1 Define the following terminology used in data handling. 1.1.1 Raw data. 1.1.2 Range. 1.2 The ages of people attending a gymnasium to stay fit are represented as follows: ... 1.3 The points scored by a basketball player are represented as follows: ...

The sorted data set is:
18, 19, 20, 23, 25, 25, 25, 29, 32, 32, 33, 34, 34, 35, 36, 37, 40, 42, 54, 55
There are $n = 20$ data points.

• 1.2.1 Write down the mode. The mode is the value that appears most frequently in the data set. In the sorted data, 25 appears 3 times, which is more than any other value. The mode is $\boxed{\text{25}}$.

• 1.2.2 Determine the median. Since there are $n=20$ (an even number) data points, the median is the average of the two middle values: the $(\frac{n}{2})^{\text{th}}$ and $(\frac{n}{2}+1)^{\text{th}}$ values. The 10th value is 32. The 11th value is 33.

  $$\text{Median} = \frac{32 + 33}{2} = \frac{65}{2} = \boxed{\text{32.5}}$$

• 1.2.3 Calculate the mean. The mean is the sum of all values divided by the number of values.

  $$\text{Sum} = 18+19+20+23+25+25+25+29+32+32+33+34+34+35+36+37+40+42+54+55 = 652$$ $$\text{Mean} = \frac{\text{Sum}}{n} = \frac{652}{20} = \boxed{\text{32.6}}$$

• 1.2.4 Construct a stem-and-leaf diagram. The stem represents the tens digit, and the leaf represents the units digit.

  Stem | Leaf
  -----|-----------------
  1    | 8, 9
  2    | 0, 3, 5, 5, 5, 9
  3    | 2, 2, 3, 4, 4, 5, 6, 7
  4    | 0, 2
  5    | 4, 5
  Key: 1 | 8 = 18
  

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The sorted data set is:
2, 3, 3, 6, 10, 11, 11, 18, 26, 29, 32, 33, 35, 42, 42, 47, 49, 49
There are $n = 18$ data points.

• 1.3.1 Find the range. The range is the difference between the maximum and minimum values.

  $$\text{Range} = \text{Maximum value} - \text{Minimum value} = 49 - 2 = \boxed{\text{47}}$$

• 1.3.2 Arrange the points in ascending order. The points arranged in ascending order are: $\boxed{\text{2, 3, 3, 6, 10, 11, 11, 18, 26, 29, 32, 33, 35, 42, 42, 47, 49, 49}}$

• 1.3.3 Determine the lower quartile (Q1). The median of the entire data set is between the 9th and 10th values (26 and 29). The lower half of the data consists of the first 9 values: 2, 3, 3, 6, 10, 11, 11, 18, 26. Q1 is the median of this lower half, which is the 5th value.

  $$Q_1 = \boxed{\text{10}}$$

• 1.3.4 Determine the upper quartile (Q3). The upper half of the data consists of the last 9 values: 29, 32, 33, 35, 42, 42, 47, 49, 49. Q3 is the median of this upper half, which is the 5th value in this half (the 14th value overall).

  $$Q_3 = \boxed{\text{42}}$$

• 1.3.5 Find the interquartile range. The interquartile range (IQR) is the difference between the upper and lower quartiles.

  $$\text{IQR} = Q_3 - Q_1 = 42 - 10 = \boxed{\text{32}}$$

• 1.3.6 Determine the semi inter-quartile range. The semi inter-quartile range is half of the interquartile range.

  $$\text{Semi-IQR} = \frac{\text{IQR}}{2} = \frac{32}{2} = \boxed{\text{16}}$$

• 1.3.7 Determine the 40th percentile. The position of the $k^{\text{th}}$ percentile is given by $P_k = \frac{k}{100}(n+1)$. For the 40th percentile ($k=40$) with $n=18$:

  $$\text{Position} = \frac{40}{100}(18+1) = 0.4 \times 19 = 7.6$$

  The 40th percentile is between the 7th and 8th values. 7th value ($x_7$) = 11 8th value ($x_8$) = 18

  $$P_{40} = x_7 + 0.6(x_8 - x_7) = 11 + 0.6(18 - 11) = 11 + 0.6(7) = 11 + 4.2 = \boxed{\text{15.2}}$$