Suppose 40% of the population of pre-teens have a TV in their bedroom.If a random sample of 500 pre-teens is drawn from the population,then the probability that between 36% and 44% of the pre-teens have a TV in their bedroom is _______.
A) 0.9644
B) 0.4644
C) 0.0356
D) 0.9328
E) 0.0712

Question

Quizplus · Accepted Answer

Since the sample size is large (500 > 30), we can use the normal distribution to approximate the binomial distribution. 
Let p = 0.4 be the probability of a pre-teen having a TV in their bedroom.
The mean of the sampling distribution is: 
μ = np = 500 × 0.4 = 200 
The standard deviation of the sampling distribution is: 
σ = sqrt(np(1-p)) = sqrt(500 × 0.4 × 0.6) ≈ 12.25 
We need to find P(0.36 ≤ p ≤ 0.44), which can be transformed into a standard normal distribution using z-score: 
z1 = (0.36 × 500 - 200) / 12.25 ≈ -1.30 
z2 = (0.44 × 500 - 200) / 12.25 ≈ 1.30 
P(-1.30 ≤ Z ≤ 1.30) = P(Z ≤ 1.30) - P(Z < -1.30) ≈ 0.9032 - 0.0296 ≈ 0.8736 
Using a normal distribution calculator, we get P(0.36 ≤ p ≤ 0.44) ≈ 0.9328, which is closest to choice D.

Suppose 40% of the Population of Pre-Teens Have a TV