Answers:
1) The first quartile (Q₁) = 11 ; 2) The median = 38.5 ;
3) The third quartile (Q₃) = 45 ;
4) The difference of the largest value and the median = 10.5 .
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Explanation:
Given this data set with 8 (eight) values: → {6, 47, 49, 15, 43, 41, 7, 36};
→Rewrite the values in increasing order; to help us find the median, first quartile (Q,) and third quartile (Q₃) : → {6, 7, 15, 36, 41, 43, 47, 49}.
→We want to find; or at least match; the following 4 (four) values [associated with the above data set] — 38.5, 11, 10, 45 ;
1) The first quartile (Q₁); 2) The median; 3) The third quartile (Q₃); &
4) The difference of the largest value and the median.
Note: Let us start by finding the "median". This will help us find the correct values for the descriptions in "Numbers 2 & 4" above.
The "median" would be the middle number within a data set, when the values are placed in smallest to largest (or, largest to smallest). However, our data set contains an EVEN number [specifically, "8" (eight)] values. In these cases , we take the 2 (two) numbers closest to the middle, and find the "mean" of those 2 (two) numbers; and that value obtained is the median. So, in our case, the 2 (two) numbers closest to the middle are:
"36 & 41". To get the "mean" of these 2 (two) numbers, we add them together to get the sum; and then, we divide that value by "2" (the number of values we are adding):
→ 36 + 41 = 77; → 77/2 = 38.5 ; → which is the median for our data set; and is a listed value.
→Now, examine Description "(#4): The difference of the largest value and the median"—(SEE ABOVE) ;
→ We can calculate this value. We examine the values within our data set to find the largest value, "49". Our calculated "median" for our dataset, "38.5". So, to find the difference, we subtract: 49 − 38.5 = 10.5 ; which is a given value".
→Now, we have 2 (two) remaining values, "11" & "45"; with only 2 (two) remaining "descriptions" to match;
→So basically we know that "11" would have to be the "first quartile (Q₁)"; & that "45" would have to be the "third quartile (Q₃)".
→Nonetheless, let us do the calculations anyway.
→Let us start with the "first quartile"; The "first quartile", also denoted as Q₁, is the median of the LOWER half of the data set (not including the median value)—which means that about 25% of the numbers in the data set lie below Q₁; & that about 75% lie above Q₁.).
→Given our data set: {6, 7, 15, 36, 41, 43, 47, 49};
We have a total of 8 (eight) values; an even number of values.
The values in the LOWEST range would be: 6, 7, 15, 36.
The values in the highest range would be: 41, 43, 47, 49.
Our calculated median is: 38.5 . →To find Q₁, we find the median of the numbers in the lower range. Since the last number of the first 4 (four) numbers in the lower range is "36"; and since "36" is LESS THAN the [calculated] median of the data set, "38.5" ; we shall include "36" as one of the numbers in the "lower range" when finding the "median" to calculate Q₁
→ So given the lower range of numbers in our data set: 6, 7, 15, 36 ;
We don't have a given "median", since we have an EVEN NUMBER of values. In this case, we calculate the MEDIAN of these 4 (four) values, by finding the "mean" of the 2 (two) numbers closest to the middle, which are "7 & 15". To find the mean of "7 & 15" ; we add them together to get a sum;
then we divide that sum by "2" (i.e. the number of values added up);
→ 7 + 15 = 22 ; → 22 ÷ 2 = 11 ; ↔ Q₁ = 11.
Now, let us calculate the third quartile; also known as "Q₃".
Q₃ is the median of the last half of the higher values in the set, not including the median itself. As explained above, we have a calculated median for our data set, of 38.5; since our data set contains an EVEN number of values. We now take the median of our higher set of values (which is Q₃). Since our higher set of values are an even number of values; we calculate the median of these 4 (four) values by taking the mean of the 2 (two) numbers closest to the center of the these 4 (four) values. This value is Q₃. →Given our higher set of values: 41, 43, 47, 49 ; → We calculate the "median" of these 4 (four) numbers; by taking the mean of the 2 (two) numbers in the middle; "43 & 47".
