Question

if someone took an lsat test and received a 143, what proportion of scores will be greater than this?
step 3: report the proportion of the distribution greater than z.
please report the final proportion after applying the correct rule procedure.
please answer this question by:

1. find the proportion based on the z - score
2. apply the rule
3. report the final proportion

what is the proportion based on the z - score computed in step 2 (before applying the rule)?
p(z > choose your answer...) = choose your answer...

Explanation:

Step1: Assume LSAT scores are normally - distributed

Typical mean ($\mu$) of LSAT scores is 150 and standard - deviation ($\sigma$) is 10.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 143$, $\mu=150$, and $\sigma = 10$. So $z=\frac{143 - 150}{10}=\frac{-7}{10}=- 0.7$.

Step3: Find the proportion from the standard normal table

The standard normal table gives $P(Z\leq z)$. For $z=-0.7$, from the standard normal table, $P(Z\leq - 0.7)=0.2420$.

Step4: Calculate the proportion greater than z

We want $P(Z > - 0.7)$. Since the total area under the normal curve is 1, $P(Z > - 0.7)=1 - P(Z\leq - 0.7)=1 - 0.2420 = 0.7580$.

Answer:

$P(Z>-0.7)=0.7580$