Chapter 7, Problem AP2.25CPT

(a)

To determine

To Calculate: the probability that the average number of pages in the sample is less than 500.

Answer to Problem AP2.25CPT

0.1894

Explanation of Solution

Given:

  $\begin{array}{l}\mu =\text{525}\\ \sigma \text{ = 200}\\ n\text{ = 50}\end{array}$

Formula used:

  $z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$

Calculation:

Sample mean is

  $\frac{25,00}{50}=500$

  $\begin{array}{c}z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}\\ =\frac{500-525}{200/\sqrt{50}}\\ =-0.88\end{array}$

  $\begin{array}{c}P\left(\overline{x}<500\right)=P\left(Z<-0.88\right)\\ =0.1894\end{array}$

Therefore the probability that the total number of pages is less than 25,000 is $0.1894$

(b)

To determine

To Calculate: the probability that at least 20 of the novels have fewer than 400 pages.

Answer to Problem AP2.25CPT

0.0618

Explanation of Solution

Given:

  $\begin{array}{c}n\text{=50}\\ p=30\%=0.30\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

  $z=\frac{x-\mu }{\sigma }$

Calculation:

For a normal approximation of the binomial distribution: $np\ge \text{and }nq\ge \text{10}$ .

  $\begin{array}{c}\text{np=50}\left(0.30\right)=15\ge \text{10}\\ nq=n\left(1-p\right)=50\left(1-0.35\right)=35\ge \text{10}\end{array}$

Then approximation the binomial distribution by the normal distribution

  ${\mu }_{\widehat{p}}=p=0.30$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.30\left(1-0.30\right)}{50}}\\ =0.0648\end{array}$

The standardized score is the value $x$ decreased by the mean and then divided by the standard deviation.

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.4-0.30}{0.0648}\\ =1.54\end{array}$

The corresponding probability using the normal probability table

  $\begin{array}{c}P\left(x\ge 20\right)=P\left(\widehat{p}-0.40\right)=P\left(Z>1.54\right)=1-P\left(Z<1.54\right)\\ =1-0.9382\\ =0.0618\end{array}$