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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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just answer d and e
![### Understanding Normal Distributions in SAT and ACT Scores
The SAT Verbal Test scores in recent years have followed an approximate normal distribution with a mean of 460 and a standard deviation of 110. The ACT English Test scores similarly follow a normal distribution with a mean of 17 and a standard deviation of 2.5.
#### a. Fill Out the Distribution Curves
For each distribution curve, label the set of numbers corresponding to both the SAT Verbal Test and the ACT English Test. Each set of values should represent the spread of scores covering approximately 3 standard deviations from the mean.
**SAT Verbal Test (Mean = 460, SD = 110):**
- The values:
- \[130, 240, 350, 460, 570, 680, 790\]
**ACT English Test (Mean = 17, SD = 2.5):**
- The values:
- \[9.5, 12, 14.5, 17, 19.5, 22, 24.5\]
Ensure each value is correctly placed according to its position relative to the mean in the normal distribution curves.
#### b. Probability of a Student Scoring Less Than 14.5 on the ACT English Test
To find this probability, calculate the z-score:
- \( z = \frac{X - \mu}{\sigma} = \frac{14.5 - 17}{2.5} = \frac{-2.5}{2.5} = -1 \)
Using the standard normal distribution table, the probability corresponding to a z-score of -1 is approximately 0.1587. Hence, there is a 15.87% chance that a randomly selected student will score less than 14.5 on the ACT English Test.
#### c. Probability of a Student Scoring More Than 650 on the SAT Verbal Test
Calculate the z-score:
- \( z = \frac{X - \mu}{\sigma} = \frac{650 - 460}{110} = \frac{190}{110} \approx 1.727 \)
Using the standard normal distribution table, the probability corresponding to a z-score of 1.727 is approximately 0.9582. The required probability is \(1 - 0.9582 = 0.0418\). Thus, there is a 4.18% chance a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9d14dc9-6899-40d3-8d6b-066b6648b231%2F26289a95-9c9d-45f3-8a14-fb0b214b3b7c%2Fg1nd3vq_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Normal Distributions in SAT and ACT Scores
The SAT Verbal Test scores in recent years have followed an approximate normal distribution with a mean of 460 and a standard deviation of 110. The ACT English Test scores similarly follow a normal distribution with a mean of 17 and a standard deviation of 2.5.
#### a. Fill Out the Distribution Curves
For each distribution curve, label the set of numbers corresponding to both the SAT Verbal Test and the ACT English Test. Each set of values should represent the spread of scores covering approximately 3 standard deviations from the mean.
**SAT Verbal Test (Mean = 460, SD = 110):**
- The values:
- \[130, 240, 350, 460, 570, 680, 790\]
**ACT English Test (Mean = 17, SD = 2.5):**
- The values:
- \[9.5, 12, 14.5, 17, 19.5, 22, 24.5\]
Ensure each value is correctly placed according to its position relative to the mean in the normal distribution curves.
#### b. Probability of a Student Scoring Less Than 14.5 on the ACT English Test
To find this probability, calculate the z-score:
- \( z = \frac{X - \mu}{\sigma} = \frac{14.5 - 17}{2.5} = \frac{-2.5}{2.5} = -1 \)
Using the standard normal distribution table, the probability corresponding to a z-score of -1 is approximately 0.1587. Hence, there is a 15.87% chance that a randomly selected student will score less than 14.5 on the ACT English Test.
#### c. Probability of a Student Scoring More Than 650 on the SAT Verbal Test
Calculate the z-score:
- \( z = \frac{X - \mu}{\sigma} = \frac{650 - 460}{110} = \frac{190}{110} \approx 1.727 \)
Using the standard normal distribution table, the probability corresponding to a z-score of 1.727 is approximately 0.9582. The required probability is \(1 - 0.9582 = 0.0418\). Thus, there is a 4.18% chance a
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