Chapter 7: Problem 6
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 Known Information
Calculate the Standard Error of the Mean
Find the Z-score for Part a (Less than 10 minutes)
Calculate the Probability for Part a
Find the Z-scores for Part b (Between 5 and 10 minutes)
Calculate the Probability for Part b
Find the Z-score for Part c (Less than 6 minutes)
Calculate the Probability for 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 mean (\( \mu \)) is the central point where most data points cluster.
- The standard deviation (\( \sigma \)) determines the spread of the data.
Standard Error
- \( \sigma \) is the standard deviation of the population.
- \( n \) is the sample size.
Z-score
- \( \bar{X} \) is the sample mean.
- \( \mu \) is the population mean.
- \( SE \) is the standard error.
Probability Calculation
For instance, to find out the probability that the average waiting time is less than 10 minutes (Z ≈ 8.400), we discover a probability very close to 1, indicating it's almost certain. Conversely, for less than 6 minutes with Z ≈ -10.267, the probability is about 0, showing such occurrence is extremely unlikely. When calculating probabilities between two points, like waiting times between 5 and 10 minutes, we evaluate Z-scores for both points and find the probability of values falling in between.
Probability guides data-driven decisions by quantifying expected outcomes, touching everyday occurrences such as client wait times, helping professionals from various disciplines predict future events with an evidence-based approach.