The time spent on Facebook per month by a user is normally d

The time spent on Facebook per month by a user is normally distributed with mean 7.5 hours and standard deviation 2.4 hours. if you select a random sample of 36 Facebook users

a. what is the probability that the sample mean is between 7 and 8 hours?

b. if you select a sample of 100 faceobok users, what is the probability that the sample mean is between 7 and 8 hours?

c. explain the diffference between the results in (a) and (b)

Solution

The time spent on Facebook per month by a user is normally distributed with mean 7.5 hours and standard deviation 2.4 hours. if you select a random sample of 36 Facebook users

a. what is the probability that the sample mean is between 7 and 8 hours?

Standard error = sd/sqrt(n)=2.4/sqrt(36) =0.4

z value of 7, z=(7-7.5)/0.4 = -1.25

z value of 8, z=(8-7.5)/0.4 = 1.25

P( 7<mean x<8)=P(-1.25<z<1.25)

P( z <1.25)-P( z < -1.25)

= 0.8944 - 0.1056

= 0.7888

b. if you select a sample of 100 faceobok users, what is the probability that the sample mean is between 7 and 8 hours?

Standard error = sd/sqrt(n)=2.4/sqrt(100) =0.24

z value of 7, z=(7-7.5)/0.24 = -2.08

z value of 8, z=(8-7.5)/0.24 = 2.08

P( 7<mean x<8)=P(-2.08<z<2.08)

P( z <2.08)-P( z < -2.08)

= 0.9812 - 0.0188

= 0.9624

c. explain the diffference between the results in (a) and (b)

Increase in the sample size decreases the standard error. This results in the increases in the required probability.