find the probability of getting

  $1200\times 15\%=180$ Hispanics or more extreme:

  $\begin{array}{c}P\left(X\ge 180\right)={\displaystyle \sum _{k=180}^{1200}\left({}^n{C}_{1200}\right)\times {0.13}^k}\times {\left(1-0.13\right)}^{1200-k}\\ =0.0235\end{array}$

Although this probability is less than 0.05 or 5% it is not likely to get 180 Hispanics by chance and therefore it should be suspicious.