4.1.1 Median 4.1.2 Upper quartile The school-based assessment (SBA) marks and percentages of the ten lowest performing learners in Mathematical Literacy of a particular school in 2016 are represented in TABLE 5 below - NSC Mathematical Literacy - Question 4 - 2017 - Paper 1

The upper quartile is the median of the upper half of the data set. After ordering the data from Table 5, the upper half consists of the values: 41, 42, 42, 46, 47. The median of this subset, which is the 3rd value in the sorted order, is 42.

To find the probability, identify how many learners wrote all the assessment tasks out of the total number of learners which is 10. From Table 5, only Learner J did not write all tasks. Therefore, the probability is:

P = rac{9}{10} imes 100 ext{%} = 90 ext{%}

The total marks attained by learners are as follows, when ordered: 109, 118, 137, 144, 162, 168. With 10 learners, the median is the average of the 5th and 6th values:

ext{Median} = rac{144 + 137}{2} = 140.5

The modal actual SBA percentage mark is the most frequently occurring percentage. From the data, we see 42 occurs twice. Thus, the mode is 42.

Based on the values retrieved from Table 5, Learner C scored the lowest actual SBA percentage mark, with a score of 34%.

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

ext{Mean} = rac{34 + 36 + 39 + 39 + 40 + 41 + 42 + 42 + 46 + 47}{10} = rac{ 392}{10} = 39.2

To calculate Learner J's adjusted SBA percentage mark:

The initial calculation is based on total marks, which are 137 out of 350. To find the new percentage:

ext{Adjusted SBA ext{ (new)}} = rac{137}{350} imes 100 ext{ ext{%}} \\ = 39.14 ext{ ext{%}} ext{(rounded)} \\ ext{which is approximately } 39 ext{ ext{%}}.

The estimated 2015 midyear total population is given as 54 957 764. Hence, the correct option is B.

From the data in Table 6, the race and age group where males and females are equal is the 'Indians/Asians' for the age group 15-19, with both having 431 779.

To find Y, sum the total males in all categories to equal Y:

$Y = 426 156 + 430 667 + 431 779 + 437 412 + 432 917 + 425 138 + 474 354 + 496 830 + 418 314 + 402 688 + 426 510$

To compute this, use the total number of coloured males, which is 2 334 819, out of a total population of 54 957 764:

P = rac{2 334 819}{54 957 764} imes 100 ext{ ext{%}} \\ ext{which results in approximately } 4.25 ext{ ext{%}}.

From Table 6, the number of Asian females is 688 118 and Asian males is 1 362 848. The ratio is:

ext{Ratio} = rac{688 118}{1 362 848} = 0.505 ext{ (or simplified to a fraction, } rac{687}{1324}).

The number of coloured females is 2 498 098 and the total population is 54 957 764. Thus, the percentage is calculated as:

P = rac{2 498 098}{54 957 764} imes 100 ext{ ext{%}} \\ ext{approximately } 4.54 ext{ ext{%}}.

Based on the table values, the age group 0-4 has the largest number of people, with a combined total of 835 172 (coloured and Indians/Asians).

Given the nature of the data, a bar graph would be the most effective graphical representation to display the distribution of the population by age and gender.