Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

$\begin{array}{c}n\times p=500\times 0.62\\ =310\end{array}$

The standard deviation $\left({\sigma }_X\right)$ of the binomial distribution is calculated as,

$\begin{array}{c}{\sigma }_X=\sqrt{np\left(1-p\right)}\\ =\sqrt{500\times 0.62\times 0.38}\\ =10.85\end{array}$

Also,

$\begin{array}{c}n(1-p)=500\times \left(1-0.62\right)\\ =190\end{array}$

As $np$ and $n(1-p)$ are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for $p_1$ and $p_2$ as shown below,

The Z score for $p_1=0.59$ is calculated as,

$\begin{array}{c}{Z}_{-}=\frac{n{p}_1-np}{\sqrt{npq}}\\ =\frac{500\times 0.59-310}{10.85}\\ =-1.38\end{array}$

The Z score for $p_2=0.65$ is calculated as,

$\begin{array}{c}{Z}_{+}=\frac{n{p}_2-np}{\sqrt{npq}}\\ =\frac{500\times 0.65-310}{10.85}\\ =1.38\end{array}$

Now, calculate the probability that the sample proportion $\widehat{p}$ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

$\begin{array}{c}P\left(0.59\le \widehat{p}\le 0.65\right)=P\left(-1.38\le Z\le 1.38\right)\\ =P\left(Z\le 1.38\right)-P\left(Z\le -1.38\right)\end{array}$

By using the standard normal table, the areas corresponding to probabilities $P\left(Z\le 1.38\right)$ and $P\left(Z\le -1.38\right)$ are 0.916 and 0.084, respectively.

So,

$\begin{array}{c}P\left(0.59\le \widehat{p}\le 0.65\right)=P\left(Z\le 1.38\right)-P\left(Z\le -1.38\right)\\ =0.916-0.084\\ =0.832\end{array}$

Hence, the required probability is 0.832.

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

$\begin{array}{c}n\times p=500\times 0.90\\ =450\end{array}$

The standard deviation $\left({\sigma }_X\right)$ of the binomial distribution is calculated as,

$\begin{array}{c}{\sigma }_X=\sqrt{np\left(1-p\right)}\\ =\sqrt{500\times 0.9\times 0.1}\\ =6.7\end{array}$

Also,

$\begin{array}{c}np(1-p)=500\times 0.9\times \left(1-0.9\right)\\ =45\end{array}$

As $np$ and $np(1-p)$ are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for $p_1$ and $p_2$ as shown below,

The Z score for $p_1=0.87$ is calculated as,

$\begin{array}{c}{Z}_{-}=\frac{n{p}_1-np}{\sqrt{npq}}\\ =\frac{500\times 0.87-450}{6.7}\\ =-2.23\end{array}$

The Z score for $p_2=0.93$ is calculated as,

$\begin{array}{c}{Z}_{+}=\frac{n{p}_2-np}{\sqrt{npq}}\\ =\frac{500\times 0.93-450}{6.7}\\ =2.23\end{array}$

Now, calculate the probability that the sample proportion $\widehat{p}$ falls between 0.87 and 0.93 as shown below,

$\begin{array}{c}P\left(0.87\le \widehat{p}\le 0.93\right)=P(-2.23\le Z\le 2.23)\\ =P\left(Z\le 2.23\right)-P\left(Z\le -2.23\right)\end{array}$

By using the standard normal table, the areas corresponding to probabilities $P\left(Z\le 2.23\right)$ and $P\left(Z\le -2.23\right)$ are 0.987 and 0.012 respectively.

So,

$\begin{array}{c}P\left(0.87\le \widehat{p}\le 0.93\right)=P\left(Z\le 2.23\right)-P\left(Z\le -2.23\right)\\ =0.987-0.012\\ =0.975\end{array}$

Hence, the required probability is 0.975.

(c)

To determine

To explain: The effect on the probability of sample proportion $\widehat{p}$ - that it falls within $\pm 3\%$ of the population proportion $\left(p\right)$ value as the value of p gets closer to 1, by using the results of parts (a) and (b).

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion $\widehat{p}$ falling within $\pm 3\%$ will increase.

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.