the average number of pushups a United States Marine does daily is 300, with a standard deviation of 50. A random sample of 36 marines is selected. What is the probability that the sample mean is at most 320 push-ups?
With a mean of 300, you are looking for a sample using a Standard Error of the Mean (SE). (The SE functions the exact same way for a distribution of means as the SD does for a distribution of raw scores.)
SE = SD/(sq.rt.of N) = 50/6
Once you can determine how many SE above the mean the value of 320 is, use the "Area Under the Normal Distribution" table in the back of your statistics book to find the probability.
I hope this helps. Thanks for asking.
To find the probability that the sample mean is at most 320 push-ups, you need to calculate the z-score and then use the "Area Under the Normal Distribution" table.
The z-score measures how many standard errors the sample mean is away from the population mean. In this case, the population mean is 300 push-ups and the standard error is 50/√36 = 8.33 push-ups.
To calculate the z-score for a sample mean of 320 push-ups:
z = (x - μ) / SE
z = (320 - 300) / 8.33
z = 2.4
Now, you need to use the z-score to find the corresponding area under the normal distribution curve. You can use the "Area Under the Normal Distribution" table to find this probability. Look up the z-score of 2.4 in the table and find the corresponding area.
Let's assume you find that the area is 0.9918 (just as an example).
Therefore, the probability that the sample mean is at most 320 push-ups is 0.9918, or 99.18% (assuming the area is 0.9918 based on the table lookup).
Remember to refer to your specific statistics book's table to get the precise value for the area corresponding to the z-score of 2.4.
I hope this explanation helps you understand how to solve this problem. Let me know if you have any further questions!
To find the probability that the sample mean is at most 320 push-ups, we first need to calculate the z-score.
Z-score = (sample mean - population mean) / (standard deviation / √sample size)
Given:
Population mean (μ) = 300
Standard deviation (σ) = 50
Sample size (n) = 36
Sample mean (x̄) = 320
Z-score = (320 - 300) / (50 / √36)
= 20 / (50 / 6)
= 20 * (6 / 50)
= 2.4
Next, we need to find the probability corresponding to the z-score of 2.4 using the z-table. Looking up the z-score in the table, we find that the probability is approximately 0.9918.
Therefore, the probability that the sample mean is at most 320 push-ups is approximately 0.9918, or 99.18%.