Answered

Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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 seconds
c. 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 given
e. The variables X and Y We not independent, therefore, neither the mean nor the standard deviation can be determined with the informantion given.

Sagot :

Answer:

D

Step-by-step explanation:

Using subtraction of normal variables, it is found that the correct option is:

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

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.

In this problem:

  • Variable X has mean 402 seconds, standard deviation of 40 seconds.
  • Variable Y has mean 368 seconds, standard deviation of 30 seconds.

They are independent, hence:

[tex]\mu = 402 - 368 = 34[/tex]

[tex]\sigma = \sqrt{40^2 + 30^2} = 50[/tex]

Option b is correct.

A similar problem is given at https://brainly.com/question/24250158

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.