91Ó°ÊÓ

Given:

$$\begin{gathered} \mathrm{μ}=525 \\ \mathrm{σ}=200 \\ n=50 \end{gathered}$$

Below is the Formula we can use:

$z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{dl\sqrt{n}}$

The required Sample mean is

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

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:

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

For a binomial distribution with a normal approximation: $npâ‰\yen$and $nqâ‰\yen 10$.

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

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

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

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

$\begin{array}{rcl}z& =& \frac{x-\mathrm{μ}}{\mathrm{σ}}\\ & =& \frac{0.4-0.30}{0.0648}\\ & =& 1.54\end{array}$

Using the normal probability table, calculate the corresponding probability.

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