4.1.1 Median 4.1.2 Upper quartile Use TABLE 5 and the information regarding school-based assessment marks and percentages of the ten lowest performing learners in Mathematical Literacy in 2016 - NSC Mathematical Literacy - Question 4 - 2017 - Paper 1

The upper quartile is the median of the upper half of the ordered dataset, indicating the threshold below which 75% of the data falls.

To find the probability of selecting a learner who wrote all tasks, identify the learner(s) who wrote the maximum number of tasks (which is 7). The calculation is:

$P = \frac{\text{Number of learners who wrote all tasks}}{\text{Total number of learners}} \times 100$

In this case, there are 2 learners who wrote all tasks, so:

$P = \frac{2}{10} \times 100 = 20\%$

To determine the median total mark, order the total marks from lowest to highest:

• Total Marks: 39, 34, 37, 41, 46, 42, 39, 41, 34, 38

The median will be the average of the 5th and 6th values:

$\text{Median} = \frac{41 + 42}{2} = 41.5$

The modal SBA percentage mark is the value that appears most frequently in the dataset. From the provided data, the percentage marks are:

• 39%, 34%, 37%, 46%, 41%, 39%, 41%, 34%, 38%

In this case, the modal mark is 41% as it occurs most often.

From TABLE 5, Learner C scored the lowest actual SBA percentage mark of 34%.

To calculate the mean percentage mark, sum all the actual SBA percentage marks and divide by the number of learners:

$\text{Mean} = \frac{46 + 38 + 34 + 39 + 41 + 39 + 41 + 39 + 34 + 38}{10} = 38.8\%$

For learner J, whose actual percentage mark was 39% but missed one task, the adjusted mark is calculated as follows:

Adjusted Mark = ( \frac{\text{Total Marks Attained}}{\text{Total Marks Possible}} \times 100 )

If the total marks possible consisted of 350 and he missed one task (i.e., 10 marks), the calculation would be:

Adjusted Mark = ( \frac{137}{350 - 10} \times 100 \approx 40.0% )

The correct answer is B: Fifty-four million, nine hundred and fifty-seven thousand, seven hundred and forty-six (54,957,746).

From TABLE 6, the Indian/Asian category between the ages of 15-19 has the same number of males (436) and females (436).

To find Y, we will sum up the male and female totals: 2,334,819 (Males) + 2,498,098 (Females) = 4,832,917

Thus, Y = 4,832,917.

Using the total male population (2,334,819) and the number of coloured males (426,510):

$P = \frac{426510}{24987564} \times 100 \approx 1.71\%$

From TABLE 6, there are 62 Asian females and 99 Asian males. The ratio is:

$\text{Ratio} = \frac{62}{99} = \frac{62}{99} (simplest form)$

The total population is 54,957,764 and the number of coloured females is 688,118:

$\text{Percentage} = \frac{688118}{54957764} \times 100 \approx 1.25\%$

From TABLE 6, the age group 0-4 has the largest number of people, totaling 2,032,721.

A bar graph will be the best representation for displaying the data trends as it allows for a clear comparison between categories.