Question

Six data points, not in ascending order, are represented as follows:
2𝑥 2𝑥 + 1 𝑥 + 3 6𝑥 𝑥 + 3 3𝑥 + 2
Table 3.1: Data points.
3.1.1 Calculate the value of 𝑥 if the mean is 9. (3)
3.1.2 Determine the interquartile range. (3)
3.2 The figures in the table below are the marks of 30 grade 11 learners
for the Mathematics examination:
52 88 85 75 72 42 58 56 33 45
36 89 42 55 66 63 68 69 35 89
52 72 72 66 55 56 89 75 72 32
Table 3.2: Marks of 30 learners.
3.2.1 Complete the frequency table. You can redraw the table or
print Annexure A and then complete it.
Age intervals Tally Frequency
30 ≤ 𝑥 < 40
40 ≤ 𝑥 < 50
50 ≤ 𝑥 < 60
60 ≤ 𝑥 < 70
70 ≤ 𝑥 < 80
80 ≤ 𝑥 < 90

Answer 1

To solve the problems you've presented, let's break them down into the respective sections.

### 3.1 Data Points

The data points are:
\[ 2x, \, 2x + 1, \, x + 3, \, 6x, \, x + 3, \, 3x + 2 \]

#### 3.1.1 Calculate the value of \( x \) if the mean is 9.

The formula for the mean (average) is given by:

\[
\text{Mean} = \frac{\text{Sum of all data points}}{\text{Number of data points}}
\]

1. First, we calculate the sum of the data points:
\[
S = 2x + (2x + 1) + (x + 3) + 6x + (x + 3) + (3x + 2)
\]
Simplifying this:
\[
S = 2x + 2x + 1 + x + 3 + 6x + x + 3 + 3x + 2 = (2x + 2x + x + 6x + x + 3x) + (1 + 3 + 3 + 2)
\]
\[
= (15x) + (9) = 15x + 9
\]

2. The number of data points is 6, so the equation for the mean becomes:
\[
\frac{15x + 9}{6} = 9
\]

3. To find \( x \), multiply both sides by 6:
\[
15x + 9 = 54
\]
Subtract 9 from both sides:
\[
15x = 45
\]
Divide by 15:
\[
x = 3
\]

#### 3.1.2 Determine the interquartile range (IQR).

Now that we have \( x = 3 \), plug it back into the data points:
- \( 2x = 6 \)
- \( 2x + 1 = 7 \)
- \( x + 3 = 6 \)
- \( 6x = 18 \)
- \( x + 3 = 6 \)
- \( 3x + 2 = 11 \)

So, the data points are:
\[ 6, 7, 6, 18, 6, 11 \]
Arranging in ascending order gives:
\[ 6, 6, 6, 7, 11, 18 \]

To find the IQR:
1. The first quartile \( Q_1 \) (the median of the first half) is:
- Data: \( 6, 6, 6 \) → median = 6.

2. The third quartile \( Q_3 \) (the median of the second half) is:
- Data: \( 7, 11, 18 \) → median = 11.

3. The interquartile range (IQR):
\[
\text{IQR} = Q_3 - Q_1 = 11 - 6 = 5
\]

### 3.2 Marks of 30 Learners

From the marks provided:
\[
52, 88, 85, 75, 72, 42, 58, 56, 33, 45,
36, 89, 42, 55, 66, 63, 68, 69, 35, 89,
52, 72, 72, 66, 55, 56, 89, 75, 72, 32
\]

#### 3.2.1 Frequency Table

| Age intervals | Tally | Frequency |
|----------------|------------------|-----------|
| \(30 \leq x < 40\) | |||| | 4 |
| \(40 \leq x < 50\) | |||| | 6 |
| \(50 \leq x < 60\) | |||| | 9 |
| \(60 \leq x < 70\) | |||| | 7 |
| \(70 \leq x < 80\) | |||| | 4 |
| \(80 \leq x < 90\) | |||| | 4 |

### Final Frequency Counts:
- \(30 \leq x < 40\) (Marks: 33, 36, 32, 35) → 4
- \(40 \leq x < 50\) (Marks: 42, 42, 45, 42, 36) → 6
- \(50 \leq x < 60\) (Marks: 52, 52, 56, 58, 55, 56, 55) → 9
- \(60 \leq x < 70\) (Marks: 66, 63, 68, 69, 66) → 7
- \(70 \leq x < 80\) (Marks: 75, 72, 72, 75) → 4
- \(80 \leq x < 90\) (Marks: 88, 85, 89, 89, 89) → 4

The table provides a clear summary of the frequency of marks within specified intervals based on the learners' performances.