Question

Assume that x is a binomial random variable with n = 1000 andp = 0.50. Use a normal approximation to find each of the following probabilities:

a. $\mathbf{P}\mathbf{(}\mathit{x}\mathbf{>}\mathbf{500}\mathbf{)}$

b.$\mathbf{P}\mathbf{(}\mathbf{490}\mathbf{\le }\mathit{x}\mathbf{<}\mathbf{500}\mathbf{)}$

c.$\mathbf{P}\mathbf{(}\mathit{x}\mathbf{>}\mathbf{550}\mathbf{)}$

Short answer

1. $P\left(x>500\right)=0.50$
2. $P\left(490\le x\le 500\right)=0.2357$

c.$P\left(x>550\right)=0.0008$

Step-by-step solution

Given information

x be binomial distribution with n = 1000 and p= 0.50

Calculation for the z value

$\begin{array}{l}\mu =np\\ =1000\times 0.50\\ =500\end{array}$

$$\begin{gathered} \sigma =\sqrt{np\left(1-p\right)} \\ =\sqrt{1000\times 0.50\times \left(1-0.50\right)} \\ =\sqrt{1000\times 0.50\times 0.50} \\ =15.8114 \end{gathered}$$

So,

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{x-500}{15.8114}\end{array}$

Probability calculation  P(x>500)

a.

$$\begin{gathered} P(x>500)=1-P(x<500) \\ =1-P(z<0) \\ =1-0.50 \\ =0.50 \\ P(x>500)=0.50 \end{gathered}$$

Therefore, the required probability is 0.50.

Probability calculation  P(490≤x<500)

b.

$$\begin{gathered} P\left(490\le x<500\right)=P\left(<500\right)-P\left(x\le 490\right) \\ =P\left(z<0\right)-P\left(z\le -0.6325\right) \\ =P\left(z<0\right)-\left(1-P\left(z\le -0.6325\right)\right) \\ =0.50-\left(1-0.7357\right) \\ =0.2357 \\ P\left(490\le x<500\right)=0.2357 \end{gathered}$$

Therefore, the probability is 0.2357.

Probability calculation  P(x>550)

c.

$$\begin{gathered} P\left(x>550\right)=1-P\left(x>550\right) \\ =1-P\left(z<3.1623\right) \\ =1-0.999217 \\ =0.000783 \\ \approx 0.0008 \\ P\left(x>550\right)=0.0008 \end{gathered}$$

Therefore, the probability is 0.0008.