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3 years ago

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The following times were recorded by the quarter-mile and mile runners of a university track team (times are in minutes).Quarter

-Mile Times: 0.93 0.99 1.06 0.91 0.97Mile Times: 4.53 4.35 4.59 4.71 4.50After viewing this sample of running times, one of the coaches commented that the quarter milers turned in the more consistent times. Use the standard deviation and the coefficient of variation to summarize the variability in the data.

1 answer:

grigory [225]3 years ago

3 0

Answer:

Quarter mile

The sample mean for this case is $\bar X = 0.972$

$s = 0.0585$

$\hat{CV}= \frac{0.0585}{0.972}= 0.0602=6.02\%$

Mile

$\bar X =4.536$

$s= 0.131$

$\hat{CV}= \frac{0.131}{4.536}= 0.0289 =2.89\%$

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:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

We can calculate the deviation with the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

The coeffcient of variation is calculated from the following formula:

$\hat{CV}= \frac{s}{\bar X}$

Quarter mile

The sample mean for this case is $\bar X = 0.972$

$s = 0.0585$

$\hat{CV}= \frac{0.0585}{0.972}= 0.0602=6.02\%$

Mile

$\bar X =4.536$

$s= 0.131$

$\hat{CV}= \frac{0.131}{4.536}= 0.0289 =2.89\%$

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.

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t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text

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a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction

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v_{0x}=2.5\,\dfrac{\text m}{\text s}v

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m

v, start subscript, 0, x, end subscript, equals, 2, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v

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y

v, start subscript, y, end subscript directly. Since both the yyy and xxx directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:

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x

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Once we know ttt, we can solve for v_yv

y

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y

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tv, start subscript, y, end subscript, equals, v, start subscript, 0, y, end subscript, plus, a, start subscript, y, end subscript, t

Hint #22 / 4

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