Question

In a certain community, it is accepted that 85% of households are two-income households. You

select a random sample of 150 households. What is the probability that more than 130 of them are

two income households?

O 0.1977

O 0.2843

O 0.8023

O 0.7157

O 0.5596

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 $n=150, p=0.85, q=1-p=0.15$

$$np=150(0.85)=127.5\geq 10$$

$$nq=150(0.15)=22.5\geq 10$$

The approximation can safely be applied.

Use Normal Distribution with Continuity Correction

$$P(X>130)\approx1-P(X<130.5)$$

$$=1-P(Z<\dfrac{130.5-127.5}{\sqrt{150(0.85)(0.15)}})$$

$$\approx1-P(Z<0.6860)\approx0.246358$$

If we don't use Continuity Correction

$$P(X>130)\approx1-P(Z<\dfrac{130-127.5}{\sqrt{150(0.85)(0.15)}})$$

$$\approx1-P(Z<0.571662)\approx0.2838$$