Social scientists discovered that 1,013 General Social Survey (GSS) respondents in 2010 watched television for an average of 3.01 hours per day, with a standard deviation of 2.65 hours per day. Answer the following questions, assuming the distribution of the number of television hours is normal.

Required:
a. What is the Z score for a person who watches more than 8 hr/day?
b. What proportion of people watch television less than 5 hr/day? How many does this correspond to in the sample?
c. What number of television hours per day corresponds to a Z score of +1?
d. What is the percentage of people who watch between 1 and 6 hr television per day?

[tex]P(X<5)=?\\\\ x = 5\\\\ Z = \frac{(X - \mu)}{\sigma} = \frac{(5 -3.01)}{2.65} = 0.7509434\\\\P(Z< 0.7509434) = 0.7736566\\\\P(X<5) = P(Z< 0.7509434) = 0.7736566[/tex]

For point c:

[tex]X=? \\\\Z = 1\\\\\text{Formula for X score:}\\\\X = \mu + Z \times \alpha\\\\X = 3.01 + 1\times 2.65 = 3.01 + 2.65 = 5.66[/tex]

For point d:

[tex]P(1 \leq X \leq 6)=? \\\\P(1 \leq X \leq 6) = P(X \leq 6) - P(X\leq 1) = P(X< 6) - P(X<1)\\\\\text{Z score for x} = 6\\\\Z = \frac{(6 - 3.01)}{2.65} = 1.128301887 \\\\P(Z< 1.128301887) = 0.870403777 \\\\P(X<6) = P(Z< 1.128301887) = 0.870403777 \\\\\text{Z score for x} = 1 \\\\[/tex]

[tex]Z = \frac{(1 -3.01)}{2.65} = -0.758490566\\\\P(Z< -0.758490566) = 0.224078679\\\\P(X<1) = P(Z< -0.758490566) = 0.224078679 \\\\P(1 \leq X \leq 6) = P(X \leq 6) - P(X\leq 1) = P(X<6) - P(X<1) \\\\ = 0.870403777 - 0.224078679 = 0.646325097 \\\\\text{Required probability} = 0.646325097[/tex]