Suppose that the mean daily viewing time of television is 8.35 hours. Use a normal probability distribution with a standard deviation of 2.5 hours to answer the following questions about daily television viewing per household
(a) What is the probability that a household views television between 3 and 12 hours a day? (Round your answer to four decimal places.)
(b) How many hours of television viewing must a household have in order to be in the top 3% of all television viewing households? (Round your answer to two decimal places.)
hrs
(c) What is the probability that a household views television more than 5 hours a day? (Round your answer to four decimal places.)

a. The probability of viewing television between 3 and 12 hours is 0.9117b. The number of hours that the household watches television is 13.05 hours.c. The probability that a household views television more than 5 hours a day is 0.9099How to solve for the probabilityGiven that ,mean = u = 8.35standard deviation = σ = 2.5a) P(3 < x < 12) = P[(3 - 8.35)/ 2.5) < (x)  < (12 - 8.35) / 2.5) ]= P(- 2.14 < z < 1.46)= P(z < 1.46) - P(z < -2.14)we have to use the z table to solve this further0.9279 - (1 - 0.9838)= 0.9279 - 0.0162= 0.9117The probability that a household would have to view the television between 3 and 12 hours everyday is 0.91172. The probability given = 3% = 0.031 - 0.03 = 0.97P(Z < 1.88 ) = 0.97z = 1.88we have to use the z score formula= z * u + sd= 1.88 x 2.5 + 8.35= 13.05 hours.The number of television hours that a family would need to have in order to be in the top 3% would be 13.05 hours.c. the probability that a family views television more than 5 hours a dayP(x > 5 ) = 1 - p( x< 5)1 - p(5 - 8.35) / 2.5= 1- P(z < -1.34)= 1 - 0.0901= 0.9099Read more on probability distribution here: https://brainly.com/question/23286309#SPJ1