Students in a class were asked how many siblings they have.
The results are shown in the table. The modal number of siblings is 1.
Calculate the greatest possible number of students in the class. Answer ___ The median number of siblings is 2.
Calculate the smallest possible value of x. Answer x =___ The mean number of siblings is 1.525.
Calculate the value of x. Answer x = ___ A dart board comprises of 3 colours as shown below. A pouch contains six 10 cents coins, three 20 cents coins and one 50 cents coin. Tim throws a dart first then picks a coin from the pouch. Draw a tree diagram to show the probabilities of the possible outcomes. Answer Calculate the probability that the dart hits the red region and a 20 cent coin is picked. Answer ___ State one assumption made. Answer ___

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1. Use the modal condition to find the greatest possible value of $x$.

The frequency for 1 sibling is $18$.

Since the modal number of siblings is $1$, this frequency must be greater than the frequency for 3 siblings, which is $x$.

So $x<18$.

The greatest possible integer value of $x$ is therefore
$x=17.$

2. Calculate the greatest possible total number of students.

Total number of students
$=5+18+10+17+2$
$=52.$

1. Count how many students have fewer than 2 siblings, then make the median position fall in the group with 2 siblings.

Number of students with $0$ or $1$ sibling
$5+18=23.$
Number of students with $2$ siblings is $10$, so the positions for value $2$ are from the $24$th to the $33$rd student.

For the median to be $2$, the middle position must be at most the $33$rd position.

Total number of students
$=5+18+10+x+2=35+x.$
For the smallest $x$, the total number should first make the median position just exceed the first $23$ values. If $35+x$ is even, the two middle positions must both lie in the $2$ group. Try $x=16$:
$35+16=51.$
The median is the $\frac{51+1}{2}=26$th value, which lies between the $24$th and $33$rd positions.

For $x=15$, total $=50$, and the median is the average of the $25$th and $26$th values, also $2$. Hence the smallest possible value is actually checked by decreasing further until the middle just remains after the 23rd value:

For total $N=46$, the middle two positions are $23$rd and $24$th, giving median $\frac{1+2}{2}=1.5$, not $2$.

Thus the smallest total for median $2$ is $N=47$, where the median is the $24$th value.
$35+x=47$
$x=12.$

1. Write the total number of siblings and total number of students in terms of $x$.

Total number of siblings
$=0(5)+1(18)+2(10)+3(x)+4(2)$
$=46+3x$

Total number of students
$=5+18+10+x+2=35+x$

2. Use the given mean to form and solve an equation for $x$.

Since the mean number of siblings is $1.525$,
$\frac{46+3x}{35+x}=1.525$
$46+3x=1.525(35+x)$
$46+3x=53.375+1.525x$
$1.475x=7.375$
$x=5$

1. Find the probability that the dart hits the red region.

The red sector has angle $120^\circ$ out of $360^\circ$.

$P(\text{Red})=\frac{120}{360}=\frac{1}{3}$

2. Find the probability of picking a 20 cent coin and multiply the independent probabilities.

There are $6+3+1=10$ coins in total, and $3$ are 20 cent coins.

$P(20\text{ cent})=\frac{3}{10}$

Therefore,

$P(\text{Red and 20 cent})=\frac{1}{3}\times\frac{3}{10}=\frac{1}{10}$

1. State a valid probability assumption used in the earlier probability calculation.

A valid assumption is that the dart throw does not affect which coin is picked, so the two events are independent.