Problem 28
Question
Suppose the height of an adult animal is normally distributed with mean \(17.2\) in. Find the standard deviation if \(10 \%\) of the animals have a height that exceeds 19 in.
Step-by-Step Solution
Verified Answer
The standard deviation is approximately 1.41 inches.
1Step 1: Understanding the Normal Distribution
The height of the animals is normally distributed with a mean \( \mu = 17.2 \) inches. We need to find the standard deviation \( \sigma \). The problem states that \(10\%\) of animals exceed 19 inches in height. This information indicates that we are seeking the standard deviation that makes 19 inches the 90th percentile of the distribution.
2Step 2: Finding the Z-score for 90th Percentile
The 90th percentile corresponds to the point where \(90\%\) of the distribution is below it. Using a standard normal distribution table or calculator, find the Z-score for the 90th percentile. The Z-score for the 90th percentile is approximately \( 1.28 \).
3Step 3: Setting Up the Z-score Formula
The Z-score formula is given by \[ Z = \frac{X - \mu}{\sigma} \] where \( X = 19 \) inches, \( \mu = 17.2 \) inches, and \( Z = 1.28 \). Substitute these values into the formula to solve for \( \sigma \).
4Step 4: Solving for the Standard Deviation
Using the Z-score formula, substitute the known values: \[ 1.28 = \frac{19 - 17.2}{\sigma} \]. Rearrange to solve for \( \sigma \): \[ \sigma = \frac{19 - 17.2}{1.28} = \frac{1.8}{1.28} \approx 1.40625 \]. Thus, the standard deviation is approximately \(1.41\) inches.
Key Concepts
Standard DeviationZ-scorePercentile
Standard Deviation
Standard deviation is an essential concept in statistics that measures the amount of variation or dispersion in a set of values. It tells us how spread out the numbers are from the mean. A smaller standard deviation indicates that the data points are closer to the mean, while a larger value suggests a wider spread.
In the context of a normal distribution, where data follows a bell-shaped curve, the standard deviation helps us understand the probability distribution's shape. - A normal distribution with a small standard deviation will be steeper and narrower. - Conversely, a larger standard deviation will result in a flatter and wider distribution.
This concept is particularly useful when we want to assess how unusual or typical a certain value is within a distribution. For example, in our animal height exercise, knowing the standard deviation gives us insights into how typical a height of 19 inches is relative to the common height around 17.2 inches.
In the context of a normal distribution, where data follows a bell-shaped curve, the standard deviation helps us understand the probability distribution's shape. - A normal distribution with a small standard deviation will be steeper and narrower. - Conversely, a larger standard deviation will result in a flatter and wider distribution.
This concept is particularly useful when we want to assess how unusual or typical a certain value is within a distribution. For example, in our animal height exercise, knowing the standard deviation gives us insights into how typical a height of 19 inches is relative to the common height around 17.2 inches.
Z-score
The Z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It indicates how many standard deviations an element is from the mean, and it is a core part of the normal distribution.
Here's how the Z-score works:- A Z-score of 0 indicates the value is exactly at the mean.- Positive Z-scores indicate values above the mean.- Negative Z-scores show values below the mean.
To calculate the Z-score, use the formula:\[ Z = \frac{X - \mu}{\sigma} \]where:- \( X \) is the value from the dataset (e.g., height).- \( \mu \) is the mean of the dataset.- \( \sigma \) is the standard deviation.
In our exercise, we found that the Z-score for the 90th percentile, where 19 inches fall, is approximately 1.28. This tells us that a height of 19 inches is 1.28 standard deviations above the mean.
Here's how the Z-score works:- A Z-score of 0 indicates the value is exactly at the mean.- Positive Z-scores indicate values above the mean.- Negative Z-scores show values below the mean.
To calculate the Z-score, use the formula:\[ Z = \frac{X - \mu}{\sigma} \]where:- \( X \) is the value from the dataset (e.g., height).- \( \mu \) is the mean of the dataset.- \( \sigma \) is the standard deviation.
In our exercise, we found that the Z-score for the 90th percentile, where 19 inches fall, is approximately 1.28. This tells us that a height of 19 inches is 1.28 standard deviations above the mean.
Percentile
A percentile indicates the relative standing of a value within a data set. It is the value below which a certain percentage of observations fall. For instance, the 90th percentile is the value below which 90% of the data points are found.
Understanding percentiles is important for comparing individual data points to a larger data set without needing to sort the data. Here's why they're useful: - They allow us to determine how a single observation compares to the rest. - Useful for interpreting standardized tests or measurements.
In the context of a normal distribution, like our animal heights, determining the 90th percentile helped us identify that 10% of the animals exceed a height of 19 inches. This percentile ranking relates directly to the Z-score and standard deviation we've discussed, combining these elements to give a comprehensive picture of the distribution.
Understanding percentiles is important for comparing individual data points to a larger data set without needing to sort the data. Here's why they're useful: - They allow us to determine how a single observation compares to the rest. - Useful for interpreting standardized tests or measurements.
In the context of a normal distribution, like our animal heights, determining the 90th percentile helped us identify that 10% of the animals exceed a height of 19 inches. This percentile ranking relates directly to the Z-score and standard deviation we've discussed, combining these elements to give a comprehensive picture of the distribution.
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