Sec 2 Maths: Statistics, Mean, Median & Mode, practice questions & worked solutions

Reading the dots: $1$ mark (1 student), $5$ (2), $6$ (1), $7$ (3), $8$ (2), $9$ (4), $10$ (2). Check: $1+2+1+3+2+4+2=15$ students.

(a)

The marks range from $1$ mark to $10$ marks. Most of the data is clustered between $5$ and $10$ marks, with the greatest cluster at $9$ marks. The score of $1$ mark is an outlier, far below all the other scores.

(b)

The tallest column of dots is at $9$, with $4$ students.

$$\text{Modal mark}=9\text{ marks}$$

Reading the dots: $1$ book (3 students), $2$ (4), $3$ (3), $4$ (2), $5$ (2), $8$ (1). Total $15$ students.

(a)

The tallest column is at $2$ books, with $4$ students.

$$\text{Mode}=2$$

(b)

With $15$ values the median is the $\dfrac{15+1}{2}=8^{\text{th}}$ value. Positions $1$ to $3$ are $1$, positions $4$ to $7$ are $2$, and positions $8$ to $10$ are $3$.

$$\text{Median}=3$$

(c)

With $16$ values the median is the mean of the $8^{\text{th}}$ and $9^{\text{th}}$ values. If the new value is $3$ or more, positions $1$ to $7$ are unchanged and the $8^{\text{th}}$ and $9^{\text{th}}$ values are both $3$, so the median stays $3$. (A new value of $1$ or $2$ would make the $8^{\text{th}}$ value $2$ and the median $2.5$.)

So the student could have read any number of books that is $3$ or more, for example $3$ books.

(d)

The value $8$ is an outlier. It pulls the mean upwards, so the mean does not represent the typical number of books read as well as the median or mode does.

(a)

Count the leaves on each row: $6+10+9$.

$$\text{Number of students}=25$$

(b)

The scores below $23$ are $11, 12, 14, 18, 19, 19, 20, 21$: that is $8$ students. So $25-8=17$ students scored at least $23$.

$$\begin{aligned}\text{Percentage}&=\frac{17}{25}\times 100\%\\ &=68\%\end{aligned}$$

(c)

The scores less than $20$ are the $6$ leaves on the stem $1$.

$$\text{Fraction}=\frac{6}{25}$$

(d)

The leaf $7$ appears three times on the stem $2$; no other score appears more than twice.

$$\text{Modal mark}=27$$

(a)

With $20$ values the median is the mean of the $10^{\text{th}}$ and $11^{\text{th}}$ values. The first $4+5=9$ values are on stems $0$ and $1$, so the $10^{\text{th}}$ value is $20 and the $11^{\text{th}}$ is $21.

$$\begin{aligned}\text{Median}&=\frac{20+21}{2}\\ &=20.5\end{aligned}$$

The median monthly savings is $20.50.

(b)

No. Both $14 and $21 appear twice, and no other amount appears more than once, so there are two modal amounts: $14 and $21.

(a)

The heights are $152, 156, 159, 160, 161, 161, 161, 162, 165, 173$ cm.

$$\begin{aligned}\text{Sum}&=152+156+159+160+3(161)+162+165+173\\ &=1610\\ \text{Mean}&=\frac{1610}{10}\\ &=161\text{ cm}\end{aligned}$$

(b)(i)

The new mean is $161+3=164$ cm for $11$ students. Let the added height be $x$ cm.

$$\begin{aligned}\frac{1610+x}{11}&=164\\ 1610+x&=1804\\ x&=194\end{aligned}$$

The student added is $194$ cm tall.

(b)(ii)

Before: the median is the mean of the $5^{\text{th}}$ and $6^{\text{th}}$ values, $\dfrac{161+161}{2}=161$ cm. After adding $194$ cm at the top, the median is the $6^{\text{th}}$ of $11$ values, which is still $161$ cm.

The median remains unchanged at $161$ cm.

(a)

For $3$ goals to be the mode, its frequency $12$ must be greater than every other frequency, so $x\lt 12$. (If $x=12$ there would be two modes, $2$ and $3$.)

$$\text{Largest possible }x=11$$

(b)

$$\begin{aligned}\frac{0(2)+1(5)+2(x)+3(12)+4(4)+5(2)}{2+5+x+12+4+2}&=2.5\\ \frac{67+2x}{25+x}&=2.5\\ 67+2x&=62.5+2.5x\\ 4.5&=0.5x\\ x&=9\end{aligned}$$

Total goals $=\sum fx$ and total matches $=\sum f$:

$$\begin{aligned}\sum fx&=0(2)+1(3)+2(8)+3r+4(4)+5(1)+6(2)\\ &=52+3r\\ \sum f&=2+3+8+r+4+1+2\\ &=20+r\end{aligned}$$
$$\begin{aligned}\frac{52+3r}{20+r}&=2.68\\ 52+3r&=53.6+2.68r\\ 0.32r&=1.6\\ r&=5\end{aligned}$$

Class $X$ has $1+4+6+4+6+1=22$ plants and Class $Y$ has $4+5+6+4+2=21$ plants. Class $X$ leaves are read from the stem outwards, so the row $7\ 5\ 3\ 1\mid 13$ means $13.1, 13.3, 13.5, 13.7$ cm.

