In: Statistics and Probability
A survey was conducted to measure the number of hours per week adults in the United States spend on home computers. In the survey, the number of hours was normally distributed, with a mean of 7 hours and a standard deviation of 1 hour. A survey participant is randomly selected. Which of the following statements is true?
(a) The probability that the hours spent on the home computer by the participant are less than 4.5 hours per week is 0.9938.
(b) The probability that the hours spent on the home computer by the participant are between 4.5 and 9.5 hours per week is 0.0124.
(c) The probability that the hours spent on the home computer by the participant are more than 9.5 hours per week is 0.9938.
(d) 0.13% of the adults spend more than 4 hours per week on a home computer. (e) If 43 adults in the United States are randomly selected, you would expect to say about 1 adult spend less than 5 hours per week on a home computer.
Solution:- (e) If 43 adults in the United States are randomly selected, you would expect to say about 1 adult spend less than 5 hours per week on a home computer. True
Mean of 7 hours and a standard deviation of 1 hour.
(a) The probability that the hours spent on the home computer by the participant are less than 4.5 hours per week is 0.9938. False
x = 4.5
By applying normal distribution:-
z = - 2.5
P(z < - 2.5) = 0.0062
(b) The probability that the hours spent on the home computer by the participant are between 4.5 and 9.5 hours per week is 0.0124.
x1 = 4.5
x2 = 9.5
By applying normal distribution:-
z1 = - 2.50
z2 = 2.50
P( -2.50 < z < 2.50) = P(z > - 2.50) - P(z > 2.50)
P( -2.50 < z < 2.50) = 0.9938 - 0.0062
P( -2.50 < z < 2.50) = 0.9876
(c) The probability that the hours spent on the home computer by the participant are more than 9.5 hours per week is 0.9938. False
x = 9.5
By applying normal distribution:-
z = 2.5
P(z > 2.5) = 0.0062
(d) 0.13% of the adults spend more than 4 hours per week on a home computer. False
x = 5.0
By applying normal distribution:-
z = - 3.0
P(z > - 3.0) = 0.9987
P(z > - 3.0) = 99.87%
(e) If 43 adults in the United States are randomly selected, you would expect to say about 1 adult spend less than 5 hours per week on a home computer. True
x = 5.0
By applying normal distribution:-
z = - 2.0
P(z < - 2.0) = 0.0228
Expected number among 43 adults who spend less than 5 hours per week on a home computer = 0.0228*43 = 0.9804 = approximately 1