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A recent study by the Greater Los Angeles Taxi Drivers Association showed that the mean fare charged for service from Hermosa Beach to Los Angeles International Airport is \(\$ 18.00\) and the standard deviation is \(\$ 3.50 .\) We select a sample of 15 fares. a. What is the likelihood that the sample mean is between \(\$ 17.00\) and \(\$ 20.00 ?\) b. What must you assume to make the above calculation?

Short Answer

Expert verified
a. The probability is approximately 0.865. b. Assume normal distribution or n large enough for CLT.

Step by step solution

01

Identify the Main Problem

We're tasked with finding the probability that the sample mean fare falls between $17.00 and $20.00 for a sample of 15 fares with certain known parameters.
02

Understand the Distribution Assumptions

Assume that the distribution of fare prices is approximately normal, or that the sample size is sufficiently large to rely on the Central Limit Theorem (CLT) to assume normality of the sample mean. This allows us to proceed using the normal distribution.
03

Define Known Variables

We have: mean (\(\mu = \\(18.00\)), standard deviation (\(\sigma = \\)3.50\)), and sample size (\(n = 15\)). We need to calculate the standard deviation of the sample mean, also known as the standard error (SE).
04

Calculate Standard Error

The standard error is calculated using the formula \(SE = \frac{\sigma}{\sqrt{n}}\). For our problem: \(SE = \frac{3.50}{\sqrt{15}} \approx 0.903\).
05

Standardize the Values

We need to standardize the values using the Z-score formula for a sample mean. For a given value \(X\), Z is calculated as \(Z = \frac{X - \mu}{SE}\).
06

Calculate Z-scores for the Boundaries

Calculate the Z-score for \(17.00:\[Z_{17} = \frac{17.00 - 18.00}{0.903} \approx -1.11\]Calculate the Z-score for \)20.00:\[Z_{20} = \frac{20.00 - 18.00}{0.903} \approx 2.21\]
07

Find the Probability Related to Z-scores

Using the standard normal distribution table, find the proportion of data lying between these Z-scores. The probability of Z between -1.11 and 2.21 is approximately 0.865.
08

Assumptions for Calculation

To make the calculation valid, we must assume either that the population is normally distributed or that the sample size is large enough for the Central Limit Theorem to apply, which here is reasonable with n=15.

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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 is an essential concept in statistics, particularly when it comes to understanding data that centers around a mean value. Imagine it as a bell-shaped curve where most of the data points cluster around the center (mean), and the probabilities of values further from the mean become progressively smaller.
This distribution is essential in many statistical methods, including those that rely on the Central Limit Theorem.
  • In the context of the taxi fare example, we may assume that individual fare charges roughly follow a normal distribution.
  • Even if the data isn't perfectly normal, with a sample of sufficient size, the average fares will be normally distributed due to the Central Limit Theorem.
The beauty of the normal distribution is its predictability, which allows for the calculation of probabilities associated with different outcomes. This predictability is why it's applicable to many real-world problems.
Sample Standard Deviation
The sample standard deviation is a measure that tells us how spread out the individual data points are around the sample mean. It gives us an idea of the variability within the sample.
  • For instance, the sample standard deviation in the taxi fare problem is given as $3.50, a measure of how much individual fares typically differ from the average fare of $18.00.
  • This helps us understand the reliability of our average fare and makes it possible to estimate the variability of the sample mean.
In practical terms, a larger standard deviation suggests more variability in fares, while a smaller one indicates that fares are more consistent. Understanding the spread is vital for making sense of varying data, especially when predicting or estimating outcomes.
Probability Calculation
Understanding probability calculation is crucial when gauging the likelihood of certain outcomes within a given range. In the taxi fare example, our task is to find the probability that average fares fall between \(17.00 and \)20.00.
  • This involves multiple steps, starting from standardizing the range using Z-scores, which tell us how many standard deviations a value is from the mean.
  • The first step is to calculate the Z-score for each boundary of your range using the formula: \(Z = \frac{X - \mu}{SE}\).
  • For \(17.00, this corresponds to a Z-score of approximately -1.11, and for \)20.00, approximately 2.21.

  • Once we have these Z-scores, we use a standard normal distribution table to find the probability that the sample mean lies between them, which is about 0.865 in this case.
    This reflects the likelihood that our estimated average fares fall in the specified range. Probability calculation hinges on knowing your data's distribution and understanding these tools and why they are used makes the calculation more intuitive.
    Standard Error
    The standard error (SE) is another fundamental concept when working with samples, as it measures the dispersion of the sample mean estimates from the true population mean.
    • In simpler terms, it tells us how much the sample mean (average of 15 fares) is expected to differ from the actual population mean.
    • For the taxi fare problem, the standard error provides insight on the estimate's accuracy, calculated as \(SE = \frac{\sigma}{\sqrt{n}}\), resulting in approximately 0.903.
    A smaller standard error indicates that the sample mean is a more precise estimate of the population mean.
    This is useful because it allows us to confidently compare the calculated average fare to the expected mean.
    The standard error becomes particularly important when the sample size isn't very large, bridging the sample data to infer properties about the population.

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    Most popular questions from this chapter

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