Question

A newspaper article reported that people spend a mean of 7.5 hours per day watching TV, with a standard deviation of 2.0 hours. A psychologist would like to conduct interviews with the 15% of the population who spend the most time watching TV. She assumes that the dally time people spend watching TV is normally distributed. At least how many hours of dally TV watching are necessary for a person to be eligible for the interview? Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.

Answer 1

Using the z-score for the 85th percentile, and the mean and standard deviation given, individuals must watch at least 9.6 hours of television daily to be in the top 15% of viewers eligible for the interview.

Step-by-step explanation:

To find the minimum number of hours of daily TV watching necessary for a person to be in the top 15% of viewers, we need to use the z-score associated with the top 15% of a standard normal distribution. The mean (μ) is 7.5 hours, and the standard deviation (σ) is 2.0 hours.

First, we find the z-score that corresponds to the 85th percentile because the top 15% is the same as the bottom 85%. Utilizing z-tables or a calculator, we find that the z-score is approximately 1.0364. Next, we use the z-score formula:

Z = (X - μ) / σ

Where Z is the z-score, X is the number of hours, μ is the mean, and σ is the standard deviation. Now we solve for X:

X = Z σ + μ

X = (1.0364)(2.0) + 7.5

X = 2.0728 + 7.5

X = 9.5728 hours

After rounding to one decimal place, a person would need to watch at least 9.6 hours of TV daily to be eligible for the psychologist's interview.