Mr. Green examines the Unit 1 Biology test scores in his class. The test scores are normally distributed, with a mean of 82. Natalie's test score is 76. If her z-score is -1.5, what is the standard deviation of the Unit 1 Biology test scores?
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
![### Unit 1 Biology Test Scores Analysis
Mr. Green examines the Unit 1 Biology test scores in his class. The test scores are normally distributed, with a mean of 82.
#### Problem Statement
Natalie's test score is 76. If her z-score is -1.5, what is the standard deviation of the Unit 1 Biology test scores?
### Explanation
To solve for the standard deviation, we will use the z-score formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
Where:
- \( z \) is the z-score
- \( X \) is the value of the score
- \( \mu \) is the mean
- \( \sigma \) is the standard deviation
Given:
- Natalie's score (\( X \)) is 76
- The z-score (\( z \)) is -1.5
- The mean (\( \mu \)) is 82
We need to solve for the standard deviation (\( \sigma \)).
Rearranging the formula to solve for \( \sigma \):
\[ \sigma = \frac{(X - \mu)}{z} \]
\[ \sigma = \frac{(76 - 82)}{-1.5} \]
\[ \sigma = \frac{-6}{-1.5} \]
\[ \sigma = 4 \]
So, the standard deviation of the Unit 1 Biology test scores is 4.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81d865ca-ed2c-4f6d-a4cc-7f843242dec1%2F49f31b7f-4105-4a4f-bc1c-5e6603a35629%2Fgnazk3q_processed.jpeg&w=3840&q=75)
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