Respuesta :
Answer:
Quarter mile
The sample mean for this case is [tex] \bar X = 0.972[/tex]
[tex] s = 0.0585[/tex]
[tex]\hat{CV}= \frac{0.0585}{0.972}= 0.0602=6.02\%[/tex]
Mile
[tex] \bar X =4.536 [/tex]
[tex] s= 0.131[/tex]
[tex]\hat{CV}= \frac{0.131}{4.536}= 0.0289 =2.89\%[/tex]
We need to analyze the following statement:
After viewing this sample of running times, one of the coaches commented that the quarter milers turned in the more consistent times.
We se that we have more variation for the case of quarter miles since the Cv is higher than for the Milers case.
So then the conclusion is: No, the coefficient doesn't show that as a percentage of the mean the quarter milers' times show more variability.
Step-by-step explanation:
For this case we have the following data:
Quarter-Mile Times:0.93 0.99 1.06 0.91 0.97
Mile Times: 4.53 4.35 4.59 4.71 4.50
The mean for each case can be calculated with the following formula:
[tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
We can calculate the deviation with the following formula:
[tex] s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}[/tex]
The coeffcient of variation is calculated from the following formula:
[tex] \hat{CV}= \frac{s}{\bar X}[/tex]
Quarter mile
The sample mean for this case is [tex] \bar X = 0.972[/tex]
[tex] s = 0.0585[/tex]
[tex]\hat{CV}= \frac{0.0585}{0.972}= 0.0602=6.02\%[/tex]
Mile
[tex] \bar X =4.536 [/tex]
[tex] s= 0.131[/tex]
[tex]\hat{CV}= \frac{0.131}{4.536}= 0.0289 =2.89\%[/tex]
We need to analyze the following statement:
After viewing this sample of running times, one of the coaches commented that the quarter milers turned in the more consistent times.
We se that we have more variation for the case of quarter miles since the Cv is higher than for the Milers case.
So then the conclusion is :No, the coefficient doesn't show that as a percentage of the mean the quarter milers' times show more variability.
