下面的直方图显示了波东巴西中学（Potong Pasir Secondary School）学生在一场时事问答比赛中得分的分布情况。

1. Find the total frequency from the histogram

Total number of participants
$=10+8+30+20+25=93$

1. Calculate the estimated mean using midpoints and frequencies from the histogram

Class midpoints: $1.5,\ 4.5,\ 7.5,\ 10.5,\ 13.5$.

Frequencies: $10,\ 8,\ 30,\ 20,\ 25$.

$\text{Estimated mean}=\frac{1.5(10)+4.5(8)+7.5(30)+10.5(20)+13.5(25)}{93}$

$=\frac{15+36+225+210+337.5}{93}$

$=\frac{823.5}{93}=8.8548\ldots$

$\approx 8.9$

1. Calculate the estimated standard deviation using midpoints and frequencies

Using the midpoints $1.5, 4.5, 7.5, 10.5, 13.5$ and frequencies $10, 8, 30, 20, 25$,

$\sum fx^2=10(1.5^2)+8(4.5^2)+30(7.5^2)+20(10.5^2)+25(13.5^2)$

$=8633.25$

From earlier parts, $\sum f=93$ and $\bar{x}=\dfrac{828}{93}$.

$\text{Standard deviation}=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}^2}$

$=\sqrt{\frac{8633.25}{93}-\left(\frac{828}{93}\right)^2}$

$=3.682\ldots$

1. State why the calculated values are not exact

The histogram gives grouped data only, so the exact scores of the participants within each interval are unknown. In calculating the mean and standard deviation, the midpoint of each class interval was used to represent all scores in that interval. Hence the mean and standard deviation are estimates.

1. Find the median position and locate it using cumulative frequencies

Total number of participants $=93$.

Median position $=\dfrac{93+1}{2}=47$th value.

Cumulative frequencies:

$0$–$3$: $10$

$3$–$6$: $10+8=18$

$6$–$9$: $18+30=48$

Since the $47$th value lies between the $19$th and $48$th values, the median mark is in the interval $6 < x \leq 9$.

1. Find the number of students who attained at least a Silver Award

At least a Silver Award means Silver or Gold.

Silver: $9 < x \leq 12$, frequency $=20$

Gold: $12 < x \leq 15$, frequency $=25$

Number who attained at least Silver
$=20+25=45$

2. Express this number as a percentage of all participants

Total number of participants $=93$.

Percentage
$=\frac{45}{93}\times 100\%$
$=48.387\ldots\%$
$\approx 48.4\%$

1. Find the probabilities for the first selection from Box A

Box A contains $3$ chocolate and $4$ strawberry cups, so the total number of cups is

$3+4=7.$

First branch probabilities:

$P(\text{Chocolate from A})=\frac{3}{7}, \qquad P(\text{Strawberry from A})=\frac{4}{7}.$

2. Update Box B and find second-selection probabilities after chocolate is transferred

If a chocolate cup is selected from Box A and placed in Box B, Box B then contains:

$3\text{ chocolate},\ 3\text{ strawberry},\ 4\text{ vanilla}.$

Total in Box B:

$3+3+4=10.$

Second branch probabilities after chocolate from A:

$P(\text{Chocolate from B})=\frac{3}{10}, \quad P(\text{Strawberry from B})=\frac{3}{10}, \quad P(\text{Vanilla from B})=\frac{4}{10}.$

3. Update Box B and find second-selection probabilities after strawberry is transferred

If a strawberry cup is selected from Box A and placed in Box B, Box B then contains:

$2\text{ chocolate},\ 4\text{ strawberry},\ 4\text{ vanilla}.$

Total in Box B:

$2+4+4=10.$

Second branch probabilities after strawberry from A:

$P(\text{Chocolate from B})=\frac{2}{10}, \quad P(\text{Strawberry from B})=\frac{4}{10}, \quad P(\text{Vanilla from B})=\frac{4}{10}.$

1. Identify the favourable same-flavour outcomes and multiply along each tree branch

Same flavour can occur in two ways:

$P(\text{chocolate then chocolate})=\dfrac{3}{7}\times\dfrac{3}{10}=\dfrac{9}{70}$

$P(\text{strawberry then strawberry})=\dfrac{4}{7}\times\dfrac{4}{10}=\dfrac{16}{70}$

2. Add the mutually exclusive favourable probabilities and simplify

$P(\text{same flavour})=\dfrac{9}{70}+\dfrac{16}{70}=\dfrac{25}{70}=\dfrac{5}{14}$

1. Identify and calculate the probabilities of all branches where the second cup is not chocolate

The second cup is not chocolate means the second cup is strawberry or vanilla.

Using the tree diagram:

$P(\text{second not chocolate})$
$=P(C_A \text{ then } S_B)+P(C_A \text{ then } V_B)+P(S_A \text{ then } S_B)+P(S_A \text{ then } V_B)$

$=\frac{3}{7}\times\frac{3}{10}+\frac{3}{7}\times\frac{4}{10}+\frac{4}{7}\times\frac{4}{10}+\frac{4}{7}\times\frac{4}{10}$

2. Add the favourable probabilities and simplify the fraction

$=\frac{9}{70}+\frac{12}{70}+\frac{16}{70}+\frac{16}{70}$

$=\frac{53}{70}$

Alternatively, using complements:

$P(\text{second not chocolate})=1-P(\text{second chocolate})$

$=1-\left(\frac{3}{7}\times\frac{3}{10}+\frac{4}{7}\times\frac{2}{10}\right)$

$=1-\left(\frac{9}{70}+\frac{8}{70}\right)$

$=\frac{53}{70}$