A final exam in Math 160 has a mean of 73 with standard deviation 7.73. Assume that a random sample of 24 students is selected and the test score of the sample is computed. Assuming the scores are normally distributed, what percentage of sample means are less than 69?
To find the percentage of sample means that are less than 69, we need to use the concept of the standard error of the mean (SEM), which is calculated by dividing the standard deviation (SD) of the population by the square root of the sample size.
First, let's calculate the SEM:
SEM = SD / √n
Given that the standard deviation (SD) is 7.73 and the sample size (n) is 24, we can compute the SEM as follows:
SEM = 7.73 / √24 ≈ 1.57
Next, we need to standardize the value of 69 by using the formula for z-score:
z = (x - μ) / σ
where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.
In this case, we want to standardize 69 with a mean of 73 and a standard deviation of 7.73:
z = (69 - 73) / 7.73 ≈ -0.52
Now that we have the z-score, we can find the corresponding area under the normal distribution curve using a standard normal table or a statistical calculator. The area to the left of the z-score represents the percentage of sample means that are less than 69.
Using a standard normal table or calculator, we find that the area to the left of -0.52 is approximately 0.3015. This means that approximately 30.15% of sample means are less than 69.
So, the answer is that approximately 30.15% of sample means are less than 69.
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 related to the Z score.