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 .
Answer:
Visible Light Absorption
Atoms and molecules contain electrons. It is often useful to think of these electrons as being attached to the atoms by springs. The electrons and their attached springs have a tendency to vibrate at specific frequencies. Similar to a tuning fork or even a musical instrument, the electrons of atoms have a natural frequency at which they tend to vibrate. When a light wave with that same natural frequency impinges upon an atom, then the electrons of that atom will be set into vibrational motion. (This is merely another example of the resonance principle introduced in Unit 11 of The Physics Classroom Tutorial.) If a light wave of a given frequency strikes a material with electrons having the same vibrational frequencies, then those electrons will absorb the energy of the light wave and transform it into vibrational motion. During its vibration, the electrons interact with neighboring atoms in such a manner as to convert its vibrational energy into thermal energy. Subsequently, the light wave with that given frequency is absorbed by the object, never again to be released in the form of light. So the selective absorption of light by a particular material occurs because the selected frequency of the light wave matches the frequency at which electrons in the atoms of that material vibrate. Since different atoms and molecules have different natural frequencies of vibration, they will selectively absorb different frequencies of visible light.
