Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15. Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?
how can i write this in a mathematical expression?
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 proportion/probability related to the Z score.
To determine if it would be unusual for the mean of a sample of 3 to be 115 or more, we can use the concept of sampling distribution and the Z-score.
The sampling distribution of the mean is the distribution of all possible sample means that could be drawn from a population. For a population that is normally distributed, the sampling distribution of the mean will also be normally distributed.
In this case, the mean of the population is 100 and the standard deviation is 15. When we take a sample of 3, the sampling distribution of the mean will have a mean equal to the population mean (100) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (15/sqrt(3)).
To determine if it would be unusual for the mean of a sample of 3 to be 115 or more, we can calculate the Z-score for this value using the formula:
Z = (X - μ) / (σ / sqrt(n))
Where:
- X is the value we want to test (115 in this case)
- μ is the mean of the sampling distribution (100 in this case)
- σ is the standard deviation of the sampling distribution (15/sqrt(3) in this case)
- n is the sample size (3 in this case)
In our case:
Z = (115 - 100) / (15 / sqrt(3))
Z = 15 / (15 / sqrt(3))
Z = sqrt(3)
The Z-score for a value of 115 is sqrt(3).
Now, we can consult a Z-table or use a calculator to find the proportion of the standard normal distribution beyond the Z-score of sqrt(3). The result will give us the probability of observing a sample mean of 115 or more.
By looking up the Z-score in the Z-table or using a calculator, we find that the proportion beyond the Z-score of sqrt(3) is approximately 0.739, or 73.9%. This means that the probability of observing a sample mean of 115 or more is approximately 0.739, or 73.9%.
Since the probability is relatively high (more than 50%), it would not be considered unusual for the mean of a sample of 3 to be 115 or more.