Question

High school students from track teams in the state participated in a training program to improve running times. Before the training, the mean running time for the students to run a mile was 402 seconds with standard deviation 40 seconds. After completing the program, the mean running time for the students to run a mile was 368 seconds with standard deviation 30 seconds. Let X represent the running time of a randomly selected student before training, and let Y represent the running time of the same student after training. Which of the following is true about the distribution of X-Y?a. The variables X and Y are independent, therefore, the meanis 34 seconds and the standard deviation is 10 seconds.b. The v ales X and Y are independent therefore, the meanis 34 seconds and the standard deviation is 50 secondsc. The variables X and Y are not independent, therefore, the standard deviation is 50 seconds and the mean cannot be determined with the information given.d. The variables and are not independent, therefore, the meanis 3 seconds and the standard deviation cannot be determined with the information givene. The variables X and Y We not independent, therefore, neither the mean nor the standard deviation can be determined with the informantion given.

Answer 1

Answer:

b. The values X and Y are independent therefore, the mean is 34 seconds and the standard deviation is 50 seconds

Step-by-step explanation:

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Before the training, the mean running time for the students to run a mile was 402 seconds with standard deviation 40 seconds.

This means that $\mu_X = 402, \sigma_X = 40$

After completing the program, the mean running time for the students to run a mile was 368 seconds with standard deviation 30 seconds.

This means that $\mu_Y = 368, \sigma_Y = 30$

Which of the following is true about the distribution of X-Y?

They are independent, so:

$\mu = \mu_X - \mu_Y = 402 - 368 = 34$

$\sigma = \sqrt{\sigma_X^2+\sigma_Y^2} = \sqrt{40^2+30^2} = 50$

This means that the correct answer is given by option b.

Answer 2

The values X and Y are independent therefore, the mean is 34 seconds and the standard deviation is 50 seconds.

What is the subtraction between normal variables?

The two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Before the training, the mean running mean  time for the students to runing a mile was 402 seconds with standard deviation 40 seconds.

That is the $\mu_x=402 , \sigma_x=40$

That is the after completing the program, the mean running time for the students to run a mile was 368 seconds with standard deviation 30 seconds.

That is $\mu_y=368,\sigma_y=30$

which of the following is true about the distribution of X-Y?

They are independent

Therefore we get,

$\mu=\mu_x-\mu_y=402-368=34$

$\sigma=\sqrt{\sigma_x^2-\sigma_y^2}\\\sigma=\sqrt{40^2-30^2}\\\sigma =50$

Therefore the option b is correct.

To learn more about the distribution visit:

https://brainly.com/question/24756209

This means that the correct answer is given by option b.