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The marks scored by 20 learners in a test were recorded as 19, 17, 19, 15, 17, 19, 22, 17, 17, 33, 17, 29, 18, 15, 19, 18, 22, and 19. (a) Prepare a frequency distribution table for the data. (b) Determine the: (i) mean mark; (ii) median mark.

Mathematics

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Answers

3 parts
i

19.56\text{19.56}

ii

18.5\text{18.5}

b

112\text{}\frac{1}{12}\text{}

Step-by-step solution

9 steps

For question 18, the problem states "20 learners" but only 18 marks are provided in the list. The solution will use the 18 provided marks.

18. (a) Prepare a frequency distribution table for the data.
The given marks are: 19, 17, 19, 15, 17, 19, 22, 17, 17, 33, 17, 29, 18, 15, 19, 18, 22, 19.

Step 1: List the unique marks and count their occurrences (frequencies).

  • Mark 15: Appears 2 times
  • Mark 17: Appears 5 times
  • Mark 18: Appears 2 times
  • Mark 19: Appears 5 times
  • Mark 22: Appears 2 times
  • Mark 29: Appears 1 time
  • Mark 33: Appears 1 time

Step 2: Construct the frequency distribution table.

MarkFrequency152175182195222291331Total18\begin{array}{|c|c|} \hline \text{Mark} & \text{Frequency} \\ \hline 15 & 2 \\ 17 & 5 \\ 18 & 2 \\ 19 & 5 \\ 22 & 2 \\ 29 & 1 \\ 33 & 1 \\ \hline \text{Total} & 18 \\ \hline \end{array}

18. (b) Determine the:

Part i· 2 steps

mean mark;

  1. Calculate the sum of all marks.

    Sum=(15×2)+(17×5)+(18×2)+(19×5)+(22×2)+(29×1)+(33×1)\text{Sum} = (15 \times 2) + (17 \times 5) + (18 \times 2) + (19 \times 5) + (22 \times 2) + (29 \times 1) + (33 \times 1) Sum=30+85+36+95+44+29+33\text{Sum} = 30 + 85 + 36 + 95 + 44 + 29 + 33 Sum=352\text{Sum} = 352

  2. Calculate the mean mark by dividing the sum of marks by the total number of marks (18).

    Mean=Sum of marksTotal number of marks=35218\text{Mean} = \frac{\text{Sum of marks}}{\text{Total number of marks}} = \frac{352}{18} Mean≈19.555...\text{Mean} \approx 19.555... Mean=19.56 (to 2 decimal places)\text{Mean} = \boxed{\text{19.56}} \text{ (to 2 decimal places)}

Part ii

18.5\text{18.5}

median mark.

  1. Arrange the marks in ascending order.

    15, 15, 17, 17, 17, 17, 17, 18, 18, 19, 19, 19, 19, 19, 22, 22, 29, 33

  2. Since there are 18 marks (an even number), the median is the average of the (182)\left(\frac{18}{2}\right)-th and (182+1)\left(\frac{18}{2} + 1\right)-th values, which are the 9th and 10th values.

    The 9th mark is 18.
    The 10th mark is 19.

    Median=9th mark+10th mark2=18+192\text{Median} = \frac{9\text{th mark} + 10\text{th mark}}{2} = \frac{18 + 19}{2} Median=372\text{Median} = \frac{37}{2} Median=18.5\text{Median} = \boxed{\text{18.5}}

    40. Rebecca tossed a coin and a die at the same time.

Part a· 2 steps

Write down the probability space to show the possible outcomes.

  1. List the possible outcomes for tossing a coin (H for Head, T for Tail) and rolling a die (1, 2, 3, 4, 5, 6).

  2. Combine these outcomes to form the sample space.

    Sample Space={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}\text{Sample Space} = \{ (H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6) \}

Part b

112\text{}\frac{1}{12}\text{}

Determine the probability that Rebecca obtained a head on the coin and a 4 on the die.

  1. Identify the favorable outcome from the sample space.

    The favorable outcome is (H,4). There is 1 such outcome.

  2. Determine the total number of possible outcomes in the sample space.

    There are 12 possible outcomes.

  3. Calculate the probability.

    P(Head and 4)=Number of favorable outcomesTotal number of outcomes=112P(\text{Head and 4}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{12} P(Head and 4)=112P(\text{Head and 4}) = \boxed{\text{}\frac{1}{12}\text{}}
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