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

Answer:

a. P=0.98

b. P=0.66

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

Step-by-step 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?

$z_1=(X_1-\mu)/\sigma=(5-8.25)/2.50=-1.3\\\\z_2=(X_2-\mu)/\sigma=(10-8.25)/2.50=0.7\\\\P(5$

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.

Answer 2

Answer:

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

Step-by-step 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) = $\frac{1}{\sqrt{2\pi}2.5}\exp[-\frac{(x-8.35)^{2}}{2(2.5)^{2}}]$, 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.