Question

4. Television viewing reached a new high when the Nielsen 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 more than 3 hours a day? b. What is the probability that a household spends 5 – 10 hours watching television more a day? c. How many hours of television viewing must a household have in order to be in the top 3% of all television viewing households?

Answer 1

(a) 0.9838 (b) 0.6553 (c) 13.05198

Explanation:

We have that the daily viewing time is a random variable normally distributed with mean and standard deviation

`\mu` = 8.35 hours and

`\sigma` = 2.5 hours

respectively. If we call the random variable X, the density function of this random variable is given by

f(x) =
`(1)/(√(2\pi)2.5)\exp[-((x-8.35)^(2))/(2(2.5)^(2))]`, and we can calculate the next probabilities using a computer or a table from a book.

(a) P(X>3)=
`\int\limits^(\infty)_3 {f(x)} \, dx`=0.9838

in the R statistical programming language we use the instruction pnorm(3, mean = 8.35, sd = 2.5, lower.tail = FALSE)

(b) P(
`5\leq X\leq 10`) =
`\int\limits^(10)_5 {f(x)} \, dx` = 0.6553

in the R statistical programming language we use the instruction

pnorm(10, mean = 8.35, sd = 2.5) - pnorm(5, mean = 8.35, sd = 2.5)

(c) You should find a value
`x_(0)` such that

`P(X\geq x_(0)) = 0.03`, this value is
`x_(0)`=13.05198

The instruction qnorm(0.03, mean = 8.35, sd = 2.5, lower.tail = FALSE) give us 13.05198 in the R statistical programming language.

Answer 2

a. P=0.98

b. P=0.66

c. The top 3% of all TV viewing households watch 12.95 hours or more.

Explanation:

We have a normal distribution with these parameters:

`\mu=8.35\\\\\sigma=2.50`

a. What is the probability that a household views television more than 3 hours a day?

To calculate this, first we calculate the z-value for X=3 and then calculate the probability according to the standard normal distribution.

`z=(X-\mu)/\sigma=(3-8.25)/2.50=-2.1\\\\P(X>3)=P(z>-2.1)=0.98214`

b. What is the probability that a household spends 5 – 10 hours watching television more a day?

`z_1=(X_1-\mu)/\sigma=(5-8.25)/2.50=-1.3\\\\z_2=(X_2-\mu)/\sigma=(10-8.25)/2.50=0.7\\\\P(5<X<10)=P(-1.3<z<0.7)\\\\P(-1.3<z<0.7)=P(z>-1.3)-P(z>0.7)=0.9032-0.2412=0.662`

c. How many hours of television viewing must a household have in order to be in the top 3% of all television viewing households?

To calculate this we have to know the z-value for
`P(z>z_1)=0.03`.

This z-value, according to the standard normal distribution is z=1.88.

Then, we can calculate the number of hours X as:

`X=\mu+z*\sigma=8.25+1.88*2.5=8.25+4.7=12.95`

The top 3% of all TV viewing households watch 12.95 hours or more.