Question

a newspaper article reported that people spend a mean of 6 hours per day watching tv, with a standard deviation of 2.1 hours. a psychologist would like to conduct interviews with the 20% 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 to find the z - score corresponding to the top 20% of the normal distribution. The area to the left of this z - score is \(1 - 0.20=0.80\). Looking up in the standard normal distribution table (z - table), the z - score \(z\) such that \(P(Z<z)=0.80\) is approximately \(z = 0.8416\) (from standard normal table).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 6\) (mean), \(\sigma = 2.1\) (standard deviation), and \(x\) is the value we want to find.
We can re - arrange the formula to solve for \(x\): \(x=\mu + z\sigma\).
Substitute \(\mu = 6\), \(z = 0.8416\), and \(\sigma = 2.1\) into the formula:
\[x=6+0.8416\times2.1\]
\[x = 6+1.76736\]
\[x=7.76736\]

Answer:

7.8