Question

In a city library, the mean number of pages in a novel is $525$with a standard deviation of 200. Approximately $30\%$of the novels have fewer than 400 pages. Suppose that you randomly select 50 novels from the library.

a. What is the probability that the average number of pages in the sample is less than $500$?

b. What is the probability that at least 20 of the novels have fewer than 400 pages?

Short answer

1. The chance of the total number of pages being fewer than $25,000$is$0.1894$.
2. the probability that at least 20 of the novels have fewer than 400 pages is$0.0618.$
   

Step-by-step solution

Part (a) Step 1: Given information

Given:

$$\begin{gathered} \mu =525 \\ \sigma =200 \\ n=50 \end{gathered}$$

Below is the Formula we can use:

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

Part (a) Step 2: Calculation 

The required Sample mean is

$\begin{array}{rcl}\frac{25,00}{50}& =& 500\\ z& =& \frac{\overline{x}-\mu }{a/\sqrt{n}}\\ & =& \frac{500-525}{200/\sqrt{50}}\\ & =& -0.88P(\overline{x}<500)\\ & =& P(Z<-0.88)\\ & =& 0.1894\end{array}$

The chance of the total number of pages being fewer than 25,000 is 0.1894.

Part (b)  Step 1: Given information

Given,

$$\begin{gathered} n=50 \\ p=30\%=0.30 \end{gathered}$$

Folllow the following Formula used:

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ z=\frac{x-\mu }{\sigma } \end{gathered}$$

Part (b)  Step 2: Calculation 

For a binomial distribution with a normal approximation: $np\ge$and $nq\ge 10$.

$$\begin{gathered} n\mathrm{p}=50(0.30)=15\ge 10 \\ nq=n(1-p)=50(1-0.35)=35\ge 10 \end{gathered}$$

The normal distribution is then used to approximate the binomial distribution.

$\begin{array}{rcl}{\mu }_{\hat{p}}& =& p=0.30\\ {\sigma }_{\hat{p}}& =& \sqrt{\frac{p(1-p)}{n}}\\ & =& \sqrt{\frac{0.30(1-030)}{50}}\\ & =& 0.0648\end{array}$

The standardised score is calculated by dividing the value x by the mean and then by the standard deviation.

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

Using the normal probability table, calculate the corresponding probability.

$\begin{array}{rcl}P(x& \ge & 20)=P(\hat{p}-0.40)\\ & =& P(Z>1.54)\\ & =& 1-P(Z<1.54)\\ & =& 1-0.9382\\ & =& 0.0618\end{array}$