4.2 The data below represent the percentage marks achieved by Grade 12 learners inLift Sciences. The class consists of 20 learners. $$ \begin{array}{cccccccccc} 64 & 57 & 58 & 62 & 59 & \mathbf{A} & 60 & 61 & 62 & 70 \ 62 & 65 & 66 & 64 & 75 & 80 & \mathbf{B} & \mathbf{B} & 92 & 85 \end{array} $$ A is the lowest percentage mark Use the information above to answer questions that follow: 4.2.1 Determine the percentage of data values that lies between the lower quartile and the upper quartile. (2) 4.2.2 Calculate the value of A , if the range of the marks obtained by the learners is 36 . (3) (2) 4.2.3 Write down the modal percentage. 4.2.4 Determine the value of $$ B $$, given that the mean mark obtained by the learners is 68 . 4.2.5 Calculate the median percentage mark. (3) 4.2.6 Determine the probability (as common fraction in simplified form) of randomly selecting a learner who achieved more than $$ 80 \% $$. (3) 4.2.7 Complete frequency table and the histogram on the ANSWER SHEET provided

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**4.2.1**  
By definition the lower quartile ($$Q_1$$) is the 25th percentile and the upper quartile ($$Q_3$$) is the 75th percentile. Thus the data values between $$Q_1$$ and $$Q_3$$ represent  
$$
75\%-25\%=50\%
$$  
of the data.

---

**4.2.2**  
The range is given by  
$$
\text{Range} = \text{Highest mark} - \text{Lowest mark}
$$  
The highest mark in the dataset is $$92$$ and we are told that $$A$$ is the lowest mark. We are also told that the range is $$36$$. Therefore,  
$$
92 - A = 36  
$$
Solving for $$A$$:  
$$
A = 92 - 36 = 56
$$

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**4.2.3**  
The mode is the percentage that occurs most frequently. From the given data the mark $$62$$ appears three times (more than any other mark), so the modal percentage is  
$$
62\%
$$

---

**4.2.4**  
We are told that the mean mark is $$68\%$$ for $$20$$ learners, so the total sum of marks is  
$$
20 \times 68 = 1360
$$  
The data (with $$A=56$$) are as follows:

Row 1:  
$$
64,\; 57,\; 58,\; 62,\; 59,\; 56,\; 60,\; 61,\; 62,\; 70
$$

Row 2:  
$$
62,\; 65,\; 66,\; 64,\; 75,\; 80,\; B,\; B,\; 92,\; 85
$$

First, compute the sum of the known marks.

*Row 1 Sum:*  
$$
64+57+58+62+59+56+60+61+62+70=609 
$$

*Row 2 Sum (excluding the two $$B$$’s):*  
$$
62+65+66+64+75+80+92+85 = 589 
$$

Including the two unknown $$B$$’s, the total sum is:  
$$
609 + 589 + 2B = 1198 + 2B
$$
Setting this equal to $$1360$$:  
$$
1198 + 2B = 1360  
$$
$$
2B = 1360 - 1198 = 162  
$$
$$
B = \frac{162}{2} = 81
$$

---

**4.2.5**  
To find the median, we first arrange all $$20$$ values in ascending order. Substituting $$A=56$$ and $$B=81$$, the ordered list is:

$$
56,\; 57,\; 58,\; 59,\; 60,\; 61,\; 62,\; 62,\; 62,\; 64,\; 64,\; 65,\; 66,\; 70,\; 75,\; 80,\; 81,\; 81,\; 85,\; 92
$$

For an even number of data values ($$20$$), the median is the average of the $$10^{\text{th}}$$ and $$11^{\text{th}}$$ values. These are both $$64$$. Thus, the median is  
$$
\frac{64+64}{2} = 64
$$

---

**4.2.6**  
We need the probability that a randomly selected learner scored more than $$80\%$$. From the ordered list, the marks that exceed $$80$$ are:  
$$
81,\; 81,\; 85,\; 92
$$  
There are $$4$$ such marks out of $$20$$, so the probability is:  
$$
\frac{4}{20} = \frac{1}{5}
$$

---

**4.2.7**  
A frequency table can be constructed by grouping the data into class intervals. One way to choose intervals is to cover the range from the lowest mark ($$56$$) to the highest mark ($$92$$). For example, we can use the following intervals with a width of $$10\%$$:

| Class Interval | Frequencies |  
|----------------|-------------|  
| $$55 \text{ to } 64$$   | Count the values: $$56,\; 57,\; 58,\; 59,\; 60,\; 61,\; 62,\; 62,\; 62,\; 64,\; 64$$ $$\to 11$$ |  
| $$65 \text{ to } 74$$   | $$65,\; 66,\; 70$$ $$\to 3$$   |  
| $$75 \text{ to } 84$$   | $$75,\; 80,\; 81,\; 81$$ $$\to 4$$   |  
| $$85 \text{ to } 94$$   | $$85,\; 92$$ $$\to 2$$   |

*Histogram:*  
Construct the histogram on your answer sheet by drawing the horizontal axis labeled with the class intervals and the vertical axis labeled with frequency. For each class, draw a bar whose height corresponds to the frequency given in the table above.
**4.2.1** 50% of the data values lie between the lower and upper quartiles. --- **4.2.2** $$ A = 56 $$ --- **4.2.3** The modal percentage is 62%. --- **4.2.4** $$ B = 81 $$ --- **4.2.5** The median percentage mark is 64%. --- **4.2.6** The probability of selecting a learner with more than 80% is $$ \frac{1}{5} $$. --- **4.2.7** Frequency table and histogram completed as follows: | Class Interval | Frequencies | |----------------|-------------| | 55-64 | 11 | | 65-74 | 3 | | 75-84 | 4 | | 85-94 | 2 | *Histogram:* Draw bars for each class interval with heights corresponding to the frequencies.

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