(a)(i)

\begin{align*} \text{No. of students who take less than 17.5 minutes} & = 200 \end{align*}

(a)(ii)

\begin{align*} \text{No. of students who take less than 27 minutes} & = 600 \\ \\ \text{No. of students who take at least 27 minutes} & = 750 - 600 \\ & = 150 \\ \\ \text{Fraction of students who take at least 27 minutes} & = {150 \over 750} \\ & = {1 \over 5} \end{align*}

(a)(iii)

\begin{align*} \text{No. of students who take at least } x \text{ minutes} & = 750 \times 40\% \\ & = 300 \\ \\ \text{No. of students who take less than } x \text{ minutes} & = 750 - 300 \\ & = 450 \\ \\ \text{From the graph, } x & = 23 \end{align*}

(b)

\begin{align*} 90\% \text{ of students} & = 750 \times 90\% \\ & = 675 \\ \\ \text{From the graph, 90th percentile} & = 31.5 \end{align*}

(i)

\begin{align*} \boxed{ 5, 9, 9, 10, 12, 13, 15} & \boxed{ 19, 20, 20, 23, 30, 84, 120} \\ \\ \text{Median} & = {15 + 19 \over 2} \\ & = 17 \end{align*}

(ii)

\begin{align*} \text{Interquartile range} & = \text{Upper quartile} - \text{Lower quartile} \\ & = 23 - 10 \\ & = 13 \end{align*}

(iii)

\begin{align*} \text{Mean} & = {\sum fx \over \sum f} \\ & = {1 \times 5 + 2 \times 9 + 1 \times 10 + 1 \times 12 + 1 \times 13 + 1 \times 15 + 1 \times 19 + 2 \times 20 + 1 \times 23 + 1 \times 30 + 1 \times 84 + 1 \times 120 \over 14} \\ & = 27{11 \over 14} \\ \\ \sum fx^2 & = 1 \times 5^2 + 2 \times 9^2 + 1 \times 10^2 + 1 \times 12^2 + 1 \times 13^2 + 1 \times 15^2 \\ & \phantom{00} + 1 \times 19^2 + 2 \times 20^2 + 1 \times 23^2 + 1 \times 30^2 + 1 \times 84^2 + 1 \times 120^2 \\ & = 24871 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {24871 \over 14} - \left(27{11 \over 14} \right)^2 } \\ & = 31.6931 \\ & \approx 31.7 \end{align*}

(a)

\begin{align*} \sum fx & = 5 \times 0 + 6 \times 1 + 4 \times 2 + 3 \times 3 + 2 \times 4 + 1 \times 5 \\ & = 36 \\ \\ \sum f & = 5 + 6 + 4 + 3 + 2 + 1 \\ & = 21 \\ \\ \sum fx^2 & = 5 \times 0^2 + 6 \times 1^2 + 4 \times 2^2 + 3 \times 3^2 + 2 \times 4^2 + 1 \times 5^2 \\ & = 106 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( {\sum fx \over \sum f} \right)^2 } \\ & = \sqrt{ {106 \over 21} - \left(36 \over 21\right)^2 } \\ & = 1.45218 \\ & \approx 1.45 \end{align*}

(b)(i) Since there are 21 children, the median occurs at the 11th children. The lower quartile is the median of the first 10 children and the upper quartile is the median of the last 10 children.

\begin{align*} x_1 & = \text{Lower quartile} \\ & = {0 + 1 \over 2} \\ & = 0.5 \\ \\ x_2 & = \text{Median} \\ & = 1 \\ \\ x_3 & = \text{Upper quartile} \\ & = {3 + 3 \over 2} \\ & = 3 \end{align*}

(b)(ii)

\begin{align*} \text{Interquartile range} & = \text{Upper quartile} - \text{Lower quartile} \\ & = 3 - 0.5 \\ & = 2.5 \end{align*}

(a)(i)

\begin{align*} 50\% \text{ of } 124 & = 124 \times 50\% \\ & = 62 \\ \\ \text{Median} & = 44 \end{align*}

(a)(ii)

\begin{align*} 25\% \text{ of } 124 & = 124 \times 25\% \\ & = 31 \\ \\ \text{Lower quartile} & = 37 \\ \\ 75\% \text{ of 124} & = 124 \times 75% \\ & = 93 \\ \\ \text{Upper quartile} & = 50 \end{align*}

(b)

\begin{align*} \text{Interquartile range} & = \text{Upper quartile} - \text{Lower quartile} \\ & = 50 - 37 \\ & = 13 \end{align*}

