Question

At the Olympic games, many events have several rounds of competition. One of these events is the men's 100100100-meter backstroke.


The upper dot plot shows the times (in seconds) of the top 888 finishers in the final round of the 201220122012 Olympics. The lower dot plot shows the times of the same 888 swimmers, but in the semifinal round.


Which pieces of information can be gathered from these dot plots? (Remember that lower swim times are faster.)

Answer 1

Information is the output from processing, structuring or organizing data

The correct option for the information that can be gathered is the option:

The times in the final vary noticeably more than the times in the semifinals

The reason for the above selection is as follows:

Question:

The data in the dot plot can be presented as follows;

• Semifinal Round

$\begin{array}{|l|c|c|c|c|}\mathbf{Time, \, (t)} &\mathbf{Number \ of \ swimmers, \ f(x)}& \mathbf{ t \times f(t)}& \mathbf{ t^2} &\mathbf{t^2\times f(t)} \\\rule {50}{0.5}&\rule {150}{0.5}&\rule {35}{0.5}&\rule {35}{0.5}&\rule {35}{0.5}\\52.7&1& 52.7&52.7^2&2,777.29\\53&1&53&53^2&2,809\\53.3&2&106.6&53.3^2&2,840.89\\53.5&3&160.5&53.5^2&2,862.25\\53.7&1&53.7&53.7^2&2,883.69\end{array}$

∑(t × f(t)) = 426.5, ∑f(t) = 8

$Average = Mean = \mathbf{ \dfrac{\sum \left(t \times f(t)\right)}{\sum f(t)} }$

Therefore;

$\mathbf{Average \ time} \ at \ the \ \mathbf{semifinal }\ round =\mathbf{ \dfrac{426.5}{8} } = 53.3125$

∑(t² × f(t)) = 22,738.51, ∑f(t) = 8

$Sample \ variance \ s^2 = \mathbf{\dfrac{\sum x^2 \cdot f(t) - \dfrac{\left(\sum x \cdot f(t) \right)^2}{\sum f(t)} }{\sum f(t) - 1}}$

$s^2 = \mathbf{\dfrac{22,738.51 - \dfrac{\left(462.5 \right)^2}{8} }{8 - 1}} \approx 0.10411$

The variance semifinal round data, s² ≈ 0.10411

• Final Round

$\begin{array}{|l|c|c|c|c|}\mathbf{Time, \, (t)} &\mathbf{Number \ of \ swimmers, \ f(x)}& \mathbf{ t \times f(t)}& \mathbf{ t^2} &\mathbf{t^2\times f(t)} \\\rule {50}{0.5}&\rule {150}{0.5}&\rule {35}{0.5}&\rule {35}{0.5}&\rule {35}{0.5}\\52.2&1&52.2&52.2^2&2,724.84\\52.9&1&52.9&52.9^2&2,798.41\\53&1&53&53^2&2,809\\53.1&1&53.1&53.1^2&2,819.61\\53.4&1&53.4&53.4^2&2,851.61\\53.5&1&53.5&53.5^2&2,862.25\\53.6&1&53.6&53.6^2&2,872.96\\53.8&1&53.8&53.8^2&2,894.44\end{array}$

∑(t × f(t)) = 425.5, ∑f(t) = 8

Therefore;

$\mathbf{Average \ time} \ at \ the \ \mathbf{final} \ round=\mathbf{ \dfrac{425.5}{8} } = 53.1875$

∑(t² × f(t)) = 22,633.07, ∑f(t) = 8

$s^2 = \mathbf{\dfrac{22,633.07 - \dfrac{\left(462.5 \right)^2}{8} }{8 - 1}} \approx 0.26$

The variance of the final round data, s² ≈ 0.26

• Comparison of the data in the semifinal and final rounds;

Average semifinal round = 426.5, average semifinal round = 425.5

∴ Average semifinal round ≈ Average semifinal round

Variance semifinal round , s² ≈ 0.10411; Variance final round , s² ≈ 0.26

∴ Variance final round > Variance semifinal round

Therefore, the information that can be gathered from the dot plot is that the variance of the final is more than double the variance at the semifinal.

The correct option is; The times in the final vary noticeably more than the times in the semifinals

Learn more about dot plots here:

https://brainly.com/question/17170054

https://brainly.com/question/17095999

https://brainly.com/question/17142810