Question

Television viewing reached a new high when the Nielsen Company reported a mean daily viewing time of 8.35 hours per household (USA Today, November 11, 2009). 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 5 and 10 hours a day?

B) How many hours of television viewing must a household have in order to be in the top 3% of all television viewing households?

C) What is the probability that a household views television more than 3 hours a day?

Answer 1

Explanation:

Using normal distribution,

z = (x -u)/s

Where u = mean = 8.35

s = standard deviation= 2.5

x = viewing rate of households

A) the probability that a household views television between 5 and 10 hours a day

P(5 lesser than or equal to x lesser than or equal to 10)

For P(5)

z = (5-8.35)/2.5 = -3.35/2.5= -1.34

Looking at the normal distribution table,

P(5) = 0.0901

For P(10)

z = (10 -8.35)/2.5 = 1.65/2.5= 0.66

Looking at the normal distribution table,

P(10) = 0.7454

P(5 lesser than or equal to x lesser than or equal to 10)

= P(10) -P(5)= 0.7454-0.0901= 0.6553

B) They must be viewing up to 97% = 0.97

Looking at the table, z value for 0.97=1.88

z= 1.88= (x -8.35)/2.5

x = 4.7+8.35 = 13.05 hours

C) P(x greater than 3) = 1-P(0 lesser than or equal to 3)

z = (3-8.35)/2.5

= -2.14

Looking at the table

P(0 lesser than or equal to 3)=0.0162

P(x greater than 3) = 1-0.0162 = 0.9838