Question

Use the z-score table to answer the question. Note: Round z-scores to the nearest hundredth and then find the required A values using the table.

A psychologist finds that the intelligence quotients of a group of patients are normally distributed, with a mean of 101 and a standard deviation of 16. Find the percent of the patients with the following IQs.

(a) above 113
%

(b) between 89 and 119
%

Answer 1

Solution :

Given ,

mean = = 101

standard deviation = = 16

P(x >113 ) = 1 - P(x< 113)

= 1 - P[(x -) / < (113 -101) / 16]

= 1 - P(z < 0.75)

Using z table

= 1 - 0.7734

=0.2266

answer=22.66%

(B)P(89< x <119 ) = P[(89 -101) / 16 < (x - ) / < (119-101) / 16 )]

= P( -0.75< Z <1.13 )

= P(Z < 1.13) - P(Z < -0.75)

Using z table   

= 0.8708-0.2266

= 0.6442

answer=64.42%