The proof TriangleABC ≅ TriangleDCB that is shown. Given: AngleA ≅ AngleD; CD||AB Prove: TriangleABC ≅ TriangleDCB Triangles C D
2 answers:
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Answer:
c. The <em>median</em> and the IQR <em>(Interquartile range)</em>.
Step-by-step explanation:
Considering the histogram for the distribution of times for this experiment (see the graph below), we can notice that this distribution is skewed positively because of "<em>a few scores creating an elongated tail at the higher end of the distribution</em>" (Urdan, 2005). There is also a probable outlier (an animal that navigates around nine minutes). An outlier is an extreme value that is more than two standard deviations below or above the mean.
In these cases, when we have <em>skewed and extreme</em> <em>values</em> in a distribution, it is better to avoid using the <em>mean and standard deviations</em> as measures of central <em>tendency</em> and <em>dispersion</em>, respectively. Instead, we can use the <em>median</em> and the <em>interquartile range</em> for those measures.
With skewed distributions, the mean is more "sensible" to extreme data than the median, that is, it tends to not represent the most appropriate measure for central position <em>(central tendency)</em> in a distribution since in a positively skewed distribution like the one of the question, the mean is greater than the <em>median</em>, that is, <em>the extreme values tend to pull the mean to them</em>, so the mean tends to not represent a "reliable" measure for the central tendency of all the values of the experiments.
We have to remember that the dispersion measures such as the <em>standard deviation</em> and <em>interquartile range</em>, as well as the <em>variance</em> and <em>range</em>, provide us of measures that tell us <em>how spread the values are in a distribution</em>.
Because the <em>standard deviation depends upon the mean</em>, i.e., to calculate it we need to subtract each value or score from the mean, square the result, divide it by the total number of scores and finally take the square root for it, we have to conclude that with an inappropriate mean, <em>the standard deviation is not a good measure for the dispersion of the data, </em>in this case (a positively skewed distribution).
Since the median represents a central tendency measure in which 50% of the values for distribution falls below and above this value, no matter if the distribution is skewed, the median is the best measure to describe the center of the distribution in this case.
Likewise, the <em>quartiles</em> are not affected by <em>skewness</em>, since they represent values of the distribution for which there is a percentage of data below and above it. For example, the first quartile (which is also the 25th percentile) splits the lowest 25% of the data from the highest 75% of them, and the third quartile, the highest 25%, and the lowest 75%. In other words, those values do not change, no matter the extreme values or skewness.
For these reasons, we can say that the median and the interquartile range (IQR) describe the center and the spread for the distribution presented, and not the most usual measures such as the mean and standard deviation.
