An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with µ = 15.5 and σ = 3.6. What is the probability that during a given week the airline will lose less than 20 suitcases? 0.3944 0.1056 0.8944 0.4040
An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with µ = 15.5 and σ = 3.6. What is the probability that during a given week the airline will lose less than 20 suitcases? 0.3944 0.1056 0.8944 0.4040
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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![### Statistical Probability in Airline Operations
**Problem Statement:**
An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with a mean (\(\mu\)) of 15.5 and a standard deviation (\(\sigma\)) of 3.6.
**Question:**
What is the probability that during a given week the airline will lose less than 20 suitcases?
**Options:**
- \( \circ \) 0.3944
- \( \circ \) 0.1056
- \( \circ \) 0.8944
- \( \circ \) 0.4040
### Explanation:
To solve this problem, we need to understand the properties of the normal distribution. Given the normal distribution parameters (mean \(\mu = 15.5\) and standard deviation \(\sigma = 3.6\)), we can determine the probability by calculating the Z-score and then using the standard normal distribution table.
1. **Calculate the Z-score:**
The Z-score for a value \(X = 20\) can be calculated using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Here, \(X = 20\), \(\mu = 15.5\), and \(\sigma = 3.6\).
\[
Z = \frac{20 - 15.5}{3.6} \approx 1.25
\]
2. **Find the Probability:**
Using the Z-score, we can find the probability that a value is less than 20 by looking up 1.25 in the standard normal distribution table. The table gives the probability that a standard normal variable is less than the Z-score.
The standard normal distribution table value for a Z-score of 1.25 is approximately 0.8944, indicating that there is an 89.44% chance that the airline will lose less than 20 suitcases during a given week.
Therefore, the answer is:
- \( \circ \) **0.8944**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8bccc44d-7782-46f2-838e-c140c2dedd7b%2F20da1678-df43-42f8-a7d6-b1b3aff0d43e%2Ff9quuj_processed.png&w=3840&q=75)
Transcribed Image Text:### Statistical Probability in Airline Operations
**Problem Statement:**
An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with a mean (\(\mu\)) of 15.5 and a standard deviation (\(\sigma\)) of 3.6.
**Question:**
What is the probability that during a given week the airline will lose less than 20 suitcases?
**Options:**
- \( \circ \) 0.3944
- \( \circ \) 0.1056
- \( \circ \) 0.8944
- \( \circ \) 0.4040
### Explanation:
To solve this problem, we need to understand the properties of the normal distribution. Given the normal distribution parameters (mean \(\mu = 15.5\) and standard deviation \(\sigma = 3.6\)), we can determine the probability by calculating the Z-score and then using the standard normal distribution table.
1. **Calculate the Z-score:**
The Z-score for a value \(X = 20\) can be calculated using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Here, \(X = 20\), \(\mu = 15.5\), and \(\sigma = 3.6\).
\[
Z = \frac{20 - 15.5}{3.6} \approx 1.25
\]
2. **Find the Probability:**
Using the Z-score, we can find the probability that a value is less than 20 by looking up 1.25 in the standard normal distribution table. The table gives the probability that a standard normal variable is less than the Z-score.
The standard normal distribution table value for a Z-score of 1.25 is approximately 0.8944, indicating that there is an 89.44% chance that the airline will lose less than 20 suitcases during a given week.
Therefore, the answer is:
- \( \circ \) **0.8944**
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