Question

4. A study conducted by an airline showed that a random sample of nine of its passengers disembarking at a certain airport took an average of 24.1 minutes to claim their luggage. From a previous survey it was willing to assume that time to claim luggage was normally distributed with a variance of 18 (minutes)2. Being 95% confident, what would be the margin of error in predicting the mean time to claim ones lu

Answer 1

The margin of error for the mean time to claim luggage is approximately 2.76 minutes with a 95% confidence level, calculated using the given variance and the Z-distribution for the sample size of nine.

Step-by-step explanation:

To calculate the margin of error with a 95% confidence level for the given scenario, we'll use the formula for the confidence interval of the mean when the population variance is known. Since the variance is given as 18 minutes2, the standard deviation is the square root of the variance, which is √18 or approximately 4.24 minutes. The formula for the margin of error (ME) when using the Z-distribution is ME = Z x (σ/√n), where Z is the Z-value corresponding to the desired confidence level, σ (sigma) is the population standard deviation, and n is the sample size.

For a 95% confidence level, the Z-value is 1.96. The sample size n is 9. Plugging these values into the formula gives ME = 1.96 x (4.24/√9) which equals approximately 1.96 x 1.41, and the margin of error is roughly 2.76 minutes. Therefore, we can be 95% confident that the mean time for all passengers to claim their luggage is within 2.76 minutes of the sample mean of 24.1 minutes.

Answer 2

The margin of error in predicting the mean time to claim luggage with 95% confidence is approximately 2.77 minutes.

To calculate the margin of error for a mean with a normal distribution, you can use the formula:

`\[ \text{Margin of Error} = Z * (\sigma)/(√(n)) \]`

In this case, the airline is 95% confident, so the Z-score for a 95% confidence level is 1.96.

Calculate
`\( \sigma \):`

`\[ \sigma = √(18) \approx 4.24 \]`

Calculate the Margin of Error:

`\[ \text{Margin of Error} = 1.96 * (4.24)/(√(9)) \]`

`\[ \text{Margin of Error} \approx 1.96 * (4.24)/(3) \]`

`\[ \text{Margin of Error} \approx 1.96 * 1.41 \]`

Margin of Error = 2.77

Therefore, the margin of error in predicting the mean time to claim luggage with 95% confidence is approximately 2.77 minutes.