Question

A study conducted by the MLC on campus found the number of hours spent at the MLC during a week. The distribution of times spent at the MLC is normally distributed with an average of 75 minutes and a standard deviation of 18 minutes.

(a) If a single student that uses the MLC is selected at random, what is the probability that this student spent between 80 and 100 minutes at the MLC that week?

(b) Now if a sample of 20 students is selected, what is the probability that the sample will have a mean time spent at the MLC that is between 80 and 100 minutes?

Answer 1

Solution :

Given that ,

mean = = 75

standard deviation = = 18

P(80< x <100 ) = P[(80 - 75) /18 < (x - ) / < (100 - 75) / 18)]

= P(0.28 < Z <1.39 )

= P(Z <1.39 ) - P(Z <0.28 )

Using z table,  

=0.9177 -0.6103

=0.3074

(B)

n = 20

= 75

=  / n = 18 / 20=4.0249

P(80<   <100 ) = P[(80 - 75) /4.0249< ( - ) /   < (100 - 75) / 4.0249]

= P( 1.24< Z < 6.21)

= P(Z <6.21 ) - P(Z <1.24 )

Using z table,  

= 1 - 0.8925

=0.1075