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The Oil Price Information Center reports the mean price per gallon of regular gasoline is \(\$ 3.26\) with a population standard deviation of \(\$ 0.18 .\) Assume a random sample of 40 gasoline stations is selected and their mean cost for regular gasoline is computed. a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between \(\$ 3.24\) and \(\$ 3.28 ?\) c. What is the probability that the difference between the sample mean and the population mean is less than \(0.01 ?\) d. What is the likelihood the sample mean is greater than \(\$ 3.34 ?\)

Short Answer

Expert verified
a) SEM ≈ 0.0285 b) P(3.24 < mean < 3.28) ≈ 0.5160 c) P(|mean - 3.26| < 0.01) ≈ 0.9100 d) P(mean > 3.34) ≈ 0.0025

Step by step solution

01

Calculate Standard Error of the Mean

The standard error of the mean (SEM) is computed using the formula \( \text{SEM} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. Given \( \sigma = 0.18 \) and \( n = 40 \), we have:\[ \text{SEM} = \frac{0.18}{\sqrt{40}} \approx 0.0285. \]
02

Determine Z-Scores for Bounds $3.24 and $3.28

To find the probability that the sample mean is between \\(3.24 and \\)3.28, calculate the z-scores using the formula \( z = \frac{\bar{x} - \mu}{\text{SEM}} \), where \( \bar{x} \) is the sample mean and \( \mu \) is the population mean. Calculate:1. For \( \bar{x} = 3.24 \), \( z = \frac{3.24 - 3.26}{0.0285} \approx -0.70. \)2. For \( \bar{x} = 3.28 \), \( z = \frac{3.28 - 3.26}{0.0285} \approx 0.70. \)
03

Calculate Probability for Range $3.24 to $3.28

Using standard normal distribution tables, find the probability for the z-scores calculated:- Probability for \( z = -0.70 \) is approximately \( 0.2420. \)- Probability for \( z = 0.70 \) is approximately \( 0.7580. \)Thus, the probability that the sample mean is between \\(3.24 and \\)3.28 is \( 0.7580 - 0.2420 = 0.5160. \)
04

Probability for Difference Less than $0.01

The difference from the population mean is less than \$0.01 implies\[ |\bar{x} - \mu| < 0.01 \]Calculate z-values for \( \bar{x} = 3.25 \) and \( \bar{x} = 3.27 \):1. For \( \bar{x} = 3.25 \), \( z = \frac{3.25 - 3.26}{0.0285} \approx -0.35. \)2. For \( \bar{x} = 3.27 \), \( z = \frac{3.27 - 3.26}{0.0285} \approx 0.35. \)Probability for z-scores between \(-0.35\) and \(0.35\) is approximately \( 0.2736 + 0.6364 = 0.9100. \)
05

Determine Likelihood of $3.34 or More

Calculate z-score for sample mean \( \bar{x} = 3.34 \):\[ z = \frac{3.34 - 3.26}{0.0285} \approx 2.81. \]Probability for z-score \( 2.81 \) refers to area to the right in standard normal distribution, which is:\[ 1 - P(z \leq 2.81) \approx 1 - 0.9975 = 0.0025. \]

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Key Concepts

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

Standard Error of the Mean
The standard error of the mean (SEM) is a crucial concept in statistics that helps us understand the variability of a sample mean from the true population mean. It essentially tells us how much we can expect our sample mean to differ from the actual population mean.
To calculate the SEM, we use the formula \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \] where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
For example, if the population standard deviation is 0.18 and you have a sample size of 40, the SEM would be:\[ \text{SEM} = \frac{0.18}{\sqrt{40}} \approx 0.0285 \]
This small value indicates that individual sample means are likely to be quite close to the population mean, reinforcing the reliability of the sample.
Normal Distribution
The normal distribution is a key concept in statistics and is often used in probability and statistical inference. It is known for its bell-shaped curve and is defined by its mean and standard deviation.
This distribution is symmetric around the mean, which means most data points cluster around the center, with fewer and fewer occurring as you move away.
  • The mean provides the peak or center of the distribution.
  • The standard deviation determines the width of the curve.
Because many natural phenomena follow a normal distribution, it has wide applications. For instance, in our gasoline example with a mean price of $3.26, we assume the distribution of sample means follows a normal pattern with predictable probabilities, which is essential for calculating z-scores and probabilities.
Z-Scores
Z-scores are standard scores that represent how many standard deviations a data point is from the mean. They allow us to compare different data points directly, even if they come from different distributions.
To calculate a z-score, use the formula: \[ z = \frac{\bar{x} - \mu}{\text{SEM}} \] where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( \text{SEM} \) is the standard error of the mean.
For example, to find the z-score for a sample mean of \(3.24 when the population mean is \)3.26 and the SEM is 0.0285, you calculate: \[ z = \frac{3.24 - 3.26}{0.0285} \approx -0.70 \]
This z-score tells us that \(3.24 is 0.70 standard deviations below the population mean of \)3.26. Understanding z-scores is fundamental for determining probabilities.
Probability Calculations
Probability calculations help us understand the likelihood of an event occurring, particularly when dealing with sample data and means. They often involve using the normal distribution and z-scores.
For instance, in the gasoline price example, we calculated the probability of the sample mean falling between $3.24 and $3.28. After computing the z-scores, these values are used with standard normal distribution tables to find probabilities.
  • The probability that the sample mean is between $3.24 and $3.28 was found to be approximately 0.5160, or 51.60%.
  • The likelihood of the sample mean being greater than $3.34 was 0.25%, indicating it's quite unlikely.
Such calculations allow businesses or researchers to make informed decisions based on statistical evidence.

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Most popular questions from this chapter

Nike's annual report says that the average American buys 6.5 pairs of sports shoes per year. Suppose the population standard deviation is 2.1 and that a sample of 81 customers will be examined next year. a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between 6 and 7 pairs of sports shoes? c. What is the probability that the difference between the sample mean and the population mean is less than 0.25 pairs? d. What is the likelihood the sample mean is greater than 7 pairs?

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A normal population has a mean of 75 and a standard deviation of \(5 .\) You select a sample of \(40 .\) Compute the probability the sample mean is: a. Less than 74 . b. Between 74 and 76 . c. Between 76 and 77 . d. Greater than 77 .

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