Show all your work using Excel functions Attach your Excel outputs. Television v
ID: 3181718 • Letter: S
Question
Show all your work using Excel functions Attach your Excel outputs. Television viewing reached a new high when the Nielsen Company reported a mean daily viewing time of 8.35 hours per household (USA Today). Use a normal probability distribution with a standard deviation of 2.5 hours to answer the following questions about daily television viewing per household. What is the probability that a household views television less than 6 hours? What is the probability that a household views television between 7 and 9 hours a day? How many hours of television viewing must a household have in order to be in the top 5% of all television viewing households? Lori Jeffery is a successful sales representative for a major publisher of college textbooks. Historically, Lori obtains a book adoption on 25% of her sales calls. Viewing calls for one month as sample of all possible sales assume that a statistical analysis of the data yields a standard error of the proportion of 0.0625 What is the probability that Lori will obtain book adoptions on 35% or more of her sales calls during a one-month period? An application of the Excel "Data Analysis Attach your Excel outputs A simple random sample often employees of a large corporation provided the following information. Determine the point estimate for the average age of all employees using Excel "Data Analysis: Descriptive Statistics" function. What is the point estimate for the standard deviation of the average age of all employees? Describe the point estimate, normal probability distribution, and standard normal probability distribution in detail (One double space typed page).Explanation / Answer
Mean ( u ) = 8.35 hours = 515 minuetes
Standard Deviation ( sd )= 2.5 = 150 minutes
Normal Distribution = Z= X- u / sd ~ N(0,1)
Q1.
P(X < 360) = (360-515)/150
= -155/150= -1.0333
= P ( Z <-1.0333) From Standard Normal Table
= 0.1507
Q2.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 420) = (420-515)/150
= -95/150 = -0.6333
= P ( Z <-0.6333) From Standard Normal Table
= 0.26326
P(X < 540) = (540-515)/150
= 25/150 = 0.1667
= P ( Z <0.1667) From Standard Normal Table
= 0.56618
P(420 < X < 540) = 0.56618-0.26326 = 0.3029
Q3.
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u/ (s.d) > x - 515/150) = 0.05
That is, ( x - 515/150) = 1.6449
--> x = 1.6449 * 150+515 = 761.728