Question

Television viewing reached a new high when the global information and measurement company reported a mean daily viewing time of hours per household. Use a normal probability distribution with a standard deviation of 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 9 hours a day (to 4 decimals)? b. How many hours of television viewing must a household have in order to be in the 2%top of all television viewing households (to 2 decimals)? hours c. What is the probability that a household views television more than hours a day (to 4 decimals)?

Answer 1

Answer:

(a) The probability that a household views television between 3 and 9 hours a day is 0.5864.

(b) The viewing hours in the top 2% is 13.49 hours.

(c) The probability that a household views television more than 5 hours a day is 0.9099.

Step-by-step explanation:

Let X = daily viewing time of of television hours per household.

The mean daily viewing time is, μ = 8.35 hours.

The standard deviation of daily viewing time is, σ = 2.5 hours.

The random variable X is Normally distributed.

To compute the probability of a Normal random variable, first we need to compute the raw scores (X) to z-scores (Z).

$z=\frac{x-\mu}{\sigma}$

(a)

Compute the probability that a household views television between 3 and 9 hours a day as follows:

$P(3$

                      $=P(-2.14$

Thus, the probability that a household views television between 3 and 9 hours a day is 0.5864.

(b)

Let the viewing hours in the top 2% be denoted by x.

Then,

P (X > x) = 0.02

⇒ P (X < x) = 1 - 0.02

    P (X < x) = 0.98

⇒ P (Z < z) = 0.98

The value of z for the above probability is:

z = 2.054

*Use a z-table for the value.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\2.054=\frac{x-8.35}{2.5}\\x=8.35+(2.054\times 2.5)\\x=13.485\\x\approx13.49$

Thus, the viewing hours in the top 2% is 13.49 hours.

(c)

Compute the probability that a household views television more than 5 hours a day as follows:

$P(X>5)=P(\frac{X-\mu}{\sigma}>\frac{5-8.35}{2.5})$

                $=P(Z>-1.34)\\=P(Z$

Thus, the probability that a household views television more than 5 hours a day is 0.9099.