Answer:
Step-by-step explanation:
Hello!
The variable of interest is "the time a fight lasts" measured in seconds
The five number summary for her first 12 UFC fights are:
Minimum: Min= 14
First Quartile: Q₁= 25
Second Quartile/Median: Me= 44
Third Quartile: Q₃= 64
Maximum: Max= 289
a)
Three of her fights lasted more than one minute: 289, 267, 66. Out of these three fights, you have to determine if they are outliers using the IQR method.
The IQR is the distance between the third quartile and the first quartile
IQR= Q₃ - Q₁= 64 - 25= 39
Remember, an outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.
Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:
X < Q₁ - 1.5 IQR
X > Q₃ + 1.5 IQR
Or extreme outliers if:
X < Q₁ - 3 IQR
X > Q₃ + 3 IQR
The limits that define if a value is an outlier or not are:
X < 25 - 1.5*39 = -33.5
X > 64 + 1.5*39= 122.5
And for extreme outliers:
X < 25 - 3*39 = -92
X > 64 + 3*39= 181
So fights that lasted less than -33.5 or more than 122.5 seconds are to be considered outliers, and those who lasted less than -92 or more than 181 seconds are extreme outliers.
Out of the three fights, the ones that lasted 289 and 267 seconds can be considered extreme values.
b) The minimum observed value for this data set is 14 seconds, values are considered to be outliers if they are less to -33.5, so there is no low outliers on the sample.
As you can see, both values that allow you to determine if the observation is an outlier or not are negative, since the variable is "the time a fight lasts" it is impossible for it to have negative values. The lowest value of time a fight can last is "zero seconds". Although mathematically correct, these values make no sense.
c)
See attachment.
d)
The average or mean is a measurement of central tendency that shows you the value around which most of the distribution will be. It is very affected by the presence of outliers, especially extreme ones. Outliers make the mean "move" towards them, this means, that if there are small outliers, then the mean will move to the lower side of the distribution. But if the outliers are big, then the mean will move to the higher side of the distribution.
For example, let's say 5 of the fight times were:
10, 37, 49, 51, 68
For these values the mean would be:
X[bar]₀= ∑X/n= (10+37+49+51+68)/5= 215/5= 43
Now let's change one of these values for an extreme one:
Small value:
0, 37, 49, 51, 68
X[bar]₁= ∑X/n= (0+37+49+51+68)/5= 205/5= 41
⇒ As you can see, one change for a smaller value reduces the mean value X[bar]₁ < X[bar]₀
Big value:
10, 37, 49, 51, 268
X[bar]₂= ∑X/n= (10+37+49+51+268)/5= 415/5= 83
⇒ In this case, changing one of the 5 values to a bigger one moved the mean to the right: X[bar]₀ < X[bar]₂
So for the given distribution, since there are at least two high outliers, you'd expect the mean to be greater than the median.
I hope this helps!
