Question

PROBLEM 2.

Based on data from Consumer Reports, replacement times of TVs is on average 3.4 years with the standard deviation of 1.2 years. Answer the following questions.

Question 2 (2 points):

A random sample of 41 TVs is selected. Find the probability that their average replacement time is between 3 and 4 years.

a) 0.6472
b) 0.9827
c) 0.3208
d) None of the above

Question 3 (2 points):

If a random sample of 29 TVs was chosen, what would be the probability that their average replacement time exceeds 3.5 years?

a) 0.6736
b) 0.3264
c) 0.4681
d) None of the above

Answer 1

Solution:

Given that,

mean =   = 3.4

standard deviation = = 1.2

a ) n = 41

= 3.4

=  ( /n) = (1.2/ 41 ) = 0.1874

C ) P ( 3 < x < 4 )

P ( 3 - 3.4 / 0.1874 ) < ( x -  / ) < ( 4 - 3.4 / 0.1874)

P ( - 0.4 / 0.1874 < z < 0.6 / 0.1874)

P (- 2.13 < z < 3.20 )

P (z < 3.20 ) - p ( z < - 2.13 )

Using z table

= 0.9993 - 0.0166

= 0.9827

Probability = 0.9827

Option b ) is correct.

b ) n = 29

= 3.4

=  ( /n) = (1.2/29 ) = 0.2228

P ( x > 3.5 )

= 1 - P ( x < 3.5 )

= 1 - P ( x -  / ) < ( 3.5 - 3.4 / 0.2228)= 1 -

= 1 - P ( z < 0.1 / 0.2228)

= 1 - P ( z < 0.45 )

Using z table

= 1 - 06736

= 0.3264

Probability = 0.3264

Option b ) is correct.