rovide an appropriate response. Use the Standard Normal Table to find the probability. 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 u = 15.5 and a=3.6. What is the probability that during a given week the airline will lose more than 20 suitcases? OA. 0.8944 OB. 0.3944 OC. 0.4040 OD. 0.1056 HE Q
rovide an appropriate response. Use the Standard Normal Table to find the probability. 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 u = 15.5 and a=3.6. What is the probability that during a given week the airline will lose more than 20 suitcases? OA. 0.8944 OB. 0.3944 OC. 0.4040 OD. 0.1056 HE Q
A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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![**Finding Probability Using Standard Normal Distribution**
**Problem:**
Provide an appropriate response. Use the Standard Normal Table to find the probability.
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 more than 20 suitcases?
**Options:**
- A. 0.8944
- B. 0.3944
- C. 0.4040
- D. 0.1056
---
**Solution:**
To determine the probability, we'll use the standard normal distribution (Z).
1. **Calculate the Z-score:**
The Z-score is calculated using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value we are interested in (20 suitcases).
- \(\mu\) is the mean (15.5 suitcases).
- \(\sigma\) is the standard deviation (3.6 suitcases).
Plugging in the values:
\[
Z = \frac{(20 - 15.5)}{3.6} = \frac{4.5}{3.6} \approx 1.25
\]
2. **Find the probability corresponding to the Z-score:**
Using the standard normal distribution table, locate the value of Z = 1.25.
The table provides the probability that a value is less than Z. For Z = 1.25, the table value is approximately 0.8944. This is the probability that \(X\) is less than 20 suitcases.
3. **Calculate the required probability:**
We are interested in the probability that the number of lost suitcases is more than 20. This is given by:
\[
P(X > 20) = 1 - P(X < 20) = 1 - 0.8944 = 0.1056
\]
**Therefore, the probability that the airline will lose more than 20 suitcases in a given week is 0.1056.**
**Answer:**
- D. 0.1056](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2a348127-0366-4d6a-8a8b-87b3575e425d%2F7de8f1d0-8fe0-42c4-83b1-b9d51fddbdc9%2Fd7memgg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding Probability Using Standard Normal Distribution**
**Problem:**
Provide an appropriate response. Use the Standard Normal Table to find the probability.
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 more than 20 suitcases?
**Options:**
- A. 0.8944
- B. 0.3944
- C. 0.4040
- D. 0.1056
---
**Solution:**
To determine the probability, we'll use the standard normal distribution (Z).
1. **Calculate the Z-score:**
The Z-score is calculated using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value we are interested in (20 suitcases).
- \(\mu\) is the mean (15.5 suitcases).
- \(\sigma\) is the standard deviation (3.6 suitcases).
Plugging in the values:
\[
Z = \frac{(20 - 15.5)}{3.6} = \frac{4.5}{3.6} \approx 1.25
\]
2. **Find the probability corresponding to the Z-score:**
Using the standard normal distribution table, locate the value of Z = 1.25.
The table provides the probability that a value is less than Z. For Z = 1.25, the table value is approximately 0.8944. This is the probability that \(X\) is less than 20 suitcases.
3. **Calculate the required probability:**
We are interested in the probability that the number of lost suitcases is more than 20. This is given by:
\[
P(X > 20) = 1 - P(X < 20) = 1 - 0.8944 = 0.1056
\]
**Therefore, the probability that the airline will lose more than 20 suitcases in a given week is 0.1056.**
**Answer:**
- D. 0.1056
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