Answer :
Answer:
The true statement about Class B is that Class B has a smaller median and the same inter quartile range.
Step-by-step explanation:
We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;
Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40
Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42
1) Firstly, we will calculate Median for Class A;
For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;
- If n is odd, then the formula for calculating median is given by;
Median = [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]
- If n is even, then the formula for calculating median is given by;
Median = [tex]\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}[/tex]
Here, number of observation is odd, i.e. n = 17.
So, Median = [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]
= [tex](\frac{17+1}{2})^{th} \text{ obs.}[/tex]
= [tex](\frac{18}{2})^{th} \text{ obs.}[/tex]
= [tex]9^{th} \text { obs.}[/tex] = 27
Hence, the median of class A is 27.
2) Now, we will calculate Median for Class B;
For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;
- If n is odd, then the formula for calculating median is given by;
Median = [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]
- If n is even, then the formula for calculating median is given by;
Median = [tex]\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}[/tex]
Here, number of observation is odd, i.e. n = 17.
So, Median = [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]
= [tex](\frac{17+1}{2})^{th} \text{ obs.}[/tex]
= [tex](\frac{18}{2})^{th} \text{ obs.}[/tex]
= [tex]9^{th} \text { obs.}[/tex] = 26
Hence, the median of class B is 26.
3) Now, we will calculate the Inter quartile range for Class A;
Inter quartile range = Upper quartile - Lower quartile
= [tex]Q_3-Q_1[/tex]
SO, [tex]Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}[/tex]
= [tex](\frac{17+1}{4})^{th} \text{ obs.}[/tex]
= [tex](\frac{18}{4})^{th} \text{ obs.}[/tex]
= [tex]4.5^{th} \text{ obs.}[/tex]
= [tex]4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}][/tex]
= [tex]21+ 0.5[22- 21][/tex]
= 21.5
Similarly, [tex]Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}[/tex]
= [tex]3(\frac{17+1}{4})^{th} \text{ obs.}[/tex]
= [tex](\frac{54}{4})^{th} \text{ obs.}[/tex]
= [tex]13.5^{th} \text{ obs.}[/tex]
= [tex]13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}][/tex]
= [tex]34+ 0.5[35- 34][/tex]
= 34.5
Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.
4) Now, we will calculate the Inter quartile range for Class B;
Inter quartile range = Upper quartile - Lower quartile
= [tex]Q_3-Q_1[/tex]
SO, [tex]Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}[/tex]
= [tex](\frac{17+1}{4})^{th} \text{ obs.}[/tex]
= [tex](\frac{18}{4})^{th} \text{ obs.}[/tex]
= [tex]4.5^{th} \text{ obs.}[/tex]
= [tex]4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}][/tex]
= [tex]18+ 0.5[20- 18][/tex]
= 19
Similarly, [tex]Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}[/tex]
= [tex]3(\frac{17+1}{4})^{th} \text{ obs.}[/tex]
= [tex](\frac{54}{4})^{th} \text{ obs.}[/tex]
= [tex]13.5^{th} \text{ obs.}[/tex]
= [tex]13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}][/tex]
= [tex]30+ 0.5[34- 30][/tex]
= 32
Therefore, Inter quartile range for Class B = 32 - 19 = 13.
Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.
Answer:
the person above me is correct
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Step-by-step explanation: If u made it this far the nur sure to ace ur next test....