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Question 3
3.1 Staff at public schools are required to adhere to stipulated percentages or fractions related to the total number of weekly periods based on their post level.... show full transcript
Step 1
Answer
To find the number of staff members at Woodhill SS, we observe the given timetables. The number of periods taught by each member should add up to the total periods, which are 40. Counting the numbers provided: the deputy principal's periods (D) and the value of 3 for the three other posts gives us the total staff summary:
(3 + 3 + D + 3 = 40)
This gives: [ D + 9 = 40 ] [ D = 31 ] Thus, there are 4 staff members at Woodhill SS.
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Step 4
Answer
First, let's find the mean and median of the periods taught at Woodhill SS. The periods are:
[3, 3, 31, 3]
The mean is calculated as: [ \text{Mean} = \frac{3 + 3 + 31 + 3}{4} = \frac{40}{4} = 10 ]
For the median, arranging values results in: [3, 3, 3, 31]
The median, being the average of the two middle values, will be: [ \text{Median} = \frac{3 + 3}{2} = 3 ]
Thus, the median is lower than the mean, which confirms that the median is a more reliable measure of central tendency given that it is less impacted by extreme values.
Step 5
Answer
To determine this probability, we first note the total number of staff (which is 16) and then count how many teach 29 or more periods. In the list, we see:
Thus, the fraction is given by: [ P = \frac{2}{16} = \frac{1}{8} ]
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Step 8
Answer
The learner whose marks can be classified as an outlier is Learner H, who scored significantly lower in Task 2 (15) compared to other learners. This score stands out when compared with the rest.
Step 9
Answer
We're told the mean for Task 1 is 66.7%. First, let's calculate the mean for Task 2:
[ \text{Mean (Task 2)} = \frac{53 + 69 + 53 + 49 + 50 + 47 + 61 + 15 + 47 + 81}{10} = \frac{ 510 }{10} = 51 ]
Now, we find the difference between these two means:
[ \text{Difference} = 66.7 - 51 = 15.7 ]
Since 15.7 is not less than 15, the teacher's claim is INVALID.
Step 10
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