Chapter 7: Problem 12
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of \(n=49\) customers is observed. Find the probability that the average time waiting in line for these customers is (a) Less than 10 minutes (b) Between 5 and 10 minutes (c) Less than 6 minutes
Short Answer
Step by step solution
Identify the Given Parameters
Determine the Sampling Distribution
Find the Z-Score for Less Than 10 Minutes
Use Normal Distribution to Find Probability (Part a)
Find the Z-Scores for Between 5 and 10 Minutes
Use Normal Distribution to Find Probability (Part b)
Find the Z-Score for Less Than 6 Minutes
Use Normal Distribution to Find Probability (Part c)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Distribution
- The data is symmetrically distributed around the mean.
- Most of the data points fall near the mean, with fewer points further away.
- The tails of the distribution continue indefinitely without touching the x-axis.
Sampling Distribution
- The sample size (\(n = 49\)) is large enough to assume that the sampling distribution of the sample mean is approximately normal, regardless of the population distribution.
- The mean of the sampling distribution (\(\mu_{\bar{X}}\)) is the same as the population mean (\(\mu = 8.2\)) minutes.
Standard Error
- \[SE = \frac{\sigma}{\sqrt{n}}\]
- For our exercise, with a standard deviation (\(\sigma = 1.5\)) and a sample size (\(n = 49\)), the standard error is approximately 0.2143 minutes.
Z-score
- \[Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}\]
- For example, to find the probability that the average waiting time (\(\bar{X}\)) is less than a certain time, we calculate the corresponding Z-score.
- In our exercise, finding a Z-score for less than 10 minutes gives us a score of 8.37, which indicates a high probability (essentially 1) as the Z-value is far from the mean.