Chapter 6.5, Problem 30E

To determine

To find:

The probability that out of 350 entering college freshmen, 100 will need to take developmental mathematics, if there is a 36% chance of a college freshman needing to take developmental mathematics.

Answer to Problem 30E

Solution:

The probability that out of 350 entering college freshmen, 100 will need to take developmental mathematics is $0.0007$.

Explanation of Solution

Given:

Total number of independent trials is 350.

Probability of an individual trial success is $36\%$.

Concept:

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 subtracting and adding $0.5$ to the given $x$-value to make it interval.

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 using a z-table.

Calculation:

Probability of a trial to be success:

$\begin{array}{l}p=36\%\\ =0.36\end{array}$

Probability of a trial to be failure:

$\begin{array}{l}q=1-p\\ =1-0.36\\ =0.64\end{array}$

Then mean is:

$\begin{array}{rll}\mu & =& n\times p\\ & =& 350\times 0.36\\ & =& 126\end{array}$

Also, standard deviation is:

$\begin{array}{rll}\sigma & =& \sqrt{n\times p\times \left(1-p\right)}\\ & =& \sqrt{\mu \times q}\\ & =& \sqrt{126\times 0.64}\\ & =& 8.9799\end{array}$

Since here is exact “100”, continuity correction gives $x$-value as:

$x_1=99.5,x_2=100.5$.

This gives the $z$-value as:

$z_1=\frac{99.5-126}{8.9799}$

$\approx -2.95$

$z_2=\frac{100.5-126}{8.9799}$

$\approx -2.84$

Now, the required probability is:

$=P\left(z_2\le -2.84\right)-P\left(z_1\le -2.95\right)$

$=0.0023-0.0016$

$=0.0007$

Conclusion:

The probability that out of 350 entering college freshmen, 100 will need to take developmental mathematics is $0.0007$.