(c)(i)

\begin{align*} \text{No. of ears of barley with length less than or equals to 55 mm} & = 109 \\ \\ \text{No. of ears of barley with length greater than 55 mm} & = 124 - 109 \\ & = 15 \end{align*}

(c)(ii)

\begin{align*} \text{No. of ears of barley with length not greater than 25 mm} & = 4 \\ \\ \text{No. of ears of barley with length not greater than 64 mm} & = 122 \\ \\ \text{No. of ears of barley with length greater than 64 mm} & = 124 - 122 \\ & = 2 \\ \\ \text{No. of ears of barley with length not greater than 25 mm or greater than 64 mm} & = 4 + 2 \\ & = 6 \end{align*}

(a)(i)

\begin{align*} \text{Mean distance (Vishal)} & = {\sum fx \over \sum f} \\ & = {1 \times 47 + 1 \times 16 + 1 \times 32 + 1 \times 1 + 1 \times 19 + 1 \times 35 \over 6} \\ & = 25 \\ \\ \text{Mean distance (Jun Wei)} & = {1 \times 20 + 1 \times 9 + 1 \times 16 + 1 \times 43 + 1 \times 13 + 1 \times 4 \over 6} \\ & = 17.5 \end{align*}

(a)(ii)

\begin{align*} \text{For Vishal, } \sum fx^2 & = 1 \times 47^2 + 1 \times 16^2 + 1 \times 32^2 + 1 \times 1^2 + 1 \times 19^2 + 1 \times 35^2 \\ & = 5076 \\ \\ \text{S.D. (Vishal)} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {5076 \over 6} - (25)^2 } \\ & = 14.866 \\ & \approx 14.9 \\ \\ \\ \text{For Jun Wei, } \sum fx^2 & = 1 \times 20^2 + 1 \times 9^2 + 1 \times 16^2 + 1 \times 43^2 + 1 \times 13^2 + 1 \times 4^2 \\ & = 2771 \\ \\ \text{S.D. (Jun Wei)} & = \sqrt{ {2771 \over 6} - (17.5)^2 } \\ & = 12.473 \\ & \approx 12.5 \end{align*}

(b)

\begin{align} & \text{Jun Wei is overall more accurate as his mean distance from the centre of the target is lower.} \\ \\ & \text{Jun Wei is also more consistent as the standard deviation is lower.} \end{align}

(a)(i)

\begin{align*} \text{Median} & = 48 \end{align*}

(a)(ii)

\begin{align*} \text{Interquartile range} & = \text{Upper quartile} - \text{Lower quartile} \\ & = 55 - 42 \\ & = 13 \end{align*}

(b)(i)

\begin{align*} \text{Median} & = 57 \end{align*}

(b)(ii)

\begin{align*} \text{Interquartile range} & = \text{Upper quartile} - \text{Lower quartile} \\ & = 72 - 40 \\ & = 32 \end{align*}

(c)

\begin{align*} \text{No. of students from School } B \text{ who scored less than or equals to 80 marks} & = 140 \\ \\ \text{No. of students from School } B \text{ who scored greater than 80 marks} & = 160 - 140 \\ & = 20 \\ \\ \text{Percentage of students from School } B \text{ who scored greater than 80 marks} & = {20 \over 160} \times 100 \\ & = 12.5 \% \end{align*}

(d)

\begin{align} & \text{Students in School B performed better as a whole as the median score is higher.} \\ \\ & \text{Students in School A have more consistent scores as the interquartile range is lower.} \end{align}

(i) The mid-values of lifespan t as to be used: 650, 750, 850, 950, 1050, 1150, 1250

\begin{align*} p & = {\sum fx \over \sum f} \\ & = {2 \times 650 + 9 \times 750 + 16 \times 850 + 21 \times 950 + 29 \times 1050 + 18 \times 1150 + 5 \times 1250 \over 100} \\ & = 990 \\ \\ \text{For Brightworks, } \sum fx^2 & = 2 \times 650^2 + 9 \times 750^2 + 16 \times 850^2 + 21 \times 950^2 + 29 \times 1050^2 + 18 \times 1150^2 + 5 \times 1250^2 \\ & = 100,010,000 \\ \\ q & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {100,010,000 \over 100} - (990)^2 } \\ & = 141.42 \\ & \approx 141 \\ \\ \\ \text{Since there are } & 100 \text{ Lumina lightbulbs,} \\ 100 & = 8 + 10 + 12 + 16 + r + 18 + 12 \\ 100 & = 76 + r \\ 100 - 76 & = r \\ 24 & = r \\ \\ \text{For Lumina, } \sum fx^2 & = 8 \times 650^2 + 10 \times 750^2 + 12 \times 850^2 + 16 \times 950^2 + 24 \times 1050^2 + 18 \times 1150^2 + 12 \times 1250^2 \\ & = 101,130,000 \\ \\ t & = \sqrt{ {101,130,000 \over 100} - (989.5)^2 } \\ & = 179.415 \\ & \approx 179 \end{align*}

