Mathematics, 24.05.2020 01:05 2022rickskyye

The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below. Determine which of the following statements is true about Class B.

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

Class B has a smaller median and the same interquartile range.

Class B has a larger median and the same interquartile range.

Class B has a larger median and a larger interquartile range.

Class B has a smaller median and a larger interquartile range.

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Answer from: lnorred2

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 = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 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 = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 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

= Q_3-Q_1

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= 21+ 0.5[22- 21]

= 21.5

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= 34+ 0.5[35- 34]

= 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

= Q_3-Q_1

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= 18+ 0.5[20- 18]

= 19

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= 30+ 0.5[34- 30]

= 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.

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