(a)

$$\begin{aligned}\text{Tallest}-\text{shortest}&=17.0-12.2\\ &=4.8\text{ cm}\end{aligned}$$

(b)

The leaf $2$ appears three times on the stem $16$ for Class $X$, more than any other height.

$$\text{Modal height}=16.2\text{ cm}$$

(c)

Class $Y$ has $21$ plants, so the median is the $11^{\text{th}}$ value. The first $4+5=9$ values are on stems $13$ and $14$; the $10^{\text{th}}$ and $11^{\text{th}}$ values are both $15.0$.

$$\text{Median}=15\text{ cm}$$

(d)

Not healthy means shorter than $14$ cm: the $4$ plants on stem $13$.

$$\begin{aligned}\text{Percentage}&=\frac{4}{21}\times 100\%\\ &=19\tfrac{1}{21}\%\ (\approx 19.0\%)\end{aligned}$$

(e)

Compare the medians. Class $X$ has $22$ plants, so its median is the mean of the $11^{\text{th}}$ and $12^{\text{th}}$ values, $14.9$ and $15.3$:

$$\begin{aligned}\text{Median of Class }X&=\frac{14.9+15.3}{2}=15.1\text{ cm}\\ \text{Median of Class }Y&=15\text{ cm}\end{aligned}$$

Class $X$ had taller plants, because its median height ($15.1$ cm) is higher than that of Class $Y$ ($15$ cm).

In order, Class A: $43, 49, 56, 58, 60, 61, 62, 62, 67, 68, 71, 72, 72, 72, 79, 81, 88, 90, 94, 95$.

Class B: $40, 41, 42, 52, 54, 54, 73, 75, 75, 76, 77, 80, 81, 83, 84, 91, 93, 96, 96, 96$.

(a)

Class A scores of at least $75$: $79, 81, 88, 90, 94, 95$, which is $6$ students.

$$\begin{aligned}\text{Percentage}&=\frac{6}{20}\times 100\%\\ &=30\%\end{aligned}$$

(b)

$72$ appears three times in Class A, more than any other mark.

$$\text{Mode}=72\text{ marks}$$

(c)

The median of $20$ values is the mean of the $10^{\text{th}}$ and $11^{\text{th}}$ values.

$$\begin{aligned}\text{Median of Class B}&=\frac{76+77}{2}\\ &=76.5\text{ marks}\end{aligned}$$

(d)

$$\begin{aligned}\text{Median of Class A}&=\frac{68+71}{2}=69.5\text{ marks}\\ \text{Median of Class B}&=76.5\text{ marks}\end{aligned}$$

Class B performed better, because its median mark ($76.5$) is higher than the median mark of Class A ($69.5$).

(e)

Adding $2$ marks to every score shifts every value up by $2$ without changing the order. So for both classes the mode and the median each increase by $2$ marks: Class A mode $74$, median $71.5$; Class B mode $98$, median $78.5$.

Each class has $20$ students. In order, Class $B$: $52, 55, 64, 65, 66, 71, 72, 72, 74, 75, 79, 82, 83, 83, 85, 87, 90, 92, 95, 95$.

(a)

$$\begin{aligned}\text{Median}&=\frac{10^{\text{th}}+11^{\text{th}}\text{ values}}{2}\\ &=\frac{75+79}{2}\\ &=77\end{aligned}$$

(b)

In Class $A$ the leaf $2$ appears four times on the stem $7$.

$$\text{Mode}=72$$

(c)

$$\begin{aligned}\text{Sum}&=(52+55)+(64+65+66)+(71+72+72+74+75+79)\\ &\quad+(82+83+83+85+87)+(90+92+95+95)\\ &=107+195+443+420+372\\ &=1537\\ \text{Mean}&=\frac{1537}{20}\\ &=76.85\end{aligned}$$

(d)

Class $A$: $50, 51, 57, 65, 65, 66, 67, 68, 72, 72, 72, 72, 75, 75, 81, 82, 82, 84, 93, 97$.

$$\begin{aligned}\text{Sum}&=158+331+438+329+190\\ &=1446\\ \text{Mean of Class }A&=\frac{1446}{20}\\ &=72.3\end{aligned}$$

Class $B$ did better, because its mean score ($76.85$) is higher than the mean score of Class $A$ ($72.3$).

Note: the school's marking scheme gives the Class $A$ mean as $72.5$; the diagram as printed gives $72.3$. The conclusion is the same.

