12 students from class A took a Mathematics test. The table below shows the test marks of the students. However, two of the students' marks are covered with ink. Given that the box-and-whisker plot below shows the distribution of the results, explain why the information may not be sufficient to find the missing marks of the two students. Given further that the modal mark is 63, find the two missing marks. Answer ... and ... (iii) 12 students from class B also took the same Mathematics test. The distribution of their test marks is shown on the cumulative frequency graph. Make a comment comparing the averages and a comment comparing the distribution of the Mathematics Test marks between the two classes. Use figures to support your answers. Answer ... (b) Alice rolled a six-sided die, X, 100 times. Bala rolled another six-sided die, Y, 80 times. The number of times they each obtained a '6' is recorded in the table below. (i) Find the probability of rolling a '6' by Alice and Bala respectively. Answer P(obtaining a '6' by Alice) = ...
   P(obtaining a '6' by Bala) = ...
(ii) One of the dice is biased. Using your answers in (b)(i), determine which die, X or Y, is likely to be the unbiased 6-sided die. Explain your answer. Answer ...

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Worked solution

4(a)(i)

1. Identify the limitation of the box-and-whisker plot

A box-and-whisker plot summarises the data using only key values such as the minimum, lower quartile, median, upper quartile and maximum. It does not show the exact individual marks or how often each mark occurs.

Answer

The box-and-whisker plot only shows the five-number summary: the minimum, lower quartile, median, upper quartile and maximum. It does not show all the individual marks, so different pairs of missing marks could give the same box-and-whisker plot.

4(a)(iii)

1. Read and compare the average marks using the medians

From the box-and-whisker plot for Class A,
medianA≈65.\text{median}_A \approx 65.
From the cumulative frequency graph for Class B, at cumulative frequency 66,
medianB≈60 to 65.\text{median}_B \approx 60\text{ to }65.
So Class A has the higher average mark, or at least slightly higher depending on the exact graph reading.

2. Read quartiles and calculate the spread for Class A

For Class A,
Q1≈55,Q3≈75.Q_1 \approx 55, \qquad Q_3 \approx 75.
Hence
IQRA=Q3−Q1≈75−55=20.\text{IQR}_A = Q_3-Q_1 \approx 75-55=20.

3. Read quartiles and compare the distributions using interquartile ranges

For Class B,
Q1≈50,Q3≈75 to 80.Q_1 \approx 50, \qquad Q_3 \approx 75\text{ to }80.
Hence
IQRB≈75−50=25to80−50=30.\text{IQR}_B \approx 75-50=25 \quad \text{to} \quad 80-50=30.
Since IQRB\text{IQR}_B is greater than IQRA\text{IQR}_A, Class B's marks are more spread out. Class A's marks are more consistent.

Answer

Class A has a higher average mark since its median is about 6565, while Class B's median is about 6060 to 6565.

Class B's marks are more spread out since its interquartile range is about 75−50=2575-50=25 to 80−50=3080-50=30, compared with Class A's interquartile range of about 75−55=2075-55=20.

4(b)((i))

1. Find Alice's experimental probability of obtaining a 6

Alice obtained a '6' 1616 times out of 100100 rolls.

P(obtaining a ’6’ by Alice)=16100=425P(\text{obtaining a '6' by Alice})=\dfrac{16}{100}=\dfrac{4}{25}

2. Find Bala's experimental probability of obtaining a 6

Bala obtained a '6' 1818 times out of 8080 rolls.

P(obtaining a ’6’ by Bala)=1880=940P(\text{obtaining a '6' by Bala})=\dfrac{18}{80}=\dfrac{9}{40}

Answer

P(obtaining a ’6’ by Alice)=16100=425P(\text{obtaining a '6' by Alice})=\dfrac{16}{100}=\dfrac{4}{25}

P(obtaining a ’6’ by Bala)=1880=940P(\text{obtaining a '6' by Bala})=\dfrac{18}{80}=\dfrac{9}{40}

4(b)((ii))

1. Compare the experimental probabilities with the theoretical probability for an unbiased die

For an unbiased six-sided die,

P(obtaining a 6)=16≈0.167.P(\text{obtaining a 6})=\frac{1}{6}\approx 0.167.

From part (b)(i),

PX=425=0.16,PY=940=0.225.P_X=\frac{4}{25}=0.16, \qquad P_Y=\frac{9}{40}=0.225.

Since 0.160.16 is closer to 0.1670.167 than 0.2250.225 is, die XX is likely to be the unbiased die.

Answer

Die XX is likely to be unbiased, since Alice's probability 425=0.16\frac{4}{25}=0.16 is closer to 16≈0.167\frac{1}{6}\approx0.167 than Bala's probability 940=0.225\frac{9}{40}=0.225.

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