Sec 1 Maths: Data Handling & Statistics — practice questions & worked solutions

(i)

$$\begin{aligned}\frac{10(20)+20x+30(5)+40(10)}{20+x+5+10}&=21.25\\ \frac{750+20x}{35+x}&=21.25\\ 750+20x&=21.25(35+x)\\ 750+20x&=743.75+21.25x\\ 6.25&=1.25x\\ x&=5\end{aligned}$$

(ii)

The mean increases as $x$ decreases (fewer students folding only $20$); for mean $\geq 21.25$ we need $\dfrac{750+20x}{35+x}\geq 21.25$, giving $6.25\geq 1.25x$, so $x\leq 5$.

(iii)

Total students $=35+x$. The median is the average of the two middle values. The values are ordered $10$ ($20$ students), $20$ ($x$), $30$ ($5$), $40$ ($10$). For the median to be $10$, both middle positions must fall within the first $20$ students. With $35+x$ students the upper middle position must be $\leq 20$. Solving $\dfrac{35+x+1}{2}\leq 20$ gives $x\leq 4$, so the maximum value of $x$ is $4$.

(i)

Counting scores in each interval. For $20<x\leq 30$: $28, 25, 26, 30, 22 \Rightarrow a=5$. For $40<x\leq 50$: $42, 44, 50, 41 \Rightarrow b=4$.

$$\begin{aligned}a=5,\qquad b=4\end{aligned}$$

(ii)

The class with the highest frequency ($10$) is $30<x\leq 40$. Modal class: $30<x\leq 40$.

(iii)

Histogram with bars of heights $5$, $5$, $10$, $4$ over the intervals $10$–$20$, $20$–$30$, $30$–$40$, $40$–$50$ respectively (equal class widths, bars touching).

(a) “At least $700$” means values $7, 8, 9, 10, 11, 16, 17$ (hundreds). Counting the dots:

$$\begin{aligned}1+3+2+2+1+1+1 &= 11\end{aligned}$$

There are $11$ regions.

(b) “From $400$ to $800$ inclusive” means values $4, 5, 6, 7, 8$. Counting the dots:

$$\begin{aligned}1+5+4+1+3 &= 14\end{aligned}$$

Percentage:

$$\begin{aligned}\frac{14}{32}\times 100\% &= 43.75\%\end{aligned}$$

(c) These regions may be largely industrial, rural or uninhabited (e.g. nature reserves or military areas), so few residents live there and no taxi rides were taken.

(a) March and April.

(b) Temperature difference (Jan – Feb)

$$\begin{aligned}\text{Temperature difference} &= 31.0 – 30.1 \\ &= 0.9^\circ\text{C}\end{aligned}$$

(c) $14$ rainy days.

(d) Percentage of rainy days

$$\begin{aligned}\text{Percentage} &= \frac{22 + 18 + 19 + 12 + 11 + 14}{181} \times 100\% \\ &= 53.0\%\end{aligned}$$

(e) Disagree. The $y$-axis of the graph starts from $10$ instead of $0$, so the height of the bars should not be used to compare the number of rainy days.

(f) Agree. In January 2023 the number of rainy days is the highest and the temperature is the lowest in 2023, while in May 2023 the number of rainy days is the lowest, resulting in the highest maximum temperature. This shows the trend of the number of rainy days is linked to the daily maximum air temperature.

(a) Value of $x$

Angles at the centre sum to $360^\circ$. The sectors are Bus $(2x+14)^\circ$, Walk $66^\circ$ and Car $84^\circ$.

$$\begin{aligned}(2x+14)+66+84 &= 360 \\ 2x+164 &= 360 \\ 2x &= 196 \\ x &= 98\end{aligned}$$

(b) Students by car

Car sector $=84^\circ$.

$$\begin{aligned}\text{Students by car} &= \frac{84}{360}\times 120 \\ &= 28\end{aligned}$$

The points lie on a straight line with gradient

$$\begin{aligned}\text{gradient} &= \frac{68-76}{6-2} \\ &= \frac{-8}{4} \\ &= -2\end{aligned}$$

Line equation: $y=80-2x$.

(b)(i) Weight at signing up

At $x=0$:

$$\begin{aligned}y &= 80-2(0) \\ &= 80 \text{ kg}\end{aligned}$$

(b)(ii) Months to lose 15 kg

A loss of $15$ kg means $y=80-15=65$:

$$\begin{aligned}65 &= 80-2x \\ 2x &= 15 \\ x &= 7.5\end{aligned}$$

From the graph, about $7.4$ months.

(c) Gradient

Gradient $=-2$; it represents Janice’s weight loss of $2$ kg per month.

(a) Value of $x$

The youth sector $=35\%$ of $360^\circ$:

$$\text{youth angle} = 0.35 \times 360^\circ = 126^\circ$$

The remaining sectors are $3x^\circ$ (Adults), $x^\circ$ (Young Children) and $2x^\circ$ (Senior Citizens):

$$3x + x + 2x + 126 = 360$$
$$6x = 234$$
$$x = 39$$

(b) Total number of visitors

Adults $=3x^\circ = 117^\circ$, Young Children $=x^\circ = 39^\circ$. The difference $=78^\circ$ represents $130$ people:

$$1^\circ \to \frac{130}{78} \text{ people}$$
$$\text{total} = \frac{130}{78} \times 360 = 600$$