SOLUTION: The distribution of scores on a standardized aptitude test is approximately normal with a mean of 520 and a standard deviation of 95.
What is the minimum score needed to be in
Algebra ->
Probability-and-statistics
-> SOLUTION: The distribution of scores on a standardized aptitude test is approximately normal with a mean of 520 and a standard deviation of 95.
What is the minimum score needed to be in
Log On
Question 1166922: The distribution of scores on a standardized aptitude test is approximately normal with a mean of 520 and a standard deviation of 95.
What is the minimum score needed to be in the top 15% on this test? Carry your intermediate computations to at least four decimal places, and round your answer to the nearest integer. Found 2 solutions by CPhill, ikleyn:Answer by CPhill(2030) (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
# Given parameters
mu = 520
sigma = 95
percentile = 0.85 # Corresponds to the top 15% (100% - 15% = 85%)
# 1. Find the z-score corresponding to the 85th percentile
# Use the Percent Point Function (inverse of CDF)
z_score = norm.ppf(percentile)
# 2. Calculate the minimum score (X)
# X = mu + Z * sigma
X = mu + z_score * sigma
# 3. Print the intermediate z-score and the final score X
print(f"Mean (mu): {mu}")
print(f"Standard Deviation (sigma): {sigma}")
print(f"Cumulative Probability: {percentile}")
print(f"Z-score (intermediate): {z_score:.4f}")
print(f"Raw Score (X): {X}")
# Round the final answer to the nearest integer
final_score = round(X)
print(f"Final score (rounded to nearest integer): {final_score}")
```
```text?code_stdout&code_event_index=2
Mean (mu): 520
Standard Deviation (sigma): 95
Cumulative Probability: 0.85
Z-score (intermediate): 1.0364
Raw Score (X): 618.46117200191
Final score (rounded to nearest integer): 618
```
The minimum score needed to be in the top $15\%$ on this test is **618**.
This is calculated by finding the score that corresponds to the **$85^{th}$ percentile** of the normal distribution, since the top $15\%$ of scores are those above $100\% - 15\% = 85\%$.
-----
## Calculation Steps
### 1\. Find the Z-score
The first step is to find the **Z-score** ($Z$) that has $85\%$ of the area under the standard normal curve to its left.
Using a Z-table or statistical software (carrying intermediate computations to at least four decimal places):
$$Z \approx \mathbf{1.0364}$$
### 2\. Calculate the Raw Score
We use the Z-score formula rearranged to solve for the raw score ($X$):
$$X = \mu + Z \cdot \sigma$$
Where:
* $\mu$ (Mean) $= 520$
* $\sigma$ (Standard Deviation) $= 95$
* $Z$ (Z-score) $\approx 1.0364$
Substituting the values:
$$X = 520 + (1.0364) \cdot 95$$
$$X = 520 + 98.458$$
$$X \approx 618.458$$
### 3\. Round to the Nearest Integer
Rounding the result to the nearest integer:
$$618.458 \approx \mathbf{618}$$
A score of **618** is the minimum score required to be in the top $15\%$ of test takers.
You can put this solution on YOUR website! .
The distribution of scores on a standardized aptitude test is approximately normal with a mean of 520 and a standard deviation of 95.
What is the minimum score needed to be in the top 15% on this test? Carry your intermediate computations to at least four decimal places, and round your answer to the nearest integer.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The other " tutor " (which is an Artificial Intelligence, I think - it is why I write
here " tutor " in quotation marks) obtained the value for the score to be 618.458
and then rounded it to the " nearest integer " 618.
In this problem, such rounding is a : to be in the top 15%,
the rounding should go to the nearest GREATER integer number, which is 619.
