A particular fruit's weights are normally distributed, with a mean of 491 grams and a standard deviation of 36 grams. If you pick one fruit at random, what is the probability that it will weigh between 499 grams and 577 grams. (give answer to 4 decimal places)
A particular fruit's weights are normally distributed, with a mean of 491 grams and a standard deviation of 36 grams. If you pick one fruit at random, what is the probability that it will weigh between 499 grams and 577 grams. (give answer to 4 decimal places)
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 Normally Distributed Data
In this lesson, we will explore the concept of normal distribution and how to calculate probabilities within a given range using the properties of the normal distribution.
#### Problem Statement
A particular fruit's weights are normally distributed, with a mean of 491 grams and a standard deviation of 36 grams.
If you pick one fruit at random, what is the probability that it will weigh between 499 grams and 577 grams? (Give the answer to 4 decimal places)
### Explanation of Concepts
**Normal Distribution:** A type of continuous probability distribution for a real-valued random variable. The mean is the central value, and the standard deviation measures the dispersion or spread of the data points around this mean.
**Mean (μ):** The average or central value of a set of data points. In this problem, the mean weight of the fruit is given as 491 grams.
**Standard Deviation (σ):** A measure of the amount of variation or dispersion in a set of values. Here, the standard deviation is given as 36 grams.
### Steps to Solve the Problem
1. **Standardize the Variables:** Convert the weights 499 grams and 577 grams into their corresponding z-scores. The z-score formula is:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where \(X\) is the value from the data set, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For 499 grams:
\[
z_1 = \frac{(499 - 491)}{36} = \frac{8}{36} \approx 0.2222
\]
For 577 grams:
\[
z_2 = \frac{(577 - 491)}{36} = \frac{86}{36} \approx 2.3889
\]
2. **Use Z-Table:** Use the standard normal distribution table (Z-table) to find the area (probability) corresponding to these z-scores.
The probability for \(z_1 = 0.2222\) is approximately 0.5871.
The probability for \(z_2 = 2.3889\) is approximately 0.9917.
3. **Calculate the Probability:** The probability that the fruit weighs between 499 grams](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5176b0c-078c-4393-8a73-67cd4f62d79f%2F00690cd5-c3f8-4c79-a600-b70d55929a1d%2Fu110gna_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability in Normally Distributed Data
In this lesson, we will explore the concept of normal distribution and how to calculate probabilities within a given range using the properties of the normal distribution.
#### Problem Statement
A particular fruit's weights are normally distributed, with a mean of 491 grams and a standard deviation of 36 grams.
If you pick one fruit at random, what is the probability that it will weigh between 499 grams and 577 grams? (Give the answer to 4 decimal places)
### Explanation of Concepts
**Normal Distribution:** A type of continuous probability distribution for a real-valued random variable. The mean is the central value, and the standard deviation measures the dispersion or spread of the data points around this mean.
**Mean (μ):** The average or central value of a set of data points. In this problem, the mean weight of the fruit is given as 491 grams.
**Standard Deviation (σ):** A measure of the amount of variation or dispersion in a set of values. Here, the standard deviation is given as 36 grams.
### Steps to Solve the Problem
1. **Standardize the Variables:** Convert the weights 499 grams and 577 grams into their corresponding z-scores. The z-score formula is:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where \(X\) is the value from the data set, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For 499 grams:
\[
z_1 = \frac{(499 - 491)}{36} = \frac{8}{36} \approx 0.2222
\]
For 577 grams:
\[
z_2 = \frac{(577 - 491)}{36} = \frac{86}{36} \approx 2.3889
\]
2. **Use Z-Table:** Use the standard normal distribution table (Z-table) to find the area (probability) corresponding to these z-scores.
The probability for \(z_1 = 0.2222\) is approximately 0.5871.
The probability for \(z_2 = 2.3889\) is approximately 0.9917.
3. **Calculate the Probability:** The probability that the fruit weighs between 499 grams
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