A particular fruit's weights are normally distributed, with a mean of 689 grams and a standard deviation of 36 grams. If you pick 4 fruits at random, then 19% 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 689 grams and a standard deviation of 36 grams. If you pick 4 fruits at random, then 19% of the time, their mean weight will be greater than how many grams? Answer in the nearest gram.
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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![### Understanding Normal Distribution Through Fruit Weights
In this exercise, we are given the weight distribution of a particular type of fruit. The weights of this fruit are assumed to be normally distributed with a mean weight of 689 grams and a standard deviation of 36 grams.
#### Problem Statement:
If you select 4 fruits randomly, 19% of the time, their mean weight will be greater than how many grams?
#### Solution:
To solve this problem, we look at the properties of the normal distribution and use the appropriate statistical tools.
Given:
- Mean (μ) = 689 grams
- Standard deviation (σ) = 36 grams
- Sample size (n) = 4 fruits
- We need to find the weight (X) such that the probability of the sample mean being greater than X is 19%.
From normal distribution properties, we know that:
- The standard error of the mean (SEM) is calculated using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{36}{\sqrt{4}} = \frac{36}{2} = 18
\]
To find the value of X corresponding to the top 19% (81st percentile), we will use Z-scores (z):
Using a Z-table or normal distribution calculator:
- The Z-score corresponding to the 81st percentile (0.81) is approximately 0.88.
Next, we use the Z-score formula to find X:
\[
X = \mu + (Z \times \text{SEM})
\]
\[
X = 689 + (0.88 \times 18)
\]
\[
X = 689 + 15.84 = 704.84 \approx 705 \text{ grams}
\]
Thus, 19% of the time, the average weight of 4 randomly picked fruits will be greater than approximately **705 grams**.
This solution demonstrates how to apply the concepts of normal distribution and Z-scores to find probabilities and thresholds within dataset distributions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5176b0c-078c-4393-8a73-67cd4f62d79f%2Fd892acbc-7667-440f-8297-f31e1439a957%2Fzgodv7a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Normal Distribution Through Fruit Weights
In this exercise, we are given the weight distribution of a particular type of fruit. The weights of this fruit are assumed to be normally distributed with a mean weight of 689 grams and a standard deviation of 36 grams.
#### Problem Statement:
If you select 4 fruits randomly, 19% of the time, their mean weight will be greater than how many grams?
#### Solution:
To solve this problem, we look at the properties of the normal distribution and use the appropriate statistical tools.
Given:
- Mean (μ) = 689 grams
- Standard deviation (σ) = 36 grams
- Sample size (n) = 4 fruits
- We need to find the weight (X) such that the probability of the sample mean being greater than X is 19%.
From normal distribution properties, we know that:
- The standard error of the mean (SEM) is calculated using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{36}{\sqrt{4}} = \frac{36}{2} = 18
\]
To find the value of X corresponding to the top 19% (81st percentile), we will use Z-scores (z):
Using a Z-table or normal distribution calculator:
- The Z-score corresponding to the 81st percentile (0.81) is approximately 0.88.
Next, we use the Z-score formula to find X:
\[
X = \mu + (Z \times \text{SEM})
\]
\[
X = 689 + (0.88 \times 18)
\]
\[
X = 689 + 15.84 = 704.84 \approx 705 \text{ grams}
\]
Thus, 19% of the time, the average weight of 4 randomly picked fruits will be greater than approximately **705 grams**.
This solution demonstrates how to apply the concepts of normal distribution and Z-scores to find probabilities and thresholds within dataset distributions.
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