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 .
Basically, Benedict's test identifies the existence of aldehydes and alpha-hydroxy-ketones, also by hemiacetal, as well as those that take place in specific ketoses. Therefore, it is an alpha-hydroxy-ketone even if the ketose fructose is not strictly a reducing sugar, and provides a positive test since it is transformed into the mannose and aldoses glucose by the base inside the reagent.
