Question

what proportion of lsat scores will be outside the interval of 125 to 175? step 3: find the proportion of the distribution below z2; find the proportion below z1 in the z - table. then apply the correct rule. please report all values of proportion to 4 decimal places. p(z1 = choose your answer... ) = choose your answer... p(z2 = choose your answer... ) = choose your answer... .9750 .0250 apply the rule and report the final propor .9938 .0062 p(zfinal) = choose your answer...

Explanation:

Step1: Assume LSAT scores are normally - distributed

Let's assume the mean of LSAT scores is $\mu$ and the standard deviation is $\sigma$. First, we need to calculate the z - scores for $x_1 = 125$ and $x_2=175$ using the formula $z=\frac{x - \mu}{\sigma}$. But since the problem seems to be focused on using the z - table directly, we'll work with the given approach.

Step2: Find proportions from z - table

We assume that the z - scores corresponding to 125 and 175 are $z_1$ and $z_2$ respectively. Looking up in the standard normal (z -) table, if we assume that the proportion below $z_2$ is $p(z_2)$ and the proportion below $z_1$ is $p(z_1)$. The proportion of scores within the interval $[125,175]$ is $p = p(z_2)-p(z_1)$.

Step3: Calculate proportion outside the interval

The proportion of scores outside the interval $[125,175]$ is $1-(p(z_2)-p(z_1))$. If we assume that $p(z_1) = 0.0250$ and $p(z_2)=0.9750$ (from the given options and the properties of the standard normal distribution, where for a 95% confidence - like interval, the tails are symmetric and each tail has a proportion of 0.0250), then the proportion outside the interval is $1-(0.9750 - 0.0250)=0.0500$. But if we assume that the correct values from the z - table are such that the proportion within the interval is 0.9938 (a more extreme case, perhaps a 99% - like interval), then the proportion outside is $1 - 0.9938=0.0062$.

Answer:

If we assume a 95% - like interval (where the tails are symmetric and each tail has proportion 0.0250), the proportion outside the interval is 0.0500. If we assume the values from the z - table as in the given options where the proportion within is 0.9938, the proportion outside is 0.0062. Based on the given options, the answer is 0.0062.