Question

Suppose x is a binomial random variable with p = .4 and n = 25.

a. Would it be appropriate to approximate the probability distribution of x with a normal distribution? Explain.

b. Assuming that a normal distribution provides an adequate approximation to the distribution of x, what are the mean and variance of the approximating normal distribution?

c. Use Table I in Appendix D to find the exact value of $\mathbf{P}\mathbf{(}\mathit{x}\mathbf{\ge }\mathbf{9}\mathbf{)}$.

d. Use the normal approximation to find $\mathbf{P}\mathbf{(}\mathit{x}\mathbf{\ge }\mathbf{9}\mathbf{)}$.

Short answer

1. Yes.
2. The mean of x is 6 and the variance of x is 6.
3. $P\left(x\ge 9\right)=0.575$
   

1. By normal approximation,$P\left(x\ge 9\right)=0.6591$

Step-by-step solution

Given information

x is a binomial random variable withp= 0.4 and n = 25

Explanation

a.

Yes. If,$np\ge 5andn(1-p)\ge 5$ , the x would be approximate the normal distribution.

So,

$$\begin{gathered} np=25\times 0.4 \\ =10 \end{gathered}$$

And,

$$\begin{gathered} n(1-p)=25\times \left(1-0.4\right) \\ =25\times 0.6 \\ =15 \end{gathered}$$

So, here$np\ge 5andn(1-p)\ge 5$

Therefore, x would be approximated by normal distribution.

Finding the value of mean and variance

b.

Assuming x is the normal distribution, then the mean and variance of approximating normal distribution is,

$$\begin{gathered} Mean=np \\ =25\times 0.4 \\ =10 \\ Variance=np\left(1-p\right) \\ =25\times 0.4\times \left(1-0.4\right) \\ =25\times 0.4\times 0.6 \\ =6 \end{gathered}$$

Therefore, the mean of x is 10 and the variance of x is 6.

Probability calculation

c.

$$\begin{gathered} P\left(x\ge 9\right)=1-p\left(x<9\right) \\ =1-0.425\left[Bybinomialtable\right] \\ =0.575 \end{gathered}$$

Therefore, $P\left(x\ge 9\right)=0.575$

Calculation when P(x≥9)d.

d.

If $x~N\left(0.4,25\right)$

Then,

$$\begin{gathered} \mu =0.4\times 25 \\ =10 \end{gathered}$$

And,

$\begin{array}{l}\sigma =\sqrt{np(\left(1-p\right)}\\ =\sqrt{25\times 0.4\times (1-0.4)}\\ =\sqrt{25\times 0.4\times 0.6}\\ =2.4495\end{array}$

Then,

role="math" localid="1660280899974" $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{9-10}{2.4495}\\ =-0.4082\end{array}$

$$\begin{gathered} P(z\ge -0.4082)=1-P(z<-0.4082) \\ =1-(1-P\left(z<0.4082\right)) \\ =1-(1-0.6591) \\ =0.6591 \end{gathered}$$

$$\begin{gathered} P(x\ge 90)=P\left(z\ge -0.4082\right) \\ =0.6591 \end{gathered}$$

Therefore, by the normal approximation,$P(x\ge 9)=0.6591$