An agricultural economist is interested in determining the average diameter of peaches produced by a particular tree. A random sample of 50 peaches is taken and the sample mean calculated. Suppose that the average diameter of peaches on this tree is known from previous years' production to be 75 millimeters with a standard deviation of 15 millimeters. What is the probability that the sample mean exceeds 78 millimeters?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To solve this problem, we need to use the Central Limit Theorem, which states that the distribution of sample means will approach a normal distribution as the sample size increases.
Given that the average diameter of peaches on this tree is known to be 75 millimeters with a standard deviation of 15 millimeters, we can calculate the standard error of the sample mean using the formula:
Standard Error = Standard Deviation / √(Sample Size)
In this case, the standard deviation is 15 millimeters and the sample size is 50, so the standard error is:
Standard Error = 15 / √(50) ≈ 2.12 millimeters
Next, we need to calculate the z-score, which is a measure of how many standard errors the sample mean is away from the population mean. The formula for calculating the z-score is:
z = (Sample Mean - Population Mean) / Standard Error
In this case, the sample mean is 78 millimeters, the population mean is 75 millimeters, and the standard error is 2.12 millimeters, so the z-score is:
z = (78 - 75) / 2.12 ≈ 1.42
Finally, we need to find the probability that the sample mean exceeds 78 millimeters. We can do this by looking up the z-score in a standard normal distribution table or using statistical software. For a z-score of 1.42, the probability is approximately 0.9236 or 92.36%. Therefore, the probability that the sample mean exceeds 78 millimeters is 0.9236 or 92.36%.
To find the probability that the sample mean exceeds 78 millimeters, we can use the concept of sampling distribution and the Central Limit Theorem.
The Central Limit Theorem states that, regardless of the shape of the population distribution, the sampling distribution of the sample mean approaches a normal distribution as the sample size increases.
In this case, the average diameter of peaches on the tree follows a normal distribution with a mean of 75 millimeters and a standard deviation of 15 millimeters.
To calculate the probability that the sample mean exceeds 78 millimeters, we need to z-score transform the value and then use a standard normal distribution table or calculator.
The z-score can be calculated using the formula:
z = (x - μ) / (σ / sqrt(n))
Where:
x = the value of interest (78 millimeters)
μ = population mean (75 millimeters)
σ = population standard deviation (15 millimeters)
n = sample size (50 peaches in this case)
Plugging in the values into the formula:
z = (78 - 75) / (15 / sqrt(50))
z = 3 / (15 / 7.071)
z = 3 / 2.121
z ≈ 1.414
Now, we can use the standard normal distribution table or a calculator to find the probability associated with the z-score of 1.414. The table or calculator will give us the probability of the sample mean exceeding 78 millimeters.
Using a standard normal distribution table, we can find that the corresponding probability is approximately 0.0788, or 7.88%.
Therefore, the probability that the sample mean exceeds 78 millimeters is approximately 0.0788, or 7.88%.