An airline company models the weight of each passengers luggage distributed random variables according to a Normal distribution wi pounds and standard deviation of a =5 pounds. What is the probability a random passenger's luggage being betwe Round to the nearest 2nd decimal place, 0.xx
An airline company models the weight of each passengers luggage distributed random variables according to a Normal distribution wi pounds and standard deviation of a =5 pounds. What is the probability a random passenger's luggage being betwe Round to the nearest 2nd decimal place, 0.xx
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 and Statistics: Calculating Probabilities in Normal Distribution
**Problem Description:**
An airline company models the weight of each passenger's luggage as independent and identically distributed random variables according to a Normal distribution with a mean (expectation) of \( \mu = 35 \) pounds and a standard deviation of \( \sigma = 5 \) pounds.
**Question:**
What is the probability that a random passenger's luggage weighs between 30 and 40 pounds?
**Instructions:**
Round your answer to the nearest second decimal place, 0.xx.
*Input Box for Answer:* [ ]
**Detailed Explanation:**
To find the probability of a random passenger's luggage weight falling between 30 and 40 pounds, we can use the properties of the Normal distribution. Given:
- Mean (\(\mu\)): 35 pounds
- Standard Deviation (\(\sigma\)): 5 pounds
We need to calculate the Z-scores for 30 and 40 pounds:
\[ Z = \frac{X - \mu}{\sigma} \]
For \(X = 30\) pounds:
\[ Z_1 = \frac{30 - 35}{5} = \frac{-5}{5} = -1 \]
For \(X = 40\) pounds:
\[ Z_2 = \frac{40 - 35}{5} = \frac{5}{5} = 1 \]
With these Z-scores, we can use the standard normal distribution table or a statistical software package to find the corresponding probabilities.
The probability corresponding to \( Z = -1 \) is approximately 0.1587 and the probability corresponding to \( Z = 1 \) is approximately 0.8413.
The probability that the luggage weight is between 30 and 40 pounds is:
\[ P(30 < X < 40) = P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) \]
\[ P(30 < X < 40) = 0.8413 - 0.1587 = 0.6826 \]
So, the probability is approximately **0.68**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11615fdb-2fe6-4d89-90b4-ed63cf4ab528%2Ff5c29c60-27dd-40d4-963b-fc947c699ef1%2F8a6icp_processed.png&w=3840&q=75)
Transcribed Image Text:### Probability and Statistics: Calculating Probabilities in Normal Distribution
**Problem Description:**
An airline company models the weight of each passenger's luggage as independent and identically distributed random variables according to a Normal distribution with a mean (expectation) of \( \mu = 35 \) pounds and a standard deviation of \( \sigma = 5 \) pounds.
**Question:**
What is the probability that a random passenger's luggage weighs between 30 and 40 pounds?
**Instructions:**
Round your answer to the nearest second decimal place, 0.xx.
*Input Box for Answer:* [ ]
**Detailed Explanation:**
To find the probability of a random passenger's luggage weight falling between 30 and 40 pounds, we can use the properties of the Normal distribution. Given:
- Mean (\(\mu\)): 35 pounds
- Standard Deviation (\(\sigma\)): 5 pounds
We need to calculate the Z-scores for 30 and 40 pounds:
\[ Z = \frac{X - \mu}{\sigma} \]
For \(X = 30\) pounds:
\[ Z_1 = \frac{30 - 35}{5} = \frac{-5}{5} = -1 \]
For \(X = 40\) pounds:
\[ Z_2 = \frac{40 - 35}{5} = \frac{5}{5} = 1 \]
With these Z-scores, we can use the standard normal distribution table or a statistical software package to find the corresponding probabilities.
The probability corresponding to \( Z = -1 \) is approximately 0.1587 and the probability corresponding to \( Z = 1 \) is approximately 0.8413.
The probability that the luggage weight is between 30 and 40 pounds is:
\[ P(30 < X < 40) = P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) \]
\[ P(30 < X < 40) = 0.8413 - 0.1587 = 0.6826 \]
So, the probability is approximately **0.68**.
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