→ Method 1): List the integers from "43 to 47" ; → 43, 44, 45, 46, 47;
→ Since this is an ODD number of integers in sequential order;
→ "45" is not only the "median"; but also the "mean" of (43 & 47);
thus, 45 = Q₃;
→ Method 2): Our higher set of values: 41, 43, 47, 49 ;
→ We calculate the "median" of these 4 (four) numbers; by taking the
"mean" of the 2 (two) numbers in the middle; "43 & 47"; We don't have a given "median", since we have an EVEN NUMBER of values. In this case, we calculate the MEDIAN of these 4 (four) values, by finding the mean of the 2 (two) numbers closest to the middle, which are "43 & 47." To find the mean of "43 & 47"; we add them together to get a sum; then we divide that sum by "2" (i.e. the number of values added);
→ 43 + 47 = 90 ; → 90 ÷ 2 = 45 ; → 45 = Q₃ .
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- Martiinez <3
Little is known about how mutualistic interactions affect the distribution of species richness on broad geographic scales. Because mutualism positively affects the fitness of all species involved in the interaction, one hypothesis is that the richness of species involved should be positively correlated across their range, especially for obligate relationships. Alternatively, if mutualisms involve multiple mutualistic partners, the distribution of mutualists should not necessarily be related, and patterns in species distributions might be more strongly correlated with environmental factors. In this study, we compared the distributions of plants and vertebrate animals involved in seed‐dispersal mutualisms across the United States and Canada. We compiled geographic distributions of plants dispersed by frugivores and scatter‐hoarding animals, and compared their distribution of richness to the distribution in disperser richness. We found that the distribution of animal dispersers shows a negative relationship to the distribution of the plants that they disperse, and this is true whether the plants dispersed by frugivores or scatter‐hoarders are considered separately or combined. In fact, the mismatch in species richness between plants and the animals that disperse their seeds is dramatic, with plants species richness greatest in the in the eastern United States and the animal species richness greatest in the southwest United States. Environmental factors were corelated with the difference in the distribution of plants and their animal mutualists and likely are more important in the distribution of both plants and animals. This study is the first to describe the broad‐scale distribution of seed‐dispersing vertebrates and compare the distributions to the plants they disperse. With these data, we can now identify locations that warrant further study to understand the factors that influence the distribution of the plants and animals involved in these mutualisms.
Introduction
A central problem in ecology is to understand the patterns and processes shaping the distribution of species. There is a preponderance of studies of species richness at broad geographic scales (Hawkins et al. 2003, Rahbek et al. 2007, Stein et al. 2014, Rabosky and Hurlbert 2015) that has facilitated our understanding of why species are found where they are, a central tenet within the domain of ecology (Scheiner and Willig 2008). Most commonly, these studies find species distributions to be correlated with resource availability and use environmental variables (e.g. temperature and productivity; Rabosky and Hurlbert 2015) to explain putative determinants of the distributions. Environmental variables are only one determinant of species’ distributions. Another, species interaction, is a key and understudied determinant of species’ distributions (Cazelles et al. 2016). In fact, in some cases species interactions may be more important for determining distribution than environmental variables (Fleming 2005).
When species interact, we expect their geographic distributions to be correlated – either positively or negatively – depending on the effect (or sign of the interaction) of one species on the other (Case et al. 2005). For pairwise interactions, where one species benefits from another species, a positive relationship is expected between the distribution and abundance due to the increase in the average fitness of the benefitting species where they overlap (Svenning et al. 2014). Furthermore, most species interactions are not simply pairwise, but diffuse, consisting of multiple interacting species, here referred to as guilds (with guilds referring to species that use the same resource). It therefore follows that where one guild benefits from another guild, a positive relationship is expected between the distribution and richness of the guids. This should be true in the case of mutualisms, where both sides of the interaction share an increase in average fitness from being together (Bronstein 2015), and there is some evidence for correlated geographic distributions of mutualists in the New World (Fleming 2005). One example of a mutualism where both sides of the interaction have a fitness advantage in each other's presence is animal‐mediated seed dispersal. Because both interacting species and guilds in seed dispersal mutualism benefit from the relationship we would predict that the richness of animal‐dispersed plants ought to be correlated with the richness of their animal dispersers and vice versa. To our knowledge, this prediction has never been tested on a large geographic scale.
