Question

a newspaper article reported that people spend a mean of 7 hours per day watching tv, with a standard deviation of 2.1 hours. a psychologist would like to conduct interviews with the 10% of the population who spend the most time watching tv. she assumes that the daily time people spend watching tv is normally distributed. at least how many hours of daily 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.

Explanation:

Step1: Find the z - score

We want the top 10% of the population. The area to the left of the z - score we want is \(1 - 0.10=0.90\). Looking up in the standard normal distribution table (z - table), the z - score corresponding to an area of 0.90 is approximately \(z = 1.2816\) (from the standard normal table, the closest value to 0.90 gives us this z - score).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value we want to find, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. We know that \(\mu = 7\), \(\sigma=2.1\), and \(z = 1.2816\). Rearranging the formula for \(x\) gives \(x=\mu + z\sigma\).
Substitute the values: \(x=7+1.2816\times2.1\).
First, calculate \(1.2816\times2.1 = 2.69136\).
Then, \(x=7 + 2.69136=9.69136\).

Answer:

9.7 hours