Consider a binomial experiment with 15 trials and probability 0.45 of success on

Step 1
Solution: It is given here that a random variable say x follows the binomial distribution with parameters ${\displaystyle n=15\text{and}p=0.45}$
The binomial probability function is:
${\displaystyle P(X=x)=\frac{n!}{(n-x)!x!}{p}^x{(1-p)}^{n-x};x=0,1,2,..,n}$
Step 2
(a) Use the binomial distribution to find the probability of exactly 10 successes.
Answer: It is required to find:
${\displaystyle P(x=10)}$
Using the binomial distribution function:
${\displaystyle P(x=10)=\frac{15!}{(15-10)!10!}{0.45}^{10}{(1-0.45)}^{15-10}}$
${\displaystyle =3003\times 0.000340506\times 0.050328438}$
${\displaystyle =0.051}$
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:
${\displaystyle \mu =np=15\times 0.45=6.75}$
${\displaystyle \sigma =\sqrt{np(1-p)}=\sqrt{15\times 0.45(1-0.45)}=1.92678}$
It is required to find:
${\displaystyle P(x=10)}$
Using the continuity correction factor, the above probability can be written as:
${\displaystyle P(x=10)=P(10-0.5<x<10+0.5)}$
${\displaystyle =P\left(9.5<x<10.5\right)}$
Using the z-score formula:
${\displaystyle P\left(9.5<x<10.5\right)=P(\frac{9.5-6.75}{1.92678}<\frac{x-\mu }{\sigma }<\frac{10.5-6.75}{1.92678}}$
${\displaystyle =P\left(1.4272<z<1.9462\right)}$
${\displaystyle =P\left(z<1.9462\right)-P\left(z<1.4272\right)}$
Now using the excel functions:
${\displaystyle P\left(9.5<x<10.5\right)=P\left(z<1.9462\right)-P\left(z<1.4272\right)=0.9742-0.9232=0.051}$
The excel functions are:
${\displaystyle =NORMSDIST\left(1.4272\right)=0.9232}$
${\displaystyle =NORMSDIST\left(1.9462\right)=0.9742}$
Therefore, Using the normal distribution to approximate the probability of exactly 10 successes is 0.051