Question 10 (12 marks):
Diagram shows a histogram which represents the mass, in kg, for a group of 100 students.

Diagram


(a) Based on the Diagram, complete the Table in the answer space.

(b) Calculate the estimated mean mass of a student.

(c) For this part of the question, use graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 10 kg on the horizontal axis and 2 cm to 10 students on the vertical axis, draw an ogive for the data.

(d) Based on the ogive drawn in 10(c), state the third quartile.

Answer:



Solution:
(a)


(b)


$\begin{array}{l}\text{Estimated mean mass}\\ =\frac{8110}{100}\\ =81.1\text{ kg}\end{array}$


(c)



(d)
Third quartile
= 75th student
= 90.0 kg

Question 9 (12 marks):
The data in the Diagram shows the mass, in g, of 30 strawberries plucked by a tourist from a farm.

Diagram

(a) Based on the Diagram, complete Table 3 in the answer space.

(b) Based on Table, calculate the estimated mean mass of a strawberry.

(c) For this part of the question, use graph paper.
By using the scale of 2 cm to 10 g on the horizontal axis and 2 cm to 1 strawberry on the vertical axis, draw a histogram for the data.

(d) Based on the histogram drawn in 14(c), state the number of strawberries with the mass of more than 50 g.

Answer:




Solution:
(a)


(b)
$\begin{array}{l}\text{Estimated mean mass}\\ =\frac{\text{Total }\left[\text{midpoint}\times \text{frequency}\right]}{\text{Total frequencies}}\\ =\frac{\begin{array}{l}2\left(14.5\right)+5\left(24.5\right)+10\left(34.5\right)+8\left(44.5\right)+\\ \text{3}\left(54.5\right)+2\left(64.5\right)\end{array}}{2+5+10+8+3+2}\\ =\frac{1145}{30}\\ =38.17\end{array}$


(c)


(d)
Number of strawberries with the mass more than 50 g
= 3 + 2
= 5 

Question 8 (12 marks):
Diagram shows the marks obtained by a group of 36 students in a Mathematics test.

Diagram

(a) Based on the data in Diagram, complete Table in the answer space.

(b) Based on the Table, calculate the estimated mean mark of a student.

(c) For this part of the question, use graph paper.
By using the scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data.

(d) Based on the frequency polygon in 8(c), state the number of students who obtained more than 40 marks.

Answer:
Table


Solution:
(a)

(b)
$\begin{array}{l}\text{Estimated mean mark}\\ =\frac{\text{Total }\left[\text{Frequency}\times \text{midpoint}\right]}{\text{Total frequency}}\\ =\frac{\begin{array}{l}2\left(28\right)+5\left(33\right)+7\left(38\right)+10\left(43\right)+\\ 8\left(48\right)+4\left(53\right)\end{array}}{2+5+7+10+8+4}\\ =\frac{1513}{36}\\ =42.03\end{array}$

(c)



(d)
Number of students who obtained more than 40 marks
= 10 + 8 + 4
= 22 students

Question 7:
Diagram below shows the marks obtained by a group of 30 students in a history test.

(a) Using data in diagram above and a class interval of 5 marks, complete the table in the answer space.

(b) Based on the completed table in part (a)
(i) find the modal class,
(ii) calculate the mean mark obtained by the students.

(c) By using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data.

Answer:



Solution:
(a)


(b)(i)
Modal class = 50 – 54


(b)(ii)


$\begin{array}{l}\text{Mean mark = }\frac{1615}{30}\\ \text{= 53}\text{.83}\end{array}$

(c)

Question 6:
The table below shows the frequency distribution of the age of 100 scouts in a camping spot.


(a) Based on table above, complete table in the answer space.

(b) Calculate the estimated age, in years, of a scout in the camping spot.

(c) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 5 years on the horizontal axis and 2 cm to 10 scouts on the vertical axis, draw an ogive for the data.

(d) Any scout which is older than 18 years old is considered senior scout.
Based on the ogive drawn in part (c), find the percentage of senior scout in the camping spot.

Answer:



Solution:
(a)


(b)
$\begin{array}{l}\text{Estimated mean age}\\ \text{= }\frac{0\times 2+12\times 7+18\times 12+26\times 17+25\times 22+13\times 27+6\times 32}{100}\\ \text{= }\frac{1835}{100}\\ =18.35\text{ years old}\end{array}$


(c)



(d)
$\begin{array}{l}\text{Number of senior scouts}\\ =100-49\\ =51\\ \text{Percentage of senior scouts}\\ =\frac{51}{100}\times 100\%\\ =51\%\end{array}$

Question 5:

Time (seconds)	Frequency
35 – 39	5
40 – 44	8
45 – 49	9
50 – 54	15
55 – 59	11
60 – 64	2

Solution:
(a) Modal class = time 50 – 54 seconds (highest frequency).

(b)

$\begin{array}{l}\text{Estimated mean}\\ \text{=}\frac{37\times 5+42\times 8+47\times 9+52\times 15+57\times 11+62\times 2}{50}\\ \text{=}\frac{2475}{50}=49.5\end{array}$

(c)
 

Upper Boundary	Cumulative Frequency
34.5	0
39.5	5
44.5	13
49.5	22
54.5	37
59.5	48
64.5	50

(d) 

Question 4:
Diagram below shows the marks obtained by a group of 30 students in a Science test.


(a) Based on the data in diagram above, complete the table in the answer space.   

(b) Based on the completed table in part (a), calculate the estimated mean mark of a student.

For this part of the question, use graph paper.
(c) By using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data.

(d) The passing mark for the test is 44. Using the frequency polygon drawn in part (c), find the number of students who passed the test.  

Answer:

Solution:
(a)


(b)


$\begin{array}{l}\text{Estimated mean mark = }\frac{1255}{30}\\ \text{= 41}\text{.83}\end{array}$

(c)


(d)
Number of students who passed the test
= 9 + 5 + 3
= 17

$\begin{array}{l}\text{(a)}\frac{1}{2}\text{of 100 students}=\frac{1}{2}\times 100=50\\ \text{From the ogive, median,}M=50.5\text{second}\\ \\ \text{(b)}\frac{1}{4}\text{of 100 students}=\frac{1}{4}\times 100=25\\ \text{From the ogive, first quartile,}{Q}_1\text{=}44.5\text{second}\\ \\ \text{(c)}\frac{3}{4}\text{of 100 students}=\frac{3}{4}\times 100=75\\ \text{From the ogive, third quartile,}{Q}_3\text{= 5}4.5\text{second}\end{array}$

Number of magazines	Frequency
220 – 229	3
230 – 239	5
240 – 249	11
250 – 259	6
260 – 269	5

(c) calculate the mean number of magazine sold per day. 

Number of magazines	Frequency (f)	Class midpoint (x)
220 – 229	3	224.5
230 – 239	5	234.5
240 – 249	11	244.5
250 – 259	6	254.5
260 – 269	5	264.5

$\begin{array}{l}\text{Midpoint}=\frac{220+229}{2}=224.5\\ \\ \text{mean, }\overline{x}=\frac{\sum fx}{\sum f}\\ \text{= }\frac{3\times 224.5+5\times 234.5+11\times 244.5+6\times 254.5+5\times 264.5}{30}\\ =\frac{7385}{30}\\ =246.2\end{array}$