Respuesta :
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
[tex]\mu[/tex] = 8.35 hours and
[tex]\sigma[/tex] = 2.5 hours
respectively. If we call the random variable X, the density function of this random variable is given by
f(x) = [tex]\frac{1}{\sqrt{2\pi}2.5}\exp[-\frac{(x-8.35)^{2}}{2(2.5)^{2}}][/tex], and we can calculate the next probabilities using a computer or a table from a book.
(a) P(X>3)=[tex]\int\limits^{\infty}_3 {f(x)} \, dx[/tex]=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([tex]5\leq X\leq 10[/tex]) = [tex]\int\limits^{10}_5 {f(x)} \, dx[/tex] = 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 [tex]x_{0}[/tex] such that
[tex]P(X\geq x_{0}) = 0.03[/tex], this value is [tex]x_{0}[/tex]=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:
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:
[tex]\mu=8.35\\\\\sigma=2.50[/tex]
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.
[tex]z=(X-\mu)/\sigma=(3-8.25)/2.50=-2.1\\\\P(X>3)=P(z>-2.1)=0.98214[/tex]
b. What is the probability that a household spends 5 – 10 hours watching television more a day?
[tex]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[/tex]
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 [tex]P(z>z_1)=0.03[/tex].
This z-value, according to the standard normal distribution is z=1.88.
Then, we can calculate the number of hours X as:
[tex]X=\mu+z*\sigma=8.25+1.88*2.5=8.25+4.7=12.95[/tex]
The top 3% of all TV viewing households watch 12.95 hours or more.
