Answer :
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).
[tex]z=\frac{x-\mu}{\sigma}[/tex]
(a)
Compute the probability that a household views television between 3 and 9 hours a day as follows:
[tex]P(3<X<9)=P(\frac{3-8.35}{2.5}<\frac{X-\mu}{\sigma}<\frac{9-8.35}{2.5})[/tex]
[tex]=P(-2.14<Z<0.26)\\=P(Z<0.26)-P(Z<-2.14)\\=0.60257-0.01618\\=0.58639\\\approx0.5864[/tex]
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:
[tex]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[/tex]
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:
[tex]P(X>5)=P(\frac{X-\mu}{\sigma}>\frac{5-8.35}{2.5})[/tex]
[tex]=P(Z>-1.34)\\=P(Z<1.34)\\=0.90988\\\approx0.9099[/tex]
Thus, the probability that a household views television more than 5 hours a day is 0.9099.