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 fewer 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 find the mean (μ) and standard deviation (σ) of the distribution. The mean is given by μ = np, where n is the sample size (500) and p is the probability of success (0.40). Thus, μ = 500 * 0.40 = 200. The standard deviation is given by σ = sqrt(np(1-p)), so σ = sqrt(500 * 0.40 * 0.60) ≈ 10.95.We are interested in the probability that the proportion is 44% or fewer, which means 220 or fewer pre-teens out of 500 have a TV in their bedroom (since 44% of 500 is 220). To find this probability, we convert the number of pre-teens to a z-score. The z-score is given by z = (X - μ) / σ, where X is the value of interest (220), μ is the mean (200), and σ is the standard deviation (10.95). Thus, z = (220 - 200) / 10.95 ≈ 1.83.Looking up the z-score of 1.83 in a standard normal distribution table or using a calculator, we find that the area to the left of z = 1.83 is approximately 0.9664. This area represents the probability that 44% or fewer of the pre-teens in a sample of 500 have a TV in their bedroom, which is closest to choice A.

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