The mean score on the Law School Admission Test (LSAT) is 152, while the standard deviation is 12. Assume LSAT score follows a normal distribution. (A z-table will be provided upon request.)
a) The average LSAT of a person attending Stanford’s law school is 168. What percentage of the LSAT test takers had a score of 168 or higher?
b) A score of 160 is considered competitive. What percentage of test takers is not considered to have a competitive score? @amistre64

I HAVE THE Z TABLE

Question

The mean score on the Law School Admission Test (LSAT) is 152, while the standard deviation is 12.  Assume LSAT score follows a normal distribution. (A z-table will be provided upon request.)
a)	The average LSAT of a person attending Stanford’s law school is 168.  What percentage of the LSAT test takers had a score of 168 or higher?
b)	A score of 160 is considered competitive.  What percentage of test takers is not considered to have a competitive score?    @amistre64
 
I HAVE THE Z TABLE

amistre64 · Answer

do you recall your z calcuations?

amistre64 · Answer

your zscore defines the number of standard deviations that can go int to the difference between the mean and a given value:$$z=\frac{x-mean}{sd}$$

anonymous · Answer

YES I DO @amistre64

anonymous · Answer

FROM my work = 160-152/12=0.67 which is equivalent to 0.2486 on the z table @amistre64

amistre64 · Answer

i assume the first one was no problem then .....

anonymous · Answer

but where do i go from here.

amistre64 · Answer

look for onthe ztable for a score of .67

anonymous · Answer

0.2486

amistre64 · Answer

sounds like a measure from the mean

amistre64 · Answer

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