Question

The time college students spend on the internet follows a Normal distribution. At Johnson University, the mean time is 6 hours per day with a standard deviation of 1.5 hours per day.

1. If 100 Johnson University students are randomly selected, what is the probability that the average time spent on the internet will be more than 6.2 hours per day?
   ____________ Round to 4 places.
   
2. If 100 Johnson University students are randomly selected, what is the probability that the average time spent on the internet is between 5.7 hour and 6.2 hours?
   _________ Round to 4 places
   
3. If 100 Johnson University students are randomly selected, what mean number of hours on the internet per day separated the bottom 33% and top 67% of internet usage for college students?
   hours per day. _______________ Round to 2 places.

Answer 1

solution:-

given that mean = 6 , standard deviation = 1.5

=> P(x > 6.2)

= P(z > (6.2-6)/(1.5/sqrt(100)))

= P(z > 1.33)

= 1 - P(z < 1.33)

= 1 - 0.9082

= 0.0918
-------------------------------
=> P(5.7 < x < 6.2)

= P((5.7-6)/(1.5/sqrt(100)) < z < (6.2-6)/(1.5/sqrt(100)))

= P(-2 < z < 1.33)

= P(z < 1.33) - P(z < -2)

= 0.9082 - 0.0228

= 0.8854
---------------------------------------------
here given that n = 100 and internet per day separated the bottom 33% and top 67% of internet usage for college students

the z-value corresponds to the probability of 0.33 is -0.44

=> x = z*(standard deviation/sqrt(n)) + mean

=> x = -0.44 *(1.5/sqrt(100))+6

=> x = 5.93