(ii)

\begin{align} & \text{The light bulbs produced by Brightworks have a longer lifespan on the whole} \\ & \text{as the mean lifespan is higher.} \\ \\ & \text{The lifespan of the life bulbs produced by Brightworks are more consistent as} \\ & \text{the standard deviation is lower.} \end{align}

(i)

\begin{align*} \text{Interquartile range for University } A & = \text{Upper quartile} - \text{Lower quartile} \\ & = (3.52 - 2.76) \times 1000 \\ & = 760 \\ \\ \text{Interquartile range for University } B & = (3.16 - 2.64) \times 1000 \\ & = 520 \end{align*}

(ii)

\begin{align*} \text{Median for University } A & = 3.16 \times 1000 \\ &= 3160 \\ \text{Median for University } B & = 3.0 \times 1000 \\ & = 3000 \\ \\ \\ \text{Yes, I agree with the statement} & \text{ since the median salary for university A is higher} \end{align*}

(iii)

\begin{align} & \text{University A. The upper quartile (top 25%) for university A is \$3520} \\ & \text{while the upper quartile (top 25%) for university B is \$3160.} \\ \\ & \text{Thus a higher proportion of fresh graduates in university A get more than} \\ & \text{\$3500 for their starting salary.} \end{align}

(a)(i)

\begin{align*} \text{Median} & = 49 \end{align*}

(a)(ii) If 60% of the students pass the exam, then 40% of the students failed

\begin{align*} 40\% \text{ of students} & = 600 \times 40\% \\ & = 240 \\ \\ \text{From the graph, passing mark} & = 44 \end{align*}

(b)

\begin{align*} \text{Upper quartile} & = 64 \\ \\ \text{Lower quartile} & = 36 \\ \\ \text{Interquartile range} & = 64 - 36 \\ & = 28 \end{align*}

(c)

\begin{align*} \text{Median mark} & = 68 \\ \\ \text{Interquartile range} & = 82 - 58 \\ & = 24 \\ \\ \\ \text{Students in Fermat High} & \text{ School performed better} \\ \text{as a whole since the med} & \text{ian mark is higher.} \\ \\ \text{The scores obtained by st} & \text{udents in Fermat High School are} \\ \text{more consistent as the int} & \text{erquartile range is lower.} \end{align*}

(a)

(b)(i) The mid-values are 190, 210, 230, 250, 270

\begin{align*} \text{Mean} & = {\sum fx \over \sum f} \\ & = {18 \times 190 + 42 \times 210 + 15 \times 230 + 4 \times 250 + 1 \times 270 \over 80} \\ & = 212 \end{align*}

(b)(ii)

\begin{align*} \sum fx^2 & = 18 \times 190^2 + 42 \times 210^2 + 15 \times 230^2 + 4 \times 250^2 + 1 \times 270^2 \\ & = 3,618,400 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {3,618,400 \over 80} - (212)^2 } \\ & = 16.911 \\ & \approx 16.9 \end{align*}

(c)

\begin{align} & \text{Since the median is the same, the new curve will pass through the point (40, 211).} \\ \\ & \text{Since the standard deviation is larger, the middle 50% of the new curve will be more} \\ & \text{gentle compared to the given curve.} \end{align}

(a)(i)

\begin{align*} \text{Median mass} & = 54.5 \end{align*}

(a)(ii)

\begin{align*} \text{Interquartile range} & = \text{Upper quartile} - \text{Lower quartile} \\ & = 59 - 51.5 \\ & = 7.5 \end{align*}

(a)(iii)

\begin{align*} \text{Percentage of eggs in Grade 3} & = {45 \over 200} \times 100 \\ & = 22.5 \% \\ \\ \text{Percentage of eggs in Grade 2} & = {165 - 45 \over 200} \times 100 \\ & = 60 \% \\ \\ \text{Percentage of eggs in Grade 1} & = 100\% - 22.5\% - 60\% \\ & = 17.5 \% \end{align*}

(b)(i)

\begin{align*} \text{Median mass} & = 50 \\ \\ \text{Interquartile range} & = 52 - 45 \\ & = 7 \end{align*}

(b)(ii)

\begