Question

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 11 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.)

Answer 1

Explanation:

The mean daily viewing time of television is 8.35 hours and the standard deviation is 2.5 hours. We can use a normal probability distribution to answer the following questions about daily television viewing per household:

(a) The probability that a household views television between 3 and 11 hours a day is 0.9772 (rounded to four decimal places).

(b) To be in the top 3% of all television viewing households, a household must have 15.68 hours of television viewing per day (rounded to two decimal places).

The probability that a household views television more than 5 hours a day is 0.8944 (rounded to four decimal places).

I hope this helps! Let me know if you have any other questions.

Answer 2

(a) To find the probability that a household views television between 3 and 11 hours a day, we need to calculate the z-scores for 3 and 11 hours using the formula z = (x - μ) / σ, where μ is the mean and σ is the standard deviation. The z-score for 3 hours is (3 - 8.35) / 2.5 = -2.14 and the z-score for 11 hours is (11 - 8.35) / 2.5 = 1.06. Using a standard normal distribution table, we find that the probability of a z-score being between -2.14 and 1.06 is approximately 0.8209.

(b) To find how many hours of television viewing a household must have in order to be in the top 3% of all television viewing households, we need to find the z-score that corresponds to the top 3% of the standard normal distribution. Using a standard normal distribution table, we find that this z-score is approximately 1.88. Using the formula x = μ + zσ, we can calculate that a household must view television for approximately 8.35 + (1.88 * 2.5) = 12.75 hours to be in the top 3% of all television viewing households.

(c) To find the probability that a household views television more than 5 hours a day, we need to calculate the z-score for 5 hours using the formula z = (x - μ) / σ, where μ is the mean and σ is the standard deviation. The z-score for 5 hours is (5 - 8.35) / 2.5 = -1.34. Using a standard normal distribution table, we find that the probability of a z-score being greater than -1.34 is approximately 0.9099.