The population of weights of a particular fruit is normally distributed, with a mean of 705 grams and a standard deviation of 5 grams. If 23 fruits are picked at random, then 17% of the time, their mean weight will be greater than how many grams? Round your answer to the nearest gram.
The population of weights of a particular fruit is normally distributed, with a mean of 705 grams and a standard deviation of 5 grams. If 23 fruits are picked at random, then 17% of the time, their mean weight will be greater than how many grams? Round your answer to 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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![The population of weights of a particular fruit is normally distributed, with a mean of 705 grams and a standard deviation of 5 grams. If 23 fruits are picked at random, then 17% of the time, their mean weight will be greater than how many grams? *Round your answer to the nearest gram.*
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In solving this problem, we are working with a normal distribution and dealing with the concept of sampling distribution of the sample mean. To find the value where 17% of the sample means are greater, we would typically use the standard normal distribution (Z-score) table or statistical software. The process involves:
1. Determining the standard error of the mean (SEM) using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\]
where \(\sigma = 5\) grams (standard deviation) and \(n = 23\) (sample size).
2. Using the desired percentile (83rd percentile, since 100% - 17% = 83%) with the standard normal distribution to find the corresponding Z-score.
3. Applying the Z-score formula:
\[
X = \mu + Z \times \text{SEM}
\]
where \(X\) is the mean weight, \(\mu = 705\) grams (population mean), and \(Z\) is the Z-score corresponding to the 83rd percentile.
By calculating, you will get the mean weight that satisfies the given condition.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4cbdcd8-71de-429b-b001-27d54b327ca5%2F43610be1-1958-4886-aea3-1edcc76c36f6%2F4ia4fa_processed.png&w=3840&q=75)
Transcribed Image Text:The population of weights of a particular fruit is normally distributed, with a mean of 705 grams and a standard deviation of 5 grams. If 23 fruits are picked at random, then 17% of the time, their mean weight will be greater than how many grams? *Round your answer to the nearest gram.*
---
In solving this problem, we are working with a normal distribution and dealing with the concept of sampling distribution of the sample mean. To find the value where 17% of the sample means are greater, we would typically use the standard normal distribution (Z-score) table or statistical software. The process involves:
1. Determining the standard error of the mean (SEM) using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\]
where \(\sigma = 5\) grams (standard deviation) and \(n = 23\) (sample size).
2. Using the desired percentile (83rd percentile, since 100% - 17% = 83%) with the standard normal distribution to find the corresponding Z-score.
3. Applying the Z-score formula:
\[
X = \mu + Z \times \text{SEM}
\]
where \(X\) is the mean weight, \(\mu = 705\) grams (population mean), and \(Z\) is the Z-score corresponding to the 83rd percentile.
By calculating, you will get the mean weight that satisfies the given condition.
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