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Fifty athletes need to access suitable training facilities - NSC Mathematics - Question 2 - 2024 - Paper 2
Question 2
Fifty athletes need to access suitable training facilities. The table below shows the distances, in km, that they need to travel to obtain access to suitable trainin... show full transcript
Worked Solution & Example Answer:Fifty athletes need to access suitable training facilities - NSC Mathematics - Question 2 - 2024 - Paper 2
Step 1
Step 2
On the grid provided in the ANSWER BOOK, draw a cumulative frequency graph (ogive) to represent the above data.
Answer
To draw the ogive, plot the cumulative frequencies against the upper limits of each class interval. The upper limits are: 5, 10, 15, 20, 25, and 30. Connect the points smoothly to form a continuous curve.
Step 3
Calculate the interquartile range (IQR) of the above data.
Answer
First, find the first quartile (Q1) and the third quartile (Q3) from the cumulative frequency table. The total number of athletes is 50:
- Q1 is located at 1/4 of 50 = 12.5, which is between the 10th and 15th data points (Q1 = 11).
- Q3 is located at 3/4 of 50 = 37.5, which is between the 30th and 42nd data points (Q3 = 17.8).
Therefore, the IQR is:
Step 4
The families of 4 of the athletes above who stay between 15 and 20 km from a suitable training facility, decide to move 10 kilometers closer to the facility. In which interval will the number of athletes increase?
Answer
The families moving from the 15 ≤ x < 20 interval, which has 12 athletes, will shift to the 5 ≤ x < 10 interval. The number of athletes in the 5 ≤ x < 10 interval will thus increase by 4.
Step 5
Calculate the estimated mean distance that the fifty athletes need to travel after the 4 families have moved 10 kilometers closer to the facility. Clearly show ALL working.
Answer
After the change, the new frequencies are:
| DISTANCE (x km) | NUMBER OF ATHLETES |
|---|---|
| 0 ≤ x < 5 | 3 |
| 5 ≤ x < 10 | 11 |
| 10 ≤ x < 15 | 20 |
| 15 ≤ x < 20 | 8 |
| 20 ≤ x < 25 | 5 |
| 25 ≤ x < 30 | 3 |
Now, calculate the estimated mean:
ext{Estimated mean} = rac{0(3) + 7(11) + 12(20) + 17(8) + 22(5) + 27(3)}{50} = rac{675}{50} = 13.5 ext{ km}