Question

You may need to use the appropriate appendix table to answer this question. Television viewing reached a new high when the Nielsen 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 11 hours a day? (Round your answer to four decimal places.)

(b) How many hours of television viewing must a household have in order to be in the top 4% of all television viewing households? (Round your answer to two decimal places.)

Answer #1

Solution :

Given that ,

mean = = 8.35

standard deviation = = 2.5

a)

P(3 < x <11 ) = P((3-8.35)/ 2.5) < (x - ) / < (11-8.35) / 2.5) )

= P( -2.14 < z < 1.06)

= P(z < 1.06 ) - P(z < -2.14)

= 0.8554 - 0.0162

= 0.8392

Probability = 0.8392

b)

The z distribution of 4% is

P(Z > z ) = 4%

1 - P(Z < z ) = 0.04

P(Z < z ) = 1-0.04

P(Z < z ) = 0.96

P(Z < 1.75 ) = 0.96

z = 1.75

Using z-score formula

X = z* +

X = 1.75 * 2.5 + 8.35

X = 12.73

12.73 hours of television viewing must a household have in order to be in the top 4% of all television viewing households

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