Question 1167512: The results of the latest writing of the LSAT (Law School Aptitude Test) showed results that were normally distributed with a mean score of 826 and a standard deviation of 50.
For part (d) enter probability answers in decimal form (i.e. 0.0003 instead of 0.0300 %
). Round the final answer to 4 decimals.
c)If a group of 59 applicants is randomly selected, what is the probability that the group average is not less than 846?
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 = 826 # Population mean score
sigma = 50 # Population standard deviation
n = 59 # Sample size
X_bar = 846 # Sample average score
# 1. Calculate the Standard Error (Standard Deviation of the Sample Mean)
sigma_x_bar = sigma / np.sqrt(n)
# 2. Calculate the Z-score for X_bar = 846
Z = (X_bar - mu) / sigma_x_bar
# 3. Calculate the probability P(X_bar >= 846)
# P(X_bar >= 846) = 1 - P(X_bar < 846) = 1 - norm.cdf(Z)
probability = 1 - norm.cdf(Z)
# Print results rounded to 4 decimal places
print(f"Population Mean (mu): {mu}")
print(f"Population Standard Deviation (sigma): {sigma}")
print(f"Sample Size (n): {n}")
print(f"Standard Error (sigma_x_bar): {sigma_x_bar:.4f}")
print(f"Z-score: {Z:.4f}")
print(f"Probability P(X_bar >= 846): {probability:.4f}")
```
```text?code_stdout&code_event_index=2
Population Mean (mu): 826
Population Standard Deviation (sigma): 50
Sample Size (n): 59
Standard Error (sigma_x_bar): 6.5094
Z-score: 3.0725
Probability P(X_bar >= 846): 0.0011
```
The probability that the group average score is not less than $846$ is **0.0011**.
Since the individual scores are normally distributed, the **Central Limit Theorem** tells us that the distribution of the sample mean ($\bar{X}$) is also normally distributed.
Here is the step-by-step calculation:
### 1\. Identify the Distribution of the Sample Mean
* Population Mean ($\mu$): $826$
* Population Standard Deviation ($\sigma$): $50$
* Sample Size ($n$): $59$
* Target Sample Mean ($\bar{X}$): $846$
The mean of the sample averages is:
$$\mu_{\bar{X}} = \mu = 826$$
The standard deviation of the sample averages (Standard Error) is:
$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{59}} \approx 6.5094$$
### 2\. Calculate the Z-score
We standardize the target average ($\bar{X} = 846$):
$$Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{846 - 826}{50/\sqrt{59}} \approx \frac{20}{6.5094} \approx 3.0725$$
### 3\. Calculate the Probability
The question asks for the probability that the group average is **not less than** $846$, which means $P(\bar{X} \ge 846)$ or $P(Z \ge 3.0725)$.
$$P(\bar{X} \ge 846) = 1 - P(\bar{X} < 846) = 1 - \Phi(3.0725)$$
Using the Z-score value:
$$1 - 0.998910 \approx 0.001090$$
Rounding the final answer to $4$ decimal places: **0.0011**.
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