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 6 and 8 hours a day (to 4 decimals)?

b. How many hours of television viewing must a household have in order to be in the top 6% of all television viewing households (to 2 decimals)?

hours

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

Answer 1

Solution :

Given that ,

mean = = 8.35

standard deviation = = 2.5

a) P( 6 < x < 8) = P[(6 - 8.35)/ 2.5) < (x - ) /  < (8 - 8.35) / 2.5) ]

= P(-0.94 < z < -0.14)

= P(z < -0.14) - P(z < -0.94)

Using z table,

= 0.4443 - 0.1736

= 0.2707

b) Using standard normal table,

P(Z > z) = 6%

= 1 - P(Z < z) = 0.06  

= P(Z < z) = 1 - 0.06

= P(Z < 1.55 ) = 0.94  

z = 1.55

Using z-score formula,

x = z * +

x = 1.55 * 2.5 + 8.35

x = 12.23 hours.

c) P(x > 4) = 1 - p( x< 4)

=1- p P[(x - ) / < (4 - 8.35) / 2.5]

=1- P(z < -1.74)

= 1 - 0.0409

= 0.9591