The percent of fat calories that a person that in America consumes each day is normallydistributed with a mean of 36 and a standard deviation of 10. Suppose that 16 individuals are randomly chosen for an experiment.
1. Completely describe the distribution of sample means for the group of 16 individuals that was chosen in the experiment above. (please show work so I know how to do it myself!)
2. For the group of 16, find the probability that the average percent of fat calories consumed is more than 30 calories.
Thank you!
1. Mean = Sample mean
Variability for distribution of means = SEm = SD/√n
2. Z = (score-mean)/SEm
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 answer the questions, we need to understand the concept of sampling distribution and how to calculate probabilities using the standard normal distribution.
Let's start with Question 1:
1. Distribution of Sample Means:
The distribution of sample means is also known as the sampling distribution. It describes how the means of different samples from the same population are distributed.
In this case, we have a sample size of 16 individuals randomly chosen for the experiment. We know that the population mean is 36 and the standard deviation is 10.
To find the distribution of sample means, we need to calculate the mean and standard deviation of the sample means.
The mean of the sample means (also known as the expected value) is equal to the population mean:
μx = μ = 36
The standard deviation of the sample means (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size:
σx = σ / √n = 10 / √16 = 10 / 4 = 2.5
Therefore, the distribution of sample means for the group of 16 individuals is normally distributed with a mean of 36 and a standard deviation of 2.5.
Now let's move on to Question 2:
2. Probability that the average percent of fat calories consumed is more than 30 calories:
To find this probability, we need to convert the values to z-scores and then use the standard normal distribution table (also known as the Z-table).
The z-score formula is:
z = (x - μ) / σ
In this case, we want to find the probability that the average percent of fat calories consumed is more than 30. We can use the sample mean of 30, the population mean of 36, and the standard deviation of 2.5 (from the previous calculation).
Using the z-score formula:
z = (30 - 36) / 2.5 = -6 / 2.5 = -2.4
We want to find the probability that the z-score is greater than -2.4.
Using the Z-table, we can find that the probability associated with a z-score of -2.4 is approximately 0.0082.
Therefore, the probability that the average percent of fat calories consumed is more than 30 calories for the group of 16 individuals is approximately 0.0082 or 0.82%.
I hope this explanation helps! Let me know if you have any further questions.