Solution
Minimum:
- The minimum is gotten from the value with a stem of 1. The smallest leaf on this stem is 4.
- Thus, the minimum is 14
Q1:
- The first quartile position is gotten using the formula below:
![Q_i=(i(n+1)^(th))/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/xpqk89g2qofekm9qe9j3bswzqup8c4u32r.png)
- Thus, the quartile position is:
![\begin{gathered} n=18 \\ Q_1=((18+1))/(4)^(th)=4.75^(th) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u8e9ox4mrea0ui337p4rxpj6niigc3plsy.png)
- Thus, the first quartile is between the 4th and 5th position. This is between 23 and 27.
- To find the correct first quartile, we use the formula below:
![\begin{gathered} 0.75*(27-23)+23 \\ =26 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8yljep5bc4ayebg5vcl7mczzbj37kcpf07.png)
- Thus, Q1 = 26
Median
- The median is gotten below:
![Q_2\text{ or }Median\text{ }position=((n+1))/(2)^(th)=(18+1)^(th))/(2)=9.5^{th\text{ }}position](https://img.qammunity.org/2023/formulas/mathematics/college/f1mnlc2oofln0dm6ne9f919lgetqmqb15o.png)
- Thus, the Median is between 41 and 44.
- The median is simply the average of both numbers done below:
![Median=(41+44)/(2)=42.5](https://img.qammunity.org/2023/formulas/mathematics/college/palmic9zccb2k0mimi0zhl2d54hgjv71qi.png)
Q3:
- This is the same process as finding Q1. This is done below:
![Q_3\text{ position}=(3(18+1)^(th))/(4)=14.25^(th)](https://img.qammunity.org/2023/formulas/mathematics/college/1albcds99681lw2re94gkgoa5jbkrrnb3b.png)
- The Q3 is between 54 and 54.
- Thus, the Q3 is 54 since the Q3 is between the same number
Maximum:
- The maximum value is gotten from the stem with the largest value. The leaf of this stem will be the maximum value of the entire dataset.
- Thus the maximum is 63