Statistical Abstracts (117th edition) reports sale price of unleaded gasoline (in cents per gallon) at the refinery. The distribution is mound-shaped with mean μ = 80.04 cents per gallon and standard deviation, σ = 4.74 cents per gallon.
(a) Are we likely to get good results if we use the normal distribution to approximate the distribution of sample means for samples of size 9? Explain.
(b) Find the probability that for a random sample of size 9, the sample mean price will be between 79 and 82 cents per gallon.
(c) Find the probability that for a random sample of size 36, the sample mean price will be between 79 and 82 cents per gallon.
(d) Compare your answers for parts (b) and (c) and give a reason for the difference.
a. A larger sample would give a better estimate.
b. Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
c. Use similar process.
d. See a.
(a) To determine whether we can use the normal distribution to approximate the distribution of sample means for samples of size 9, we need to check if the sample size is sufficiently large and if the population distribution is approximately normal.
The Central Limit Theorem states that for sample sizes that are sufficiently large (around 30 or more), the sampling distribution of the sample mean will be approximately normally distributed, even if the population distribution is not normal.
In this case, the sample size is 9, which is relatively small. When the sample size is small, we need the population distribution to be approximately normal for the sampling distribution of the sample mean to also be approximately normal.
Given that the distribution of prices at the refinery is mound-shaped (indicating a distribution close to normal), we can expect that for samples of size 9, the sample means will be approximately normally distributed. Therefore, we can use the normal distribution to approximate the distribution of sample means in this case.
(b) To find the probability that for a random sample of size 9, the sample mean price will be between 79 and 82 cents per gallon, we can use the properties of the normal distribution.
First, we need to standardize the values using the formula:
Z = (X - μ) / (σ / √n),
where X is the sample mean price, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, X = 79, μ = 80.04, σ = 4.74, and n = 9. Plugging these values into the formula, we get:
Z1 = (79 - 80.04) / (4.74 / √9) = -0.4438
Z2 = (82 - 80.04) / (4.74 / √9) = 0.8876
Now, we can find the probability by subtracting the standardized values from the corresponding percentile values in the standard normal distribution table (also known as the z-table).
P(79 < X < 82) = P(-0.4438 < Z < 0.8876)
Using the z-table, we can find the corresponding probabilities for these z-values. Subtracting the area in the table for the lower z-value from the area for the higher z-value will give us the desired probability.
(c) Similarly, for a random sample of size 36, we can use the same process as in part (b) to find the probability that the sample mean price will be between 79 and 82 cents per gallon.
Let's assume X = 79, μ = 80.04, σ = 4.74, and n = 36. Plugging these values into the formula, we get:
Z1 = (79 - 80.04) / (4.74 / √36) = -1.9325
Z2 = (82 - 80.04) / (4.74 / √36) = 1.9325
Again, we need to use the z-table to find the probabilities for these z-values:
P(79 < X < 82) = P(-1.9325 < Z < 1.9325)
(d) In comparing the answers for parts (b) and (c), we expect the probability in part (c) (sample size 36) to be larger than the probability in part (b) (sample size 9). This is because as the sample size increases, the standard error of the sample mean decreases. As a result, the distribution of sample means becomes narrower, and the probability of obtaining a sample mean within a certain range increases.
Therefore, the larger sample size (36) in part (c) leads to a higher probability compared to the smaller sample size (9) in part (b).