The stem and leaf diagram below shows the marks obtained by 30 students in a Science test marked out of 80 marks.

1. Find the positions of the two middle marks

There are $30$ students.

For an even number of values, the median is the average of the $\frac{30}{2}=15$th and $16$th values.

2. Identify the 15th and 16th values from the ordered stem-and-leaf diagram and average them

Counting from the smallest mark:

$18$ is the 1st value.

The 40s give the 2nd to 9th values.

The 50s are $50,51,51,52,54,58,59$, which are the 10th to 16th values.

So the 15th value is $58$ and the 16th value is $59$.

$\text{Median}=\frac{58+59}{2}=58.5$

1. Explain why the mean is not a suitable average for this data set

From the stem-and-leaf diagram, most marks are from the $40$s to $70$s, but there is one very low mark, $18$. This outlier would pull the mean down, so the mean may not be representative of the marks obtained by the students.

1. Find how many students are awarded distinction and identify the cut-off position

Top $30\%$ of $30$ students:
$0.30 \times 30 = 9$
So the top $9$ students get distinction.

In ascending order, these are the $22$nd to $30$th marks, since
$30-9+1=22.$

2. Read the 22nd mark from the stem-and-leaf diagram

Counting from the lowest mark:

• $18$ is the 1st mark,
• the $40$-marks are the 2nd to 9th marks,
• the $50$-marks are the 10th to 16th marks,
• the $60$-marks are the 17th to 23rd marks:

$63,\ 63,\ 63,\ 63,\ 66,\ 67,\ 67$

Hence the $22$nd mark is $67$.

1. Subtract the probabilities of NETS and credit card from 1

$P(\text{cash})=1-P(\text{NETS})-P(\text{credit card})$
$=1-\frac{2}{3}-\frac{1}{4}$
$=\frac{12}{12}-\frac{8}{12}-\frac{3}{12}=\frac{1}{12}$

1. Find the probability that both customers pay by credit card

For one customer,
$P(\text{credit card})=\dfrac{1}{4}$.

For two customers,
$P(\text{both credit card})=\dfrac{1}{4}\times \dfrac{1}{4}=\dfrac{1}{16}.$

1. Find the probability that one customer does not pay by cash

From part (b)(i),
$P(\text{cash})=\dfrac{1}{12}$.

So
$P(\text{no cash})=1-\dfrac{1}{12}=\dfrac{11}{12}$.

2. Find the probability that both customers do not use cash

For two customers,
$P(\text{no cash for both})=\dfrac{11}{12}\times\dfrac{11}{12}=\dfrac{121}{144}.$

1. Identify the probability of not paying by cash and paying by cash

From the earlier part,
$P(\text{cash})=\frac{1}{12}.$
So
$P(\text{not cash})=1-\frac{1}{12}=\frac{11}{12}.$

2. Multiply the probabilities for the sequence not cash, not cash, cash

For the first cash transaction to occur with the third customer, the first two customers must not pay by cash and the third customer must pay by cash.
P=\frac{11}{12}\times\frac{11}{12}\times\frac{1}{12}=rac{121}{1728}.