Question

Suppose that the mean daily viewing time of television is 8.35 hours. 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 11 hours a day? (Round your answer to four decimal places.) Incorrect: Your answer is incorrect. (b) How many hours of television viewing must a household have in order to be in the top 3% of all television viewing households? (Round your answer to two decimal places.) Incorrect: Your answer is incorrect. hrs (c) What is the probability that a household views television more than 3 hours a day? (Round your answer to four decimal places.) Incorrect: Your answer is incorrect.

Answer 1

Solution :

Given that ,

mean = = 8.35

standard deviation = = 2.5

a) P(5 < x < 11 ) = P[(5 - 8.35)/ 2.5) < (x - ) /  < (11 - 8.35) / 2.5) ]

= P(-1.34 < z < 1.06)

= P(z < 1.06) - P(z < -1.34)

Using z table,

= 0.8554 - 0.0901

= 0.7653

b) Using standard normal table,

P(Z > z) = 3%

= 1 - P(Z < z) = 0.03

= P(Z < z) = 1 - 0.03

= P(Z < z ) = 0.97

= P(Z < 1.88 ) = 0.97

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 2.5 + 8.35

x = 13.05 hours

c) P(x > 3 ) = 1 - p( x< 3)

=1- p P[(x - ) / < (3 - 8.35) / 2.5]

=1- P(z < -2.14)

= 1 - 0.0162

= 0.9838