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 44% or more of the pre-teens have a TV in their bedroom is closest to _______.
A)0.9664
B)0.4644
C)0.0336
D)0.0400
E)0.9600

Question

Quizplus · Accepted Answer

Using the normal approximation to the binomial distribution, we first calculate the mean (μ) and standard deviation (σ) for the number of pre-teens with a TV in their bedroom. The mean is given by μ = np = 500 * 0.40 = 200. The standard deviation is σ = sqrt(np(1-p)) = sqrt(500 * 0.40 * 0.60) ≈ 10.95. To find the probability of having 44% or more with a TV, we convert 44% of 500 to a number, which is 220. We then use the z-score formula: z = (X - μ) / σ = (220 - 200) / 10.95 ≈ 1.83. Looking up the z-score of 1.83 in the z-table gives us the area to the left, which is approximately 0.9664. Since we want the area to the right, we subtract this from 1, giving us 1 - 0.9664 = 0.0336.

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