Question

. 25% of all Ethiopian households had VCR in 2002. If 500 households are randomly selected from the population, what is the probability that 100 or more households have VCR?

Answer 1

If $X$ is a binomial random variable with mean $μ = np$ and variance $σ^2 = npq,$ then the limiting form of the distribution of $Z=\dfrac{X-np}{\sqrt{npq}},$ as $n\to \infin,$ is the standard normal distribution $n(z;0,1).$

In practice, the approximation is adequate provided that both $np\geq 10$ and

$nq\geq 10.$

Given $n=500, p=0.25, q=1-p=0.75$

$$np=500(0.25)=125\geq 10$$

$$nq=500(0.75)=375\geq 10$$

The approximation can safely be applied.

Use Normal Distribution with Continuity Correction

$$P(X\geq100)\approx P(X>99.5)=1-P(X\leq99.5)$$

$$=1-P(Z\leq\dfrac{99.5-125}{\sqrt{500(0.25)(0.75)}})$$

$$\approx1-P(Z\leq-2.63363)\approx0.995776$$