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The amount of calories in a small serving of French Fries at McDonalds is Normally distributed with a mean of 230 calories and a standard deviation of 15. Suppose that a small random sample of 10 small servings was collected. What is the probability that the sample mean is greater than 240 calories? O 0.0175 O 0.2524 O 0.0274 O 0.7475

The amount of calories in a small serving of French Fries at McDonalds is Normally distributed with a mean of 230 calories and a standard deviation of 15. Suppose that a small random sample of 10 small servings was collected. What is the probability that the sample mean is greater than 240 calories? O 0.0175 O 0.2524 O 0.0274 O 0.7475

MATLAB: An Introduction with Applications
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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**Probability in Normal Distribution: Sample Mean** The amount of calories in a small serving of French Fries at McDonald's is normally distributed with a mean of 230 calories and a standard deviation of 15. Suppose that a small random sample of 10 small servings was collected. What is the probability that the sample mean is greater than 240 calories? - 0.0175 - 0.2524 - 0.0274 - 0.7475 *Explanation:* This problem involves the concept of the distribution of a sample mean. To solve this, we can use the properties of the normal distribution and the central limit theorem, which states that the distribution of the sample mean will be normal or nearly normal if the sample size is large enough. In this case, the sample size is 10. We need to standardize the sample mean to find the corresponding probability using the Z-score formula: \[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \] Where: - \( \bar{X} \) is the sample mean - \( \mu \) is the population mean - \( \sigma \) is the population standard deviation - \( n \) is the sample size Using the given values: - \( \mu = 230 \) - \( \sigma = 15 \) - \( n = 10 \) - \( \bar{X} = 240 \) Calculate the Z-score: \[ Z = \frac{240 - 230}{15 / \sqrt{10}} \] \[ Z = \frac{10}{15 / \sqrt{10}} \] \[ Z = \frac{10}{4.74} \] \[ Z \approx 2.11 \] Now, we use the Z-score to find the probability. Using Z-tables or normal distribution calculators, the area to the right of Z = 2.11 gives the probability of the sample mean being greater than 240 calories. Thus, the probability that the sample mean exceeds 240 calories is approximately 0.0175. **Conclusion:** Among the choices given, the answer is: - **0.0175** This example illustrates the application of statistical concepts to calculate probabilities using the normal distribution in practical scenarios like quality control and nutrition analysis.
Transcribed Image Text:**Probability in Normal Distribution: Sample Mean** The amount of calories in a small serving of French Fries at McDonald's is normally distributed with a mean of 230 calories and a standard deviation of 15. Suppose that a small random sample of 10 small servings was collected. What is the probability that the sample mean is greater than 240 calories? - 0.0175 - 0.2524 - 0.0274 - 0.7475 *Explanation:* This problem involves the concept of the distribution of a sample mean. To solve this, we can use the properties of the normal distribution and the central limit theorem, which states that the distribution of the sample mean will be normal or nearly normal if the sample size is large enough. In this case, the sample size is 10. We need to standardize the sample mean to find the corresponding probability using the Z-score formula: \[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \] Where: - \( \bar{X} \) is the sample mean - \( \mu \) is the population mean - \( \sigma \) is the population standard deviation - \( n \) is the sample size Using the given values: - \( \mu = 230 \) - \( \sigma = 15 \) - \( n = 10 \) - \( \bar{X} = 240 \) Calculate the Z-score: \[ Z = \frac{240 - 230}{15 / \sqrt{10}} \] \[ Z = \frac{10}{15 / \sqrt{10}} \] \[ Z = \frac{10}{4.74} \] \[ Z \approx 2.11 \] Now, we use the Z-score to find the probability. Using Z-tables or normal distribution calculators, the area to the right of Z = 2.11 gives the probability of the sample mean being greater than 240 calories. Thus, the probability that the sample mean exceeds 240 calories is approximately 0.0175. **Conclusion:** Among the choices given, the answer is: - **0.0175** This example illustrates the application of statistical concepts to calculate probabilities using the normal distribution in practical scenarios like quality control and nutrition analysis.
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