The results of the latest writing of the LSAT (Law School Aptitude Test) showed results that were normally distributed with a mean score of 837 and a standard deviation of 55.

(a) What percent of students scored between 795 and 949?

 What percent of students got 951 or more on the test? 

Solution:

Let x be the score obtained on the LSAT; we have a normal distribution with the following parameters:

μ=837

σ=55

We must calculate:

a)

P(795<x<949)=?

We standardize our values:

z=σx−μ?

For x=795__________z=55795−837?=−0.76

For x=949__________z=55949−837?=2.04

Then:

P(795<x<949)=P(−0.76<z<2.04)=P(z<2.04)−P(z<−0.76)

Using our cumulative normal distribution table we have:

P(z<2.04)−P(z<−0.76)=0.9793−0.2236=0.7557__________75.57%

The percentage of students obtained between 795 and 949 is 75.57%

b)

P(x≥951)=?

Applying the complement property in probabilities we have:

P(x≥951)=1−P(x<951)

Then:

For x=951_____________z=55951−837?=2.07

Then:

P(x<951)=P(z<2.07)=0.9808 (normal table value)

Thus:

P(x≥951)=1−0.9808=0.0192________________ 1.92%

The percentage of students who scored 951 or higher on the test is 1.92%

Step-by-step explanation

Suggested Bibliography:
George Canavos Probability and Statistics.