Question 1167636: The results of the latest writing of the LSAT (Law School Aptitude Test) showed results that were normally distributed with a mean score of 878 and a standard deviation of 51.
Enter percentage answers in percent form (i.e. 3.00
%
instead of 0.03). Round all final answers to 2 decimals.
(a) What percent of students scored between 810 and 909?
%
(b) What percent of students got 952 or more on the test?
%
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import numpy as np
from scipy.stats import norm
# Define parameters
mu = 878
sigma = 51
# --- Part (a): P(810 <= X <= 909) ---
X1_a = 810
X2_a = 909
# Calculate Z-scores
Z1_a = (X1_a - mu) / sigma
Z2_a = (X2_a - mu) / sigma
# Calculate probability
prob_a = norm.cdf(Z2_a) - norm.cdf(Z1_a)
percent_a = prob_a * 100
# --- Part (b): P(X >= 952) ---
X1_b = 952
# Calculate Z-score
Z1_b = (X1_b - mu) / sigma
# Calculate probability
# P(X >= 952) = 1 - P(X < 952) = 1 - norm.cdf(Z1_b)
prob_b = 1 - norm.cdf(Z1_b)
percent_b = prob_b * 100
# Print results rounded to 2 decimal places
print(f"Z1_a for 810: {Z1_a:.4f}")
print(f"Z2_a for 909: {Z2_a:.4f}")
print(f"Probability (a): {prob_a:.4f}")
print(f"Percent (a): {percent_a:.2f}")
print(f"Z1_b for 952: {Z1_b:.4f}")
print(f"Probability (b): {prob_b:.4f}")
print(f"Percent (b): {percent_b:.2f}")
```
```text?code_stdout&code_event_index=2
Z1_a for 810: -1.3333
Z2_a for 909: 0.6078
Probability (a): 0.6371
Percent (a): 63.71
Z1_b for 952: 1.4510
Probability (b): 0.0734
Percent (b): 7.34
```
This problem uses the properties of the **normal distribution** with a mean ($\mu$) of $878$ and a standard deviation ($\sigma$) of $51$. To solve it, we calculate the **Z-score** ($Z = \frac{X - \mu}{\sigma}$) for each score and use the standard normal cumulative distribution function (CDF).
## (a) What percent of students scored between 810 and 909?
We first find the Z-scores for the boundaries:
1. **For $X_1 = 810$:**
$$Z_1 = \frac{810 - 878}{51} = \frac{-68}{51} \approx -1.33$$
2. **For $X_2 = 909$:**
$$Z_2 = \frac{909 - 878}{51} = \frac{31}{51} \approx 0.61$$
The percentage of students who scored between $810$ and $909$ is the area under the curve between $Z_1$ and $Z_2$:
$$P(810 \le X \le 909) = P(-1.33 \le Z \le 0.61)$$
$$P(-1.33 \le Z \le 0.61) = \Phi(0.61) - \Phi(-1.33)$$
Using the standard normal CDF, we get:
$$0.7291 - 0.0912 \approx 0.6379$$
Converting to a percentage and rounding to $2$ decimals:
$$0.6379 \times 100\% \approx 63.71\%$$
The percent of students who scored between $810$ and $909$ is $\mathbf{63.71\%}$.
-----
## (b) What percent of students got 952 or more on the test?
We find the Z-score for $X = 952$:
$$Z = \frac{952 - 878}{51} = \frac{74}{51} \approx 1.45$$
The percentage of students who scored $952$ or more is the area in the right tail:
$$P(X \ge 952) = P(Z \ge 1.45) = 1 - P(Z < 1.45)$$
Using the standard normal CDF, $\Phi(1.45) \approx 0.9265$:
$$1 - 0.9265 \approx 0.0735$$
Converting to a percentage and rounding to $2$ decimals:
$$0.0735 \times 100\% \approx 7.35\%$$
(The computer-calculated value provides a slightly more precise result of $7.34\%$)
The percent of students who scored $952$ or more is $\mathbf{7.34\%}$.
-----
(a) What percent of students scored between 810 and 909? **63.71** %
(b) What percent of students got 952 or more on the test? **7.34** %
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