Unit 4 Exam Review – Sampling Distributions, Confidence Intervals, and Probability

Q1. The cost of unleaded gasoline in Syracuse once followed an unknown distribution with a mean of $3.19 and a standard deviation of $0.15. Twenty-five gas stations from Syracuse are randomly chosen. We are interested in the average cost of gasoline for the 25 gas stations.

Background

Topic: Sampling Distribution of the Sample Mean

This question explores the properties of the sampling distribution of the sample mean, including its standard deviation (standard error), probability calculations using the normal distribution, and percentiles.

Key Terms and Formulas

• Population Mean (\(\mu\)): The average value in the population.

• Population Standard Deviation (\(\sigma\)): The spread of values in the population.

• Sample Size (\(n\)): The number of observations in the sample.

• Standard Error (SE): The standard deviation of the sample mean.

Key formula for the standard error of the mean:

$SE = \frac{\sigma}{\sqrt{n}}$

Step-by-Step Guidance

1. Identify the population standard deviation (\(\sigma = 0.15\)) and the sample size (\(n = 25\)).

2. Recall the formula for the standard error of the sample mean: $SE = \frac{\sigma}{\sqrt{n}}$

3. Substitute the values into the formula: $SE = \frac{0.15}{\sqrt{25}}$

4. Simplify the denominator (\(\sqrt{25} = 5\)), then set up the division.

Try solving on your own before revealing the answer!

Final Answer: 0.03

$SE = \frac{0.15}{5} = 0.03$

The standard deviation for the sample mean is 0.03, which means the average cost of gasoline for samples of 25 stations will typically vary by about $0.03 from the population mean.

Q1(b). Sketch of the graph.

Background

Topic: Sampling Distribution Shape

This part asks you to visualize the sampling distribution of the sample mean. For large enough samples, the Central Limit Theorem tells us the distribution will be approximately normal, even if the population distribution is not normal.

Key Terms

• Normal Distribution: A symmetric, bell-shaped curve that describes the distribution of many types of data.

• Central Limit Theorem: States that the sampling distribution of the sample mean approaches normality as the sample size increases.

Step-by-Step Guidance

1. Draw a horizontal axis and mark the population mean ($\mu = 3.19$) at the center.

2. Sketch a bell-shaped curve centered at $3.19$ to represent the sampling distribution of the sample mean.

3. Label the spread of the curve using the standard error ($SE = 0.03$) calculated previously.

Try sketching the graph on your own before checking the example!

Q1(c). What's the probability that the average price for 25 gas stations is between $3.15 and $3.24?

Background

Topic: Probability Using the Normal Distribution

This question asks you to find the probability that the sample mean falls within a certain range. This involves standardizing the values (finding z-scores) and using the standard normal table.

Key Terms and Formulas

• Z-score: The number of standard errors a value is from the mean.

Key formula for z-score:

$z = \frac{\bar{x} - \mu}{SE}$

Step-by-Step Guidance

1. Calculate the z-score for $3.15$ using $z = \frac{3.15 - 3.19}{0.03}$.

2. Calculate the z-score for $3.24$ using $z = \frac{3.24 - 3.19}{0.03}$.

3. Use the standard normal table to find the probability between these two z-scores.

Try solving on your own before revealing the answer!

Final Answer: 0.9522

The probability that the average price is between $3.15 and $3.24 is approximately 0.9522, or 95.22%.

Q1(d). Find the probability that the average gas price for 25 gas stations is less than $3.16.

Background

Topic: Probability Using the Normal Distribution (Left Tail)

This question asks for the probability that the sample mean is less than a specific value, which involves finding the left-tail probability using the z-score.

Key Terms and Formulas

• Z-score: $z = \frac{\bar{x} - \mu}{SE}$

Step-by-Step Guidance

1. Calculate the z-score for $3.16$ using $z = \frac{3.16 - 3.19}{0.03}$.

2. Use the standard normal table to find the probability to the left of this z-score.

Try solving on your own before revealing the answer!

Final Answer: 0.1587

The probability that the average price is less than $3.16 is approximately 0.1587, or 15.87%.

Q1(e). Find the probability that the average gas price for 25 gas stations is greater than $3.27.

Background

Topic: Probability Using the Normal Distribution (Right Tail)

This question asks for the probability that the sample mean is greater than a specific value, which involves finding the right-tail probability using the z-score.

Key Terms and Formulas

• Z-score: $z = \frac{\bar{x} - \mu}{SE}$

Step-by-Step Guidance

1. Calculate the z-score for $3.27$ using $z = \frac{3.27 - 3.19}{0.03}$.

2. Use the standard normal table to find the probability to the right of this z-score (1 minus the cumulative probability).

Try solving on your own before revealing the answer!

Final Answer: 0.0004

The probability that the average price is greater than $3.27 is approximately 0.0004, or 0.04%.

Q1(f). Find the 90th percentile for the average gas price for the 25 gas stations. Interpret the results.

Background

Topic: Percentiles of the Sampling Distribution

This question asks you to find the value of the sample mean that separates the lowest 90% from the highest 10% of the sampling distribution.

Key Terms and Formulas

• Percentile: The value below which a given percentage of observations fall.

• Z-score for 90th percentile: Use the standard normal table to find the z-score corresponding to the 90th percentile (approximately 1.28).

• Formula: $\bar{x} = \mu + z \times SE$

Step-by-Step Guidance

1. Find the z-score for the 90th percentile (about 1.28).

2. Use the formula $\bar{x} = \mu + z \times SE$ with $\mu = 3.19$, $z = 1.28$, and $SE = 0.03$.

3. Multiply $z$ by $SE$ and add to $\mu$ to find the 90th percentile value.

Try solving on your own before revealing the answer!

Final Answer: 3.2284

The 90th percentile for the average gas price is approximately $3.23. This means that 90% of sample means will be below this value.