Question

Television viewing reached a new high when the global information and measurement 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 between 4 and 10 hours a day? (to 4 decimals)

b. how many hours of television viewing must a household have in order to be in the top 7% of all television viewing household? (to 2 decimals)

c. what is the probability that a household views television more than 4 hours a day? (to 4 decimals)

Answer 1

Let X denote the daily viewing time

We are given :

u 8.35

=2.5)

a.

Required probability = Probability that household views television between 4 and 10 hours a day = P(4<X<10)

$P(4<X<10) = P\left ( \frac{4-\mu}{\sigma}<\frac{X-\mu}{\sigma}< \frac{10-\mu}{\sigma}\right )$

X -8.35 10 8.35 4 8.35 P 2.5 2.5 2.5

=P(-1.74 Z< 0.66)

$= P\left ( Z< 0.66\right )-P\left ( Z<-1.74\right )$

We use Excel function "NORMSDIST()" as :

=P(Z<0.66)-P (Z-1.74 NORMSDIST(0.66)-NORMSDIST-1.74)

0.7454 0.0409 0.7044

Hence,

Probability that household views television between 4 and 10 hours a day = 0.7044

b.

We need to find the 93^rd Percentile

Let P₉₃ denote the 93^rd Percentile

We have :

P93H 0.93 P(X < P3)P

X-8.35 P3-8.35 - 0.93 P 2.5 2.5

$\left (Z<\frac{P_{93}-8.35}{2.5} \right ) = 0.93$

We use the Excel function "NORMSINV()" as :

$\frac{P_{93}-8.35}{2.5} = NORMSINV(0.93) = 1.476$

P3-8.35 1.476 2.5

Number of hours of television viewing must a household have in order to be in the top 7% of all television viewing household = 12.04

c.

Required probability = Probability that a household views television more than 4 hours a day = P(X>4)

$P(X>4) = P\left ( \frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma} \right )$

X-8.35 P 4 8.35 2.5 2.5

$= P\left (Z>-1.74 \right )$

$=1- P\left (Z<-1.74 \right )$

We use Excel function "NORMSDIST()" as:

$=1- P\left (Z<-1.74 \right ) = 1-NORMSINV(-1.74) = 1-0.0409 = 0.9591$

Hence,

Probability that a household views television more than 4 hours a day = 0.9591