Problem 45 - Basic Statistics for Business and Economics

Question

The goal at U.S. airports handling international flights is to clear these flights within 45 minutes. Let's interpret this to mean that 95 percent of the flights are cleared in 45 minutes, so 5 percent of the flights take longer to clear. Let's also assume that the distribution is approximately normal. a. If the standard deviation of the time to clear an international flight is 5 minutes, what is the mean time to clear a flight? b. Suppose the standard deviation is 10 minutes, not the 5 minutes suggested in part a. What is the new mean? c. A customer has 30 minutes from the time her flight landed to catch her limousine. Assuming a standard deviation of 10 minutes, what is the likelihood that she will be cleared in time?

Step-by-Step Solution

Answer

The mean time with 5-minute SD is 36.775 minutes; with 10-minute SD, it's 28.55 minutes. There's a 55.76% chance to clear in 30 minutes with 10-minute SD.

Step 1: Understanding the Problem

We want 95% of flights to clear in 45 minutes, assuming a normal distribution. This means that 45 minutes is the 95th percentile of the distribution.

Step 2: Finding the Z-score for 95th percentile

From the Z-table, a 95th percentile corresponds to a Z-score of approximately 1.645 for a normal distribution.

Step 3a: Calculate Mean with Standard Deviation of 5 Minutes

Using the Z-score formula: \( Z = \frac{X - \mu}{\sigma} \). Here, \( X = 45 \), \( Z = 1.645 \), and \( \sigma = 5 \). Plug these values into the formula:\( 1.645 = \frac{45 - \mu}{5} \).Solve for \( \mu \):\( \mu = 45 - 1.645 \times 5 = 45 - 8.225 = 36.775 \).

Step 3b: Calculate Mean with Standard Deviation of 10 Minutes

Using the same Z-score formula \( Z = \frac{X - \mu}{\sigma} \). Here, \( X = 45 \), \( Z = 1.645 \), and \( \sigma = 10 \). Substitute again:\( 1.645 = \frac{45 - \mu}{10} \).Solve for \( \mu \):\( \mu = 45 - 1.645 \times 10 = 45 - 16.45 = 28.55 \).

Step 4: Calculating the Likelihood for 30-minute Clearance

With a mean of 28.55 minutes (from part b) and \( \sigma = 10 \), find the Z-score for 30 minutes:\( Z = \frac{30 - 28.55}{10} = \frac{1.45}{10} = 0.145 \).Using the Z-table, a Z-score of 0.145 corresponds to approximately 0.5576, which implies about 55.76% likelihood of clearing in time.

Key Concepts

Z-score

A Z-score is a measure that describes a value's position relative to the mean of a group of values. Simply put, it tells us how many standard deviations away a particular data point is from the mean. In our exercise, knowing the Z-score is crucial to finding out more about the distribution of the flight clearance times.

• The Z-score formula is: \[ Z = \frac{X - \mu}{\sigma} \] where:
  • \( X \) is the value in question (for example, 45 minutes in this exercise).
  • \( \mu \) is the mean of the dataset (which we are trying to find).
  • \( \sigma \) is the standard deviation.

By understanding the Z-score, you can pinpoint where your value lies in the context of a distribution, especially if the distribution is normal. Here, it helped us determine at what point 95% of flight clearances fall under 45 minutes.

Standard Deviation

Standard deviation is a crucial measure in statistics as it quantifies the amount of variation or dispersion in a set of numbers. It tells us how spread out the numbers are in a dataset. The higher the standard deviation, the more spread out the values are. In the exercise, the standard deviation is mentioned as either 5 or 10 minutes. This directly impacts the calculation of our mean for flight clearance time.

• Standard deviation in the context of normal distribution helps to shape the bell curve, showing us how much the actual data points deviate from the mean.
• If the standard deviation is small, the data points tend to be clustered closely around the mean.
• If it is large, data points are more spread out.

This measure is fundamental as it affects the calculations of probabilities in normal distribution, as seen in the example calculations for varying standard deviations.

Percentile

The percentile is a value below which a given percentage of observations fall. In the example, the 95th percentile is used, which indicates that 95% of the data points are below 45 minutes, the time specified for clearing flights.

• Percentiles are helpful in understanding the relative standing of a particular observation within a larger dataset.
• In a normal distribution, percentiles correspond to Z-scores which provide insights into the probability and prediction about the occurrences of specific events.

Using percentiles effectively requires understanding how they relate to the data distribution. Here, the 95th percentile helps us assess whether a majority of flights meet the clearance time goal and how changes to the standard deviation affect the mean.

Mean Calculation

Mean, often referred to as the average, is a central measure in statistics that gives you the central value of a data set. It is calculated by summing all individual values and then dividing by the number of values. In this particular problem, we're interested in finding the mean time for flight clearance based on given Z-score and standard deviation.

• The need to calculate the mean arises from the requirement to see at what time, on average, flights are cleared, given they follow a normal distribution.
• The calculations involve solving for the mean using the Z-score formula: \[ \mu = X - Z \times \sigma \]

By manipulating this formula, the exercise shows us how different values of standard deviation affect our calculation of the mean. For instance, the mean changes as the standard deviation shifts from 5 minutes to 10 minutes, demonstrating a direct relationship where a larger standard deviation requires a much earlier average time to meet the 95% clearance within the recommendation.