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American Airlines claims that the average number of people who pay for in- flight moves, when the plane is fully loaded, is 42 with a standard deviation of 8. A sample of 36 fully loaded planes is taken. What is the probability that fewer than 38 people paid for the in- flight moves?
- a)0.0013
- b)0.125
- c)0.012
- d)0.2311
Correct answer is option 'A'. Can you explain this answer?
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American Airlines claims that the average number of people who pay for...
Understanding the Problem
American Airlines claims that the average number of people paying for in-flight movies is 42, with a standard deviation of 8. We want to find the probability that fewer than 38 people pay when a sample of 36 planes is taken.
Key Parameters
- Population Mean (μ): 42
- Population Standard Deviation (σ): 8
- Sample Size (n): 36
Calculating the Sample Mean and Standard Deviation
- Sample Mean (x̄): 38
- Sample Standard Deviation (σx̄): σ / √n = 8 / √36 = 8 / 6 = 1.33
Finding the Z-Score
To calculate the probability, we first need to find the Z-score for the sample mean of 38:
- Z = (x̄ - μ) / σx̄
- Z = (38 - 42) / 1.33 = -3.01
Using the Z-Score
Now, we will look up the Z-score of -3.01 in the standard normal distribution table or use a calculator.
- The Z-score of -3.01 corresponds to a probability of approximately 0.0013.
Conclusion
Thus, the probability that fewer than 38 people paid for in-flight movies is:
- P(X < 38)="" ≈="" />
The correct answer is option 'A'. This low probability indicates that it is quite rare for fewer than 38 passengers to purchase in-flight movies on a fully loaded plane.
American Airlines claims that the average number of people paying for in-flight movies is 42, with a standard deviation of 8. We want to find the probability that fewer than 38 people pay when a sample of 36 planes is taken.
Key Parameters
- Population Mean (μ): 42
- Population Standard Deviation (σ): 8
- Sample Size (n): 36
Calculating the Sample Mean and Standard Deviation
- Sample Mean (x̄): 38
- Sample Standard Deviation (σx̄): σ / √n = 8 / √36 = 8 / 6 = 1.33
Finding the Z-Score
To calculate the probability, we first need to find the Z-score for the sample mean of 38:
- Z = (x̄ - μ) / σx̄
- Z = (38 - 42) / 1.33 = -3.01
Using the Z-Score
Now, we will look up the Z-score of -3.01 in the standard normal distribution table or use a calculator.
- The Z-score of -3.01 corresponds to a probability of approximately 0.0013.
Conclusion
Thus, the probability that fewer than 38 people paid for in-flight movies is:
- P(X < 38)="" ≈="" />
The correct answer is option 'A'. This low probability indicates that it is quite rare for fewer than 38 passengers to purchase in-flight movies on a fully loaded plane.
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American Airlines claims that the average number of people who pay for in- flight moves, when the plane is fully loaded, is 42 with a standard deviation of 8. A sample of 36 fully loaded planes is taken. What is the probability that fewer than 38 people paid for the in- flight moves?a)0.0013b)0.125c)0.012d)0.2311Correct answer is option 'A'. Can you explain this answer?
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