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A particular fruit's weights are normally distributed, with a mean of 580 grams and a standard deviation of 39 grams. f you pick 4 fruits at random, then 13% of the time, their mean weight will be greater than how many grams? Answer in the nearest gram.

A particular fruit's weights are normally distributed, with a mean of 580 grams and a standard deviation of 39 grams. f you pick 4 fruits at random, then 13% of the time, their mean weight will be greater than how many grams? Answer in the nearest gram.

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
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**Normal Distribution and Sampling Problem** A particular fruit’s weights are normally distributed, with a mean of 580 grams and a standard deviation of 39 grams. - **Problem Statement:** If you pick 4 fruits at random, then 13% of the time, their mean weight will be greater than how many grams? **Answer in the nearest gram.** --- **Explanation on Topic:** In a normal distribution scenario, we often deal with a particular set of data that needs statistical analysis. Here, the weights of a specific fruit follow a normal distribution. To determine how the average of a smaller sample (4 fruits in this case) deviates, we use properties of the normal distribution and sampling distribution of the mean. **Key Components of the Problem:** 1. **Normal Distribution:** - The fruit weights are normally distributed. - Mean (μ) = 580 grams. - Standard Deviation (σ) = 39 grams. 2. **Sampling Distribution of the Mean:** - Sample Size (n) = 4 fruits. - Standard Error (SE) = σ / √n. For our case, SE = 39 / √4 = 39 / 2 = 19.5 grams. 3. **Finding Specific Percentile:** - We need to determine the weight such that 13% of the time the mean of the sample will be greater than this weight. - Since the problem is given in terms of a percentage, we look for the value corresponding to the 87th percentile (since 100% - 13% = 87%) in a normal distribution curve. Using standard Z-tables or normal distribution calculators, the 87th percentile corresponds approximately to a Z-score of 1.13. 4. **Calculating the Specific Mean Weight:** Using the Z-score formula for a sample mean: \( Z = \frac{(X - μ)}{SE} \) Rearranging for X (the sample mean we need to find), \( X = Z \cdot SE + μ \) Putting in the values, \( X = 1.13 \cdot 19.5 + 580 \) \( X ≈ 22.035 + 580 \) \( X ≈ 602.035 \) Rounded to the nearest gram,
Transcribed Image Text:**Normal Distribution and Sampling Problem** A particular fruit’s weights are normally distributed, with a mean of 580 grams and a standard deviation of 39 grams. - **Problem Statement:** If you pick 4 fruits at random, then 13% of the time, their mean weight will be greater than how many grams? **Answer in the nearest gram.** --- **Explanation on Topic:** In a normal distribution scenario, we often deal with a particular set of data that needs statistical analysis. Here, the weights of a specific fruit follow a normal distribution. To determine how the average of a smaller sample (4 fruits in this case) deviates, we use properties of the normal distribution and sampling distribution of the mean. **Key Components of the Problem:** 1. **Normal Distribution:** - The fruit weights are normally distributed. - Mean (μ) = 580 grams. - Standard Deviation (σ) = 39 grams. 2. **Sampling Distribution of the Mean:** - Sample Size (n) = 4 fruits. - Standard Error (SE) = σ / √n. For our case, SE = 39 / √4 = 39 / 2 = 19.5 grams. 3. **Finding Specific Percentile:** - We need to determine the weight such that 13% of the time the mean of the sample will be greater than this weight. - Since the problem is given in terms of a percentage, we look for the value corresponding to the 87th percentile (since 100% - 13% = 87%) in a normal distribution curve. Using standard Z-tables or normal distribution calculators, the 87th percentile corresponds approximately to a Z-score of 1.13. 4. **Calculating the Specific Mean Weight:** Using the Z-score formula for a sample mean: \( Z = \frac{(X - μ)}{SE} \) Rearranging for X (the sample mean we need to find), \( X = Z \cdot SE + μ \) Putting in the values, \( X = 1.13 \cdot 19.5 + 580 \) \( X ≈ 22.035 + 580 \) \( X ≈ 602.035 \) Rounded to the nearest gram,
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