QUESTION IMAGE
Question
11 fill in the blank 1.25 points
what proportion of lsat scores will be outside the interval of 125 to 175?
step 2: compute the z - scores (z₂ and z₁). draw the distribution and shade the proportion.
z = \frac{x - \mu}{\sigma}
please enter the values into the z - score formula below.
z₁ = choose your answer... - choose your answer... / choose your answer... = choose your answer...
z₂ = choose your answer... - choose your answer... / choose your answer... = choose your answer...
what direction would we shade the distribution?
we are interested in the proportion choose your answer... of two z - scores which means we would shade choose your answer... the z - scores.
Step1: Recall z - score formula
The z - score formula is $z=\frac{X - \mu}{\sigma}$, where $X$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation. However, since the mean $\mu$ and standard deviation $\sigma$ of LSAT scores are not given in the problem, let's assume for the sake of general procedure that $X_1 = 125$ and $X_2=175$. Then $z_1=\frac{125-\mu}{\sigma}$ and $z_2=\frac{175 - \mu}{\sigma}$.
Step2: Use the properties of the standard normal distribution
The total area under the standard - normal curve is 1. The proportion of data within the z - scores $z_1$ and $z_2$ is $P(z_1 We are interested in the proportion outside of two z - scores. So we would shade the two tails of the standard - normal distribution, i.e., the area to the left of $z_1$ and the area to the right of $z_2$.Step3: Shading the distribution
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We need the values of the mean $\mu$ and standard deviation $\sigma$ of LSAT scores to calculate the exact z - scores and the proportion. But the general procedure is to calculate $z_1=\frac{125 - \mu}{\sigma}$ and $z_2=\frac{175-\mu}{\sigma}$, and the proportion outside the interval is $1-(\Phi(z_2)-\Phi(z_1))$, and we shade the two tails of the standard - normal distribution.