According to one survey, the mean serum cholesterol level for U.S. adults was 195.1, with a standard deviation of 41.5. A simple random sample of 95 adults is chosen. Find the 46 percentile for the sample mean.
Write only a number as your answer. Round to one decimal places.
What exactly is it asking you?
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
(.46) and its Z score. Insert data into above equations and calculate.
To find the 46th percentile for the sample mean, we need to use the concept of the standard normal distribution.
1. Begin by calculating the standard error of the mean (SEM), which represents the average distance between the sample means and the population mean.
The formula for the standard error of the mean is: SEM = standard deviation / sqrt(sample size)
In this case, the standard deviation is 41.5 and the sample size is 95.
SEM = 41.5 / sqrt(95) ≈ 4.26
2. Convert the desired percentile (46th percentile) to a z-score. The z-score represents the number of standard deviations from the mean.
To find the z-score, we can use a standard normal distribution table or a statistical calculator. In this case, we want the z-score that corresponds to a cumulative probability of 0.46.
Using a standard normal distribution table or a statistical calculator, the z-score for a cumulative probability of 0.46 is approximately -0.11.
3. Use the z-score formula to calculate the sample mean.
Sample Mean = population mean + (z-score * SEM)
In this case, the population mean is 195.1, the z-score is -0.11, and the SEM is 4.26.
Sample Mean = 195.1 + (-0.11 * 4.26) ≈ 194.6
Therefore, the 46th percentile for the sample mean is approximately 194.6.
Answer: 194.6