91Ó°ÊÓ

In \(2015,\) the on-time arrival rate of all major domestic and regional airlines operating between Australian airports has a bell-shaped distribution roughly with mean 0.86 and standard deviation 0.1 . a. Let \(X\) denote the number of flights arriving on time when you observe one flight. State the probability distribution of \(X\). (This also represents the population distribution you would get if you could observe an infinite number of flights.) b. You decide to observe the airport arrival tables for one day. At the end of the day, you were able to check the arrival times of 100 flights. Show that the sampling distribution of your sample mean number of flights on time has mean \(=0.86\) and standard deviation \(=0.01\). c. Refer to part b. Using the central limit theorem, find the probability that the mean number of flights on time is at least 0.88 , so that you have a gain of at least \(2 \%\) with regard to the rate of on-time flights in the population.

Step by step solution

01

Determine Probability Distribution of X

Since we are dealing with a percentage (rate) of flights arriving on time, which is expressed as a proportion, the variable \(X\) follows a normal distribution given that it is stated to be bell-shaped. Despite being based on proportions, this implicitly assumes that \(X\) could be binomial but follows a normal distribution due to the sample size being potentially large. Therefore, \(X\) is normally distributed with mean \(\mu = 0.86\) and standard deviation \(\sigma = 0.1\).

02

Identify Sample Size and Mean

We're observing \(n = 100\) flights. The sample mean (average) of these flights' on-time ratio is the same as the population mean, which is 0.86 due to the properties of expected value. Thus, \(\mu_{\bar{X}} = 0.86\).

03

Calculate Sampling Distribution Standard Deviation

The standard deviation of the sample mean, known as the standard error (SE), is calculated by dividing the population standard deviation by the square root of the sample size: \[SE = \frac{\sigma}{\sqrt{n}} = \frac{0.1}{\sqrt{100}} = 0.01.\]

04

Use Central Limit Theorem for Probability Calculation

To find the probability that the sample mean is at least 0.88, we calculate the z-score using the formula: \[z = \frac{\bar{X} - \mu_{\bar{X}}}{SE} = \frac{0.88 - 0.86}{0.01} = 2.\]The z-score tells us how many standard deviations our sample mean is away from the population mean.

05

Refer to Z-Table for Probability

Using the standard normal distribution table (z-table), we find the probability of \(Z \geq 2\). From the z-table, the probability of \(Z < 2\) is approximately 0.9772. Therefore, the probability that \(Z \geq 2\) is:\[ P(Z \geq 2) = 1 - 0.9772 = 0.0228.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution

A normal distribution, often referred to as a bell curve, is a probability distribution that is symmetrical around its mean. In simple terms, most of the data points in a normal distribution cluster around the mean, and fewer and fewer appear as you move away. This pattern creates the iconic bell shape.
For the problem of airline on-time arrivals, the normal distribution implies that the percentage of flights that arrive on time tends to cluster around the mean value of 0.86. This is a common characteristic in many real-life distributions, such as heights, test scores, and many other measurable attributes.

• Mean (): The average of all data points. Here it is 0.86, meaning on average, 86% of flights arrive on time.
• Standard Deviation (): A measure of spread or dispersion around the mean. In this case,  is 0.1, indicating how concentrated the data is around the mean.

Understanding the normal distribution is key because it is the foundation for determining probabilities about how data spreads around the mean.

Sampling Distribution

When we take samples from a population, the sample mean will not always equal the population mean due to variability. However, as per the Central Limit Theorem, the distribution of the sample means tends towards a normal distribution as the sample size gets larger, even if the original population distribution is not normal.

In our specific task, we observed 100 flights, making our sample size quite large (0), and calculated the sample mean x, which is expected to be very close to 0.86, the population mean.
Even though each individual set of 100 flights may vary, overall the sample means form a bell-shaped distribution centered around the true mean.

• Mean of the Sampling Distribution (): The mean of the sample means, which equals the population mean.
• Standard Error (SE): The standard deviation of the sampling distribution of the sample mean. This is calculated as SE = /0 = 0.01.

Sampling distributions allow us to make inferences about the population from which the samples were drawn.

Probability Calculation

Probability calculation in this context involves determining how likely it is for the sample mean to fall above or below a certain value. With our sample size of 100 flights and using the Central Limit Theorem, we calculate probabilities related to the sample mean.

To find the probability of achieving an average on-time arrival rate of at least 0.88, we compute the z-score. The z-score measures the deviation of our sample mean from the population mean in terms of standard errors.

• Z-score: Calculated as z = (x - ) / SE, where x is 0.88,  is 0.86, and SE is 0.01.
• For this case, z = (0.88 - 0.86) / 0.01 = 2.

Looking up 2 on the standard normal distribution table shows us that this is at the upper end of the distribution, giving us the probability of this event.

Standard Deviation

Standard deviation is a key concept in statistics, symbolized as . It quantifies how much individual data points in a set differ from the mean or average of that set. In simpler terms, it tells us how spread out the numbers in our data are.

For the airline problem:

• The standard deviation is 0.1, which means most flight arrival proportions are within 0.1 of the mean value of 0.86. This gives a clear picture of the variability in flight arrivals.

To put it in perspective, in a normal distribution approximately 68% of the data falls within one standard deviation (from 0.76 to 0.96 in this case), and about 95% falls within two standard deviations. This is crucial for understanding how data spreads and is used in probability calculations and predictions.