A school gives an entry exam for admission. Suppose the score of this exam follows a normal distribution N(400, 60). This year, the school decides to admit students who score in the top 30%. Suppose a student scored 428 on the test. Will the student be admitted? Explain your reasoning.
A school gives an entry exam for admission. Suppose the score of this exam follows a normal distribution N(400, 60). This year, the school decides to admit students who score in the top 30%. Suppose a student scored 428 on the test. Will the student be admitted? Explain your reasoning.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Admission Criteria Based on Entry Exam Scores**
A school gives an entry exam for admission. Suppose the score of this exam follows a normal distribution N(400, 60). This year, the school decides to admit students who score in the top 30%. Suppose a student scored 428 on the test. Will the student be admitted? Explain your reasoning.
---
In this context, we are dealing with a normal distribution with a mean (μ) of 400 and a standard deviation (σ) of 60. We need to find out if a score of 428 is within the top 30% of this distribution.
### Steps to Determine Admission:
1. **Convert the Score to a Z-Score:**
The Z-score formula is given by:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \( X \) is the raw score (428 in this case).
- \( \mu \) is the mean score (400).
- \( \sigma \) is the standard deviation (60).
Substituting the values:
\[
Z = \frac{(428 - 400)}{60} \approx 0.467
\]
2. **Find the Percentile:**
The Z-score of 0.467 corresponds to a percentile in the standard normal distribution table. Using a Z-table or statistical software, a Z-score of 0.467 corresponds approximately to the 68th percentile.
3. **Determine Admission:**
The school admits students who score in the top 30%. This means that these students need to be in the 70th percentile or above (since 100% - 30% = 70%).
Since a score of 428 corresponds to the 68th percentile, which is below the 70th percentile threshold, the student will not be admitted.
### Final Answer:
No, the student will not be admitted, as their score of 428 does not fall within the top 30% of the distribution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9ed366a-05c2-4728-8c6d-70ad10b89152%2F68d17750-c095-496d-90bd-33e25b6716fd%2F12fkysy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Admission Criteria Based on Entry Exam Scores**
A school gives an entry exam for admission. Suppose the score of this exam follows a normal distribution N(400, 60). This year, the school decides to admit students who score in the top 30%. Suppose a student scored 428 on the test. Will the student be admitted? Explain your reasoning.
---
In this context, we are dealing with a normal distribution with a mean (μ) of 400 and a standard deviation (σ) of 60. We need to find out if a score of 428 is within the top 30% of this distribution.
### Steps to Determine Admission:
1. **Convert the Score to a Z-Score:**
The Z-score formula is given by:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \( X \) is the raw score (428 in this case).
- \( \mu \) is the mean score (400).
- \( \sigma \) is the standard deviation (60).
Substituting the values:
\[
Z = \frac{(428 - 400)}{60} \approx 0.467
\]
2. **Find the Percentile:**
The Z-score of 0.467 corresponds to a percentile in the standard normal distribution table. Using a Z-table or statistical software, a Z-score of 0.467 corresponds approximately to the 68th percentile.
3. **Determine Admission:**
The school admits students who score in the top 30%. This means that these students need to be in the 70th percentile or above (since 100% - 30% = 70%).
Since a score of 428 corresponds to the 68th percentile, which is below the 70th percentile threshold, the student will not be admitted.
### Final Answer:
No, the student will not be admitted, as their score of 428 does not fall within the top 30% of the distribution.
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