Answer to Question #341313 in Statistics and Probability for shiela

Question

According to Nielsen Media Research, the (population) average number of hours of TV viewing per household per week in the United States is 50.4 hours. Suppose the (population) standard deviation is known to be 11.8 hours. A random sample of 42 U.S. households is taken. What is the probability that the sample average is less thank 47.5 hours

Answer 1

We have a normal distribution, $μ=50.4,σ=11.8,n=42.$

Let's convert it to the standard normal distribution.

$\bar{z}=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\cfrac{47.5-50.4}{11.8/\sqrt{42}}=-1.59,\\ P(\bar{X}<47.5)=P(\bar{Z}<-1.59)=\\ =0.0559\text{ (from z-table).}$