Respuesta :
Answer:
At least 8.26 hours of daily TV watching is necessary for a person to be eligible for the interview.
Step-by-step explanation:
We are given that a newspaper article reported that people spend a mean of 6.5 hours per day watching TV, with a standard deviation of 1.7 hours.
A psychologist would like to conduct interviews with the 15% of the population who spend the most time watching TV.
Let X = daily time people spend watching TV
So, X ~ Normal([tex]\mu=6.5,\sigma^{2} =1.7^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean time watching TV = 6.5 hours
[tex]\sigma[/tex] = standard deviation = 1.7 hours
Now, we have to find that at least how many hours of daily TV watching is necessary for a person to be eligible for the interview, that means;
P(X [tex]\geq[/tex] x) = 0.15 {where x is required number of hours}
P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{x-6.5}{1.7}[/tex] ) = 0.15
P(Z [tex]\geq[/tex] [tex]\frac{x-6.5}{1.7}[/tex] ) = 0.15
Since, in the z table the critical value of x which represents the top 15% probability area is given by x = 1.0364, that is;
[tex]\frac{x-6.5}{1.7}[/tex] = 1.0364
[tex]x - 6.5 = 1.0364 \times 1.7[/tex]
x = 6.5 + 1.762 = 8.26 hours per day
Therefore, at least 8.26 hours of daily TV watching is necessary for a person to be eligible for the interview.
