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

(a)the probability that the total number of pages is less than 25,000 is 0.1894

(b)Then approximation the binomial distribution by the normal distribution${\mu }_{\hat{p}}=p=0.30$

Step-by-step solution

Part (a) Step 1: Given Information

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

Formula used:

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

Part (a) Step 2: Simplification

Sample mean is

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

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

Part (b) Step 1: Given Information

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

Formula used:

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

Part (b) Step 2: Simplification

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

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

Then approximation the binomial distribution by the normal distribution

${\mu }_{\hat{p}}=p=0.30$