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 3 and 10 hours a day (to 4 decimals)

b. how many hours of television must a household have in order to be in the top 4% of all television viewing households (to 2 decimals)

c. what is the probability that a household views television more than 5 hours a day (to 4 decimals)

Answer 1

Solution :

Given that ,

mean = = 8.35

standard deviation = = 2.5

a.

P(3 < x < 10) = P((3 - 8.35)/ 2.5) < (x - ) / < (10 - 8.35) / 2.5) )

= P(-2.14 < z < 0.75)

= P(z < 0.75) - P(z < -2.14)

= 0.7734 - 0.0162

= 0.7572

Probability = 0.7572

b.

Using standard normal table,

P(Z > z) = 4%

1 - P(Z < z) = 0.04

P(Z < z) = 1 - 0.04

P(Z < 1.75) = 0.96

z = 1.75

Using z-score formula,

x = z * +

x = 1.75 * 2.5 + 8.35 = 12.73

12.73 hours

c.

P(x > 5) = 1 - P(x < 5)

= 1 - P((x - ) / < (5 - 8.35) / 2.5)

= 1 - P(z < -1.34)

= 1 - 0.0901

= 0.9099

Probability = 0.9099