Step 1

Solution: It is given here that a random variable say x follows the binomial distribution with parameters $n=15\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}p=0.45$

The binomial probability function is:

$P(X=x)=\frac{n!}{(n-x)!x!}{p}^{x}{(1-p)}^{n-x};x=0,1,2,..,n$

Step 2

(a) Use the binomial distribution to find the probability of exactly 10 successes.

Answer: It is required to find:

$P(x=10)$

Using the binomial distribution function:

$P(x=10)=\frac{15!}{(15-10)!10!}{0.45}^{10}{(1-0.45)}^{15-10}$

$=3003\times 0.000340506\times 0.050328438$

$=0.051$

Therefore, the probability of exactly 10 successes is 0.051

Step 3

(b) Use the normal distribution to approximate the probability of exactly 10 successes.

Answer:

The mean and standard deviation of the random variable x is:

$\mu =np=15\times 0.45=6.75$

$\sigma =\sqrt{np(1-p)}=\sqrt{15\times 0.45(1-0.45)}=1.92678$

It is required to find:

$P(x=10)$

Using the continuity correction factor, the above probability can be written as:

$P(x=10)=P(10-0.5<x<10+0.5)$

$=P\left(9.5<x<10.5\right)$

Using the z-score formula:

$P\left(9.5<x<10.5\right)=P(\frac{9.5-6.75}{1.92678}<\frac{x-\mu}{\sigma}<\frac{10.5-6.75}{1.92678}$

$=P\left(1.4272<z<1.9462\right)$

$=P\left(z<1.9462\right)-P\left(z<1.4272\right)$

Now using the excel functions:

$P\left(9.5<x<10.5\right)=P\left(z<1.9462\right)-P\left(z<1.4272\right)=0.9742-0.9232=0.051$

The excel functions are:

$=NORMSDIST\left(1.4272\right)=0.9232$

$=NORMSDIST\left(1.9462\right)=0.9742$

Therefore, Using the normal distribution to approximate the probability of exactly 10 successes is 0.051