Question

This is a sampling distribution question in intro statistics.

During the last week of the semester, students at a certain college spend on the average 4.2 hours using the school's computer terminals with a standard deviation of 1.8 hours. For a random sample of 36 students at that college, find the probabilities that the average time spent using the computer terminals during the last week for the semester is

a) at least 4.8 hours

b) between 4.1 and 4.5 hours

Please show your work.

Answer 1

Solution:

Given that,

mean =  = 4.2 hours

standard deviation =  = 1.8 hours

Sample = 36

= 4.2

=  ( /n) = (1.8 / 36 ) = 0.3

a ) p (   4.8)

= 1 -p (   4.8 )

= 1 - p ( - /) (4.8 - 4.2 / 0.3)

= 1 - p ( z 0.6 / 0.3 )

= 1 - p ( z 2 )

Using z table

= 1 - 0.9772

= 0.0228

Probability = 0.0228

b ) p ( 4.1 <   <  4.5)

= p (4.1 - 4.2 / 0.3) p ( - /) <  (4.5 - 4.2 / 0.3)

= p (- 0.1 / 0.3 < z < 0.3 / 0.3 )

= p (0.33 < z < 1 )

= p ( z < 1 ) - p ( z < 0 .33 )

Using z table

= 0.8413 - 0.6293

= 0.2120

Probability = 0.2120