The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 34 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X=percent of fat calories.
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 34 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X=percent of fat calories.
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![Title: Understanding the Distribution of Fat Calorie Percentages in America
Introduction:
This exercise explores the distribution of the percentage of fat calories that a person in America consumes each day. It assumes that this percentage is normally distributed with a mean (μ) of 34 and a standard deviation (σ) of 10.
Exercise:
Let's analyze the distribution and calculate probabilities based on this normal distribution.
1. Distribution of X:
What is the distribution of \( X \), where \( X \) is the percent of fat calories consumed?
\[
X \sim N(\mu, \sigma^2)
\]
Here, the mean \( \mu = 34 \) and the standard deviation \( \sigma = 10 \).
Answer: \( X \sim N(34, 10^2) \)
2. Probability Calculation:
Find the probability that a randomly selected fat calorie percent is more than 39. Round your answer to 4 decimal places.
To find this probability, calculate the Z-score using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where \( X = 39 \), \( \mu = 34 \), and \( \sigma = 10 \).
\[
Z = \frac{39 - 34}{10} = \frac{5}{10} = 0.5
\]
Use the Z-table to find the area to the right of Z=0.5.
Answer: The probability is approximately 0.3085 (rounded to 4 decimal places).
3. Lower Quartile Calculation:
Find the minimum number for the lower quartile of fat calories. Round your answer to 2 decimal places.
To find the lower quartile (Q1), use the Z-score that corresponds to the 25th percentile. The Z-score for the 25th percentile is approximately -0.675.
Convert this Z-score to an X value using the formula:
\[
X = \mu + Z \cdot \sigma
\]
\[
X = 34 + (-0.675) \cdot 10 = 34 - 6.75 = 27.25
\]
Answer: The lower quartile is approximately 27.25](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fadb58edc-76a8-4215-8abd-8e37e0700ca4%2F2029520d-319f-48b7-9f63-97bc8f6d3f57%2Fuxczsqy.jpeg&w=3840&q=75)
Transcribed Image Text:Title: Understanding the Distribution of Fat Calorie Percentages in America
Introduction:
This exercise explores the distribution of the percentage of fat calories that a person in America consumes each day. It assumes that this percentage is normally distributed with a mean (μ) of 34 and a standard deviation (σ) of 10.
Exercise:
Let's analyze the distribution and calculate probabilities based on this normal distribution.
1. Distribution of X:
What is the distribution of \( X \), where \( X \) is the percent of fat calories consumed?
\[
X \sim N(\mu, \sigma^2)
\]
Here, the mean \( \mu = 34 \) and the standard deviation \( \sigma = 10 \).
Answer: \( X \sim N(34, 10^2) \)
2. Probability Calculation:
Find the probability that a randomly selected fat calorie percent is more than 39. Round your answer to 4 decimal places.
To find this probability, calculate the Z-score using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where \( X = 39 \), \( \mu = 34 \), and \( \sigma = 10 \).
\[
Z = \frac{39 - 34}{10} = \frac{5}{10} = 0.5
\]
Use the Z-table to find the area to the right of Z=0.5.
Answer: The probability is approximately 0.3085 (rounded to 4 decimal places).
3. Lower Quartile Calculation:
Find the minimum number for the lower quartile of fat calories. Round your answer to 2 decimal places.
To find the lower quartile (Q1), use the Z-score that corresponds to the 25th percentile. The Z-score for the 25th percentile is approximately -0.675.
Convert this Z-score to an X value using the formula:
\[
X = \mu + Z \cdot \sigma
\]
\[
X = 34 + (-0.675) \cdot 10 = 34 - 6.75 = 27.25
\]
Answer: The lower quartile is approximately 27.25
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