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 3 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 8% of all television viewing households (to 2 decimals)? hours

Answer 1

a)

Given :- = 8.35 , = 2.5 )
We convet this to Standard Normal as
P(X < x) = P( Z < ( X - ) / )
P ( 3 < X < 8 ) = P ( Z < ( 8 - 8.35 ) / 2.5 ) - P ( Z < ( 3 - 8.35 ) / 2.5 )
= P ( Z < -0.14) - P ( Z < -2.14 )
= 0.4443 - 0.0162
= 0.4282

b)

X ~ N ( µ = 8.35 , σ = 2.5 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.08 = 0.92
To find the value of x
Looking for the probability 0.92 in standard normal table to calculate critical value Z = 1.4051
Z = ( X - µ ) / σ
1.4051 = ( X - 8.35 ) / 2.5
X = 11.86