The following stem-and-leaf diagram shows the marks obtained by 20 pupils in a class test that has a total mark of 50. State the median score. Answer ... Find the standard deviation. Answer ... If distinction is awarded to pupils who scored at least 80%, find the percentage of pupils in the class who scored distinction. Answer ... % Bell has six 10-cent coins and two 50-cent coins. She takes 2 coins at random from her purse, one after the other. Complete the probability tree diagram shown in the answer space, giving the answers in fraction in the simplest form. Find the probability that the total value of the two coins is 60 cents. Answer ... If a third coin is taken out, calculate the probability that the total amount is 30 cents. Answer ...

1. Identify the middle two values and find their average

There are $20$ pupils, so the median is the average of the $10$th and $11$th marks.

From the ordered stem-and-leaf diagram, the $10$th and $11$th marks are both $35$.

$\text{Median}=\frac{35+35}{2}=35$

1. Find the total and the mean of the 20 marks

The marks are
$10,12,23,23,24,27,27,32,35,35,35,36,37,39,41,46,47,48,48,50$.

$\sum x=675,\quad n=20$

$\bar{x}=\frac{675}{20}=33.75$

2. Find the variance using the standard deviation formula

$\sum x^2=25335$

Using
$\sigma^2=\frac{\sum x^2}{n}-\bar{x}^2,$

$\sigma^2=\frac{25335}{20}-(33.75)^2$

$=1266.75-1139.0625$

$=127.6875$

3. Take the square root of the variance

$\sigma=\sqrt{127.6875}$

$=11.3008\ldots$

So the standard deviation is $11.3$ to 3 significant figures.

1. Find the minimum mark needed for distinction

Distinction is at least $80\%$ of $50$ marks.

$80\% \times 50 = 0.8 \times 50 = 40$

So pupils who scored $40$ marks or more get distinction.

2. Count the pupils who scored at least 40 and express as a percentage

Marks at least $40$ are:

$41,\ 46,\ 47,\ 48,\ 48,\ 50$

Number of pupils with distinction $=6$.

$\frac{6}{20}\times 100\% = 30\%$

1. Find the probabilities for the first coin drawn

There are $6+2=8$ coins altogether.

$P(10\text{c})=\frac68=\frac34$

$P(50\text{c})=\frac28=\frac14$

2. Find the second-draw probabilities after a 10-cent coin is drawn first

If the first coin is $10\text{c}$, the remaining coins are $5$ ten-cent coins and $2$ fifty-cent coins, out of $7$ coins.

$P(10\text{c second}\mid 10\text{c first})=\frac57$

$P(50\text{c second}\mid 10\text{c first})=\frac27$

3. Find the second-draw probabilities after a 50-cent coin is drawn first

If the first coin is $50\text{c}$, the remaining coins are $6$ ten-cent coins and $1$ fifty-cent coin, out of $7$ coins.

$P(10\text{c second}\mid 50\text{c first})=\frac67$

$P(50\text{c second}\mid 50\text{c first})=\frac17$

1. Identify the outcomes that give a total of 60 cents

A total of $60$ cents is obtained from:

$10\text{c}$ then $50\text{c}$, or $50\text{c}$ then $10\text{c}$.

2. Multiply along each branch and add the two probabilities

$P(60\text{c})=\frac{3}{4}\times\frac{2}{7}+\frac{1}{4}\times\frac{6}{7}$

$=\frac{6}{28}+\frac{6}{28}$

$=\frac{12}{28}=\frac{3}{7}$

1. Identify the only coin combination that gives 30 cents

To get a total of $30$ cents from $3$ coins, the coins must be

$10\text{c},\ 10\text{c},\ 10\text{c}.$

2. Multiply the probabilities of drawing three 10-cent coins without replacement

There are initially $6$ ten-cent coins out of $8$ coins.

$P(10\text{c},10\text{c},10\text{c})=\frac{6}{8}\times\frac{5}{7}\times\frac{4}{6}$

$=\frac{5}{14}.$