(a)

For Grade A, the leaf $8$ appears three times on the stem $34$; no other mass appears more than twice.

$$\text{Modal mass}=348\text{ g}$$

(b)

Grade A has $29$ peaches, so its median is the $15^{\text{th}}$ value. Stems $34$ and $35$ hold $4+8=12$ values; the stem $36$ read from the stem outwards is $360, 361, 362, \ldots$, so the $15^{\text{th}}$ value is $362$ g.

Grade B has $30$ peaches, so its median is the mean of the $15^{\text{th}}$ and $16^{\text{th}}$ values. Stems $34$ and $35$ hold $6+8=14$ values, so these are $360$ and $36m$.

$$\begin{aligned}\frac{360+(360+m)}{2}&=362\\ 720+m&=724\\ m&=4\end{aligned}$$

(c)

Subtracting $10$ g from every mass keeps the masses in the same order, so the middle value is also $10$ g less. The recorded median mass was $10$ g less than the true median.

(a)

Tally the timings. For $10\lt x\leq 20$: $12.9, 15, 13.6, 14.4, 15.8, 15.8$, which is $6$. For $30\lt x\leq 40$: $39, 35, 31, 35, 36.8, 33.6, 35.7, 34, 35, 40, 37.2$, which is $11$ (note that $40$ belongs here, since $40\leq 40$).

Check: $6+8+11+5=30$.

(b)(i)

No timing is exactly $20$ or $30$, so $8$ runners were between $20$ and $30$ minutes inclusive.

$$\begin{aligned}P(\text{not between 20 and 30})&=1-\frac{8}{30}\\ &=\frac{22}{30}\\ &=\frac{11}{15}\end{aligned}$$

(b)(ii)

$$\begin{aligned}P(\text{more than 30 minutes})&=\frac{11+5}{30}\\ &=\frac{16}{30}\\ &=\frac{8}{15}\end{aligned}$$

(a)

Use the mid-value of each class: $1, 3, 5, 7, 9$.

$$\begin{aligned}\text{Mean}&=\frac{1(10)+3(14)+5(8)+7(6)+9(4)}{10+14+8+6+4}\\ &=\frac{170}{42}\\ &=4.05\text{ books (3 s.f.)}\end{aligned}$$

(b)

The classes all have width $2$, so each bar’s height is its frequency: $10, 14, 8, 6, 4$, with no gaps between bars.

(c)

$4$ satisfies $2\lt 4\leq 4$, so it belongs to the class $2\lt x\leq 4$.

(d)

$$\begin{aligned}P(\text{more than 6 books})&=\frac{6+4}{42}\\ &=\frac{5}{21}\end{aligned}$$

(a)

Use the mid-values $3.5, 4.5, 5.5, 6.5, 7.5, 8.5$. The total frequency is $30+48+54+31+20+7=190$.

$$\begin{aligned}\bar{x}&=\frac{3.5(30)+4.5(48)+5.5(54)+6.5(31)+7.5(20)+8.5(7)}{190}\\ &=\frac{105+216+297+201.5+150+59.5}{190}\\ &=\frac{1029}{190}\\ &=5.4157\ldots\end{aligned}$$

The estimated mean is $5.42 (to the nearest cent).

(b)

With $190$ adults the median lies between the $95^{\text{th}}$ and $96^{\text{th}}$ values. The cumulative frequencies are $30$, $78$, $132, \ldots$, so both values lie in the third class.

The median amount lies in the class $5 $\leq x\lt$ $6.

(c)

$$\begin{aligned}P(\text{spends 6 dollars or more})&=\frac{31+20+7}{190}\\ &=\frac{58}{190}\\ &=\frac{29}{95}\end{aligned}$$

(a)(i)

In order: $15, 18, 25, 25, 28, 36, 38, 42, 42, 42$. With $10$ values the median is the mean of the $5^{\text{th}}$ and $6^{\text{th}}$ values.

$$\begin{aligned}\text{Median}&=\frac{28+36}{2}\\ &=32\text{ drinks}\end{aligned}$$

(a)(ii)

Peak hours are the hours with more than $32$ drinks: $36, 38, 42, 42, 42$.

$$\begin{aligned}\text{Mean}&=\frac{36+38+42+42+42}{5}\\ &=\frac{200}{5}\\ &=40\text{ drinks}\end{aligned}$$

(b)

There are $5$ peak hours with $200$ drinks sold in total. Compare the extra profit from the extra drinks with the extra staff cost.

$$\begin{aligned}\text{Extra drinks}&=25\%\times 200=50\\ \text{Extra profit}&=50\times 3=150\text{ dollars}\\ \text{Cost of extra staff}&=15\times 5=75\text{ dollars}\\ \text{Net gain}&=150-75=75\text{ dollars}\end{aligned}$$

Yes. Hiring an additional staff member during peak hours brings in $150 of extra profit for a cost of $75, a net gain of $75 (or $15 per peak hour), so the proposal is worth making.