Question

If the average time to finish an IQ test is 85 minutes with standard deviation of 17 minutes and assuming the data are normally distributed,

1. what percentage of test takers finish less than 117 minutes?

2. what percentage of test takers finishes between one hour to hour and half?

3. find the time that separates the slowest 15% who finish the test?

4. find the time that separates the fastest 5% who finish the test?

Answer 1

Solution:- Given that mean = 85, sd = 17,

1. P(X < 117) = P((X-mean)/sd < (117-85)/17)
= P(Z < 1.8824)
= 0.9699

Use the standard normal table to find this area:

2. one hour = 60 min

one hour and half = 60+30 = 90 min

=> P(60 < X < 90) = P((60-85)/17 < (x-mean)/sd < (90-85)/17)
= P(-1.4706 < Z < 0.2941)
= P(Z < 0.2941) − P(Z < −1.4706)
= 0.6141 - 0.0708
= 0.5433

P(Z < 0.2941) can be found by using the following standard normal table :

P(Z < 1.4706) can be found by using the following standard normal table :

3. For Z = -1.04

=> X = mean + Z*Sd = 85 - (1.04*17) = 67.32

4. for Z = 1.645

=> 85 + (1.645*17) = 112.965 = 112.97