In a certain normal distribution, the mean is 50. If an observation of 30 is randomly selected, the z-score for this observation will be O Positive O Not enough information to tell O 50 O Negative
In a certain normal distribution, the mean is 50. If an observation of 30 is randomly selected, the z-score for this observation will be O Positive O Not enough information to tell O 50 O Negative
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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![### Question
In a certain normal distribution, the mean is 50. If an observation of 30 is randomly selected, the z-score for this observation will be ________.
- [ ] Positive
- [ ] Not enough information to tell
- [ ] 50
- [ ] Negative
### Explanation
This question requires an understanding of z-scores in a normal distribution. The z-score measures how many standard deviations an observation is from the mean. To calculate the z-score, you need the formula:
\[ \text{z-score} = \frac{(X - \mu)}{\sigma} \]
Where:
- \( X \) is the observation.
- \( \mu \) is the mean of the distribution.
- \( \sigma \) is the standard deviation.
**Details:**
- Given: \( \mu = 50 \) and \( X = 30 \).
- The standard deviation (\( \sigma \)) is not provided in this question.
To determine if the z-score is positive or negative:
- If \( X < \mu \), the z-score will be negative.
- If \( X > \mu \), the z-score will be positive.
Since 30 is less than 50, regardless of the standard deviation, the z-score will be negative.
Therefore, the correct answer is:
- [ ] Negative](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F02ac992c-59f5-4674-9c6f-05ab0e575bae%2Fe372e64b-5930-4973-b108-7fbc4d5ad8a6%2Ff0er66k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question
In a certain normal distribution, the mean is 50. If an observation of 30 is randomly selected, the z-score for this observation will be ________.
- [ ] Positive
- [ ] Not enough information to tell
- [ ] 50
- [ ] Negative
### Explanation
This question requires an understanding of z-scores in a normal distribution. The z-score measures how many standard deviations an observation is from the mean. To calculate the z-score, you need the formula:
\[ \text{z-score} = \frac{(X - \mu)}{\sigma} \]
Where:
- \( X \) is the observation.
- \( \mu \) is the mean of the distribution.
- \( \sigma \) is the standard deviation.
**Details:**
- Given: \( \mu = 50 \) and \( X = 30 \).
- The standard deviation (\( \sigma \)) is not provided in this question.
To determine if the z-score is positive or negative:
- If \( X < \mu \), the z-score will be negative.
- If \( X > \mu \), the z-score will be positive.
Since 30 is less than 50, regardless of the standard deviation, the z-score will be negative.
Therefore, the correct answer is:
- [ ] Negative
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