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At the Olympic games, many events have several rounds of competition. One of these
events is the men's 100-meter backstroke.
The upper dot plot shows the times (in seconds) of the top 8 finishers in the final round of
the 2012 Olympics. The lower dot plot shows the times of the same 8 swimmers, but in the
semifinal round.
Final round
Semifinal round
distribution.
56
56.5
The center of the semifinal round distribution is
distribution.
57
Time (seconds)
The variability in the semifinal round distribution is
V
57.5
v
58
the center of the final round
the variability in the final round.

At the Olympic games many events have several rounds of competition One of these events is the mens 100meter backstroke The upper dot plot shows the times in se class=

Respuesta :

The 100–meter backstroke swimming times of the 8 swimmers during the semifinal and the finals of the Olympic games gives;

First part;

The center of the semifinal round is approximately 57.3625

The variability in the semifinal round is 0.0998

Second part;

The center of the final round is approximately 57.0125

The variability in the final round is 0.3336

How can the center and variability of the data in the dot plot be found?

The times of the swimmers in the semifinal round are;

56.7, 57.2, 57.3, 57.3, 57.4, 57.5, 57.7, 57.8

The central value is given by the mean of the data, is found as follows;

(56.7+57.2+57.3+57.3+57.4+57.5+57.7+57.8)/8 = 57.3625

The variability is given by the standard deviation which is found by the formula;

[tex] \sigma = \sqrt{ \frac{ x_{i} - \mu}{n} } [/tex]

Where;

[tex]x _{i} = the \: ith \: value[/tex]

[tex] \mu = the \: mean[/tex]

The standard deviation found using the given values above and an online tool is presented as follows;

[tex] \sigma = \sqrt{ \frac{ x_{i} - 57.3625}{8} \approx 0.316 } [/tex]

  • The variance ≈ √(0.316) ≈ 0.0998

Second part;

In the final, we have the following values from the dot plot;

56, 56.6, 56.6, 57, 57.1, 57.2, 57.7, 57.9

From the above values, using an online calculator, the center, which is taken as the mean is 57.0125

The variance for the final, [tex] \sigma^2 [/tex] ≈ 0.3336

Learn more about finding the variance of a set of data here:

https://brainly.com/question/25639778

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