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.
(a) No.
Sample means have a narrower distribution that is inversely proportional to the square root of the number of samples in the group.
(b) Assume the distribution mean of samples of 9 has a mean of 80.04 with a standard deviation of 1.58
I get 63.7% for the probability.
(a) To determine whether it is appropriate to use the normal distribution to approximate the distribution of sample means for samples of size 9, we can apply the Central Limit Theorem. According to the Central Limit Theorem, when the sample size is sufficiently large (usually greater than 30), the sampling distribution of the sample means will be approximately normal regardless of the shape of the original population distribution.
In this case, the sample size is 9, which is relatively small. Therefore, we need to check if the population distribution is approximately normal. Since it is mentioned that the distribution of sale price of unleaded gasoline at the refinery is mound-shaped (implying a bell-shaped distribution), we can assume that the population distribution is approximately normal.
Hence, it is likely to get good results if we use the normal distribution to approximate the distribution of sample means for samples of size 9.
(b) To find the probability that the sample mean price will be between 79 and 82 cents per gallon for a random sample of size 9, we will use the standard normal distribution.
First, we need to standardize the values using the formula:
Z = (x - μ) / (σ / sqrt(n))
where:
x = value (sample mean) we want to find the probability for
μ = population mean
σ = population standard deviation
n = sample size
In this case:
x1 = 79 (lower limit)
x2 = 82 (upper limit)
μ = 80.04
σ = 4.74
n = 9
Now we can calculate the z-scores for the lower and upper limits:
z1 = (x1 - μ) / (σ / sqrt(n))
z1 = (79 - 80.04) / (4.74 / sqrt(9))
z2 = (x2 - μ) / (σ / sqrt(n))
z2 = (82 - 80.04) / (4.74 / sqrt(9))
After calculating the respective z-scores, we can look up the corresponding probabilities from the standard normal distribution table, or use a calculator or software that provides the functionality to find the probability between these z-scores. The difference between these probabilities will give us the desired probability.
Using this approach, we can find the probability that the sample mean price will be between 79 and 82 cents per gallon for a random sample of size 9.