Chapter 6.5, Problem 15E

To determine

To find:

The probability that more than 150 out of 230 eight graders at a local middle school have been exposed to drugs assuming that a previous study at this school reported that the probability of an individual eighth-grade student having been exposed to drugs is $63\%$.

Answer to Problem 15E

Solution:

The probability that more than 150 out of 230 eight graders have been exposed to drugs is $0.2236$.

Explanation of Solution

Given:

Total number of eighth-grade students is 230.

Probability of an individual having exposed to drugs is $63\%$.

Continuity correction factor approximates discrete distributions to continuous ones by converting a random variable value of the former to a range of random variable values of the latter.

Binomial distribution is discrete and the probability of $x$ successes out of $n$ identical trials of the experiment is given by:

$P\left(x\right)=C{}_x{p}^x{(1-p)}^{n-x}$,

with mean $\mu =np$ and standard deviation $\sigma =\sqrt{np(1-p)}$.

Normal distribution is continuous and probability of obtaining the value $z$ is expressed as:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}$,

with mean 0 and standard deviation 1.

The continuity correction is done by changing the whole number $x$-value into an interval range:

$\left[x-0.5,x+0.5\right]$,

and then the probability $P\left(x\le a\right)$ is approximated using the probability:

$P\left(z\le z_a\right)={\displaystyle \underset{-\infty }{\overset{z_a}{\int }}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}dz}$,

where $z_a$ is the appropriate $z$-value.

In this question, discrete binomial distribution is approximated using the normal distribution, which is continuous.

Initially, the values of $n$ and $p$ are extracted after which the mean $\left(\mu \right)$ and the standard deviation $\left(\sigma \right)$ are computed from binomial distribution.

Also, continuity correction is done, according to requirements of the question, that is by adding $0.5$ to the given $x$-value.

Thereafter, the $z$-value is obtained using above $x$-value and the relation:

$z=\text{}\frac{x-\mu }{\sigma }$,

following which, area under curve is computed by subtracting corresponding value at the standard normal table from one.

Calculation:

$p=63\%$

which gets converted to:

$p=0.63$.

Then mean is:

$\begin{array}{rll}\mu & =& 230\times 0.63\\ & =& 144.9\end{array}$

Also, standard deviation is:

$\begin{array}{rll}\sigma & =& \sqrt{230\times 0.63\times \left(1-0.63\right)}\\ & =& 7.322\end{array}$

Since the inequality is “more than”, continuity correction gives $x$-value as:

$x=150.5$.

This gives the $z$-value as:

$\begin{array}{rll}{z}_a& =& \frac{150.5-144.9}{7.322}\\ & =& \frac{5.6}{7.322}\\ & =& \mathrm{0.76481...}\end{array}$

Since the $3^{\text{rd}}$ decimal place is 4 and $4\le 5$, the $z$-value is approximated as:

$z_a\approx 0.76$.

Now, the required probability is:

$P\left(z\ge 0.76\right)=1-P\left(z\le 0.76\right)$

the latter of which is obtained from table as:

$\begin{array}{rll}P\left(z\ge 0.765\right)& =& 1-0.77637\\ & =& 0.22363\end{array}$

Since the $5^{\text{th}}$ decimal digit is 3 and $3\le 5$, the required probability is:

$P\left(\text{more}\text{than}150\text{out}\text{of}230\text{eigth-graders}\text{have}\text{been}\text{exposed}\text{to}\text{drugs}\right)=0.2236$

Conclusion:

The probability that more than 150 out of 230 eight graders have been exposed to drugs is $0.2236$.