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

General
1

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 ?

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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 ?

Answer:

To determine the distribution of x - y, we need to use the rules for combining means and standard deviations. Since x represents the running time before training and y represents the running time after training, their difference (x - y) represents the change in running time due to the training program.

The mean of x - y can be found by subtracting the mean of y from the mean of x:

mean of x - y = mean of x - mean of y = 402 - 368 = 34 seconds

The standard deviation of x - y can be found using the formula:

  • standard deviation of x - y = sqrt((standard deviation of x)^2 + (standard deviation of y)^2)

  • standard deviation of x - y = sqrt((40)^2 + (30)^2) = 50 seconds

So, the distribution of x - y has a mean of 34 seconds and a standard deviation of 50 seconds. We can conclude that the distribution is normal because it is the difference between two normally distributed variables.

Therefore, we can say that x - y follows a normal distribution with mean 34 seconds and standard deviation 50 seconds.