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?

Given:

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

Below is the Formula we can use:

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

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.

Given,

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

Folllow the following Formula used:

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

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.

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