Question

television viewing reached a new high when the global information and measurement company reported a mean daily viewing time of 8.35 hours per household. 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 6 and 10 hours a day (to 4 decimals)?

b. How many hours of television viewing must a household have in order to be in the top 5% of all television viewing households (to 2 decimals)?

c.What is the probability that a household views television more than 2 hours a day (to 4 decimals)?

Answer 1

We are given the distribution here as:

a) The probability here is computed as:
P( 6 < X < 10)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we get here:

Therefore 0.5718 is the required probability here.

b) From standard normal tables, we have:
P(Z < 1.645) = 0.95

Therefore P(Z > 1.645) = 0.05

Therefore the number of hours is computed here as:
= Mean + 1.645*Std Dev

= 8.35 + 1.645*2.5 = 12.4625

Therefore 12.4625 is the required hours value here.

c) The probability here is computed as:

P(X > 2)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal variable, we get here:

Therefore 0.0055 is the required probability here.