A group of statistics students decided to conduct a survey a

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spend studying per week. Based on a simple random sample, they surveyed 144 students. The statistics showed that students studied an average of 20 hours per week with a standard deviation of 10 hours. What is the probability that average student study time is between 18 and 22 hours?

Solution

Given n=144, xbar=20, s=10

(a) What is the standard error of the mean?

standard error of the mean= s/n=10/sqrt(144)= 0.83

(b) What is the probability that a sample mean would exceed 20 hours per week?

P(xbar>20) = P((xbar-mean)/s/n > (20-20)/0.83)

=P(Z>0)

=0.5 (check standard normal table)

(c) What is the probability of finding a sample mean less than 18 hours?

P(xbar<18) = P(Z<(18-20)/(10/sqrt(144)))

=P(Z< -2.4)

= 0.0082 (check standard normal table)

(d) What is the probability that average student study time is between 18 and 22 hours?

P(18<xbar<22) = P(-2.4<Z<(22-20)/(10/sqrt(144)))

=P(-2.4<Z< 2.4)

= 0.9836(check standard normal table)