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

Let be the daily television viewing per house hold.

The mean daily television viewing time,

The standard deviation,

(a)

The probability that a household views television between 5 and 10 hours a day is,

Therefore, the probability that a household views television between 5 and 10 hours a day is .

(b)

To determine the number of hours of television viewing must a household have in order to be in the top 3% of all television viewing households. This implies find the value of such that.

From the normal table, the critical value corresponding the value 0.97 is 1.881.

Therefore, the number of hours of television viewing must be .

(c)

The probability that a household views television more than 3 hours a day is,

Therefore, the probability that a household views television more than 3 hours a day is .

Answer 2

A) P(Between 5 and 10 hours)

= P((5 - 8.35)/2.5 < z < (10 - 8.35)/2.5)

= P(-1.34 < z < 0.66)

= P(z < 0.66) - P(z < -1.34)

= 0.7454 - 0.0901

= 0.6553

B) z score corresponding to top 3% area = 1.88

Hence,

Required hours = 8.35 + 1.88*2.5 = 13.05 hours

C) P(More than 3 hours)

= P(X > 3)

= P(z > (3 - 8.35)/2.5)

= P(z > -0.14)

= 0.5557

Answer 3

C) P( X > 3 )

z = (3 - 8.35) / 2.5 = -2.14

P(z > -2.14) = 1 - cummulative @(-2.14)

= 1 - 0.0162 = 0.9838