at a certain law school scores on the law school admission test follow a normal distribution with mean =30 and standard deviation = 8.

what percentage of the students at this law school score below 36?

What percentage score between 36 and 44

find the value of the 90th percentile of the distribution of law school admission test scores at this school.

Guest Jul 10, 2017

#1**0 **

Let X the random variable of the school scores \(X- N(30,8^2)\)

The first question is to calculate:

\(P(X≤36)=P(\frac{X-30}{8}≤\frac{36-30}{8})=P(Z≤\frac{6}{8})=P(Z≤\frac{3}{4})=P(Z≤0.75)\)

\(Φ(0.75)=0.7734\)

Thus the percentage is 77.34%

note that \(Z-N(0,1)\)

The second question asks to calculate

\(P(36≤X≤44)=P(\frac{36-30}{8}≤\frac{X-30}{8}≤\frac{44-30}{8})\)

\(=P(\frac{3}{4}≤Z≤\frac{7}{4})=Φ(\frac{7}{4})-Φ(\frac{3}{4})\)

\(Φ(1.75)-Φ(0.75)=0.9599-0.7734=0.1865\)

Therefore the request percentage is 18.65%

The third question refers to the 90th percentile. You have to define \(x_{0}\)

that

\( P(X≤x_{o})=0.9\Rightarrow P(\frac{X-30}{8}≤\frac{x_{0}-30}{8})=0.9\)

\(⇒P(\frac{X-30}{8}≤\frac{x_{0}-30}{8})=0.9⇒Φ(\frac{x_{0}-30}{8})≤0.9⇒\frac{x_{0}-30}{8}=1.28\)

\(⇒x_{o}=40.24\)

Guest Jul 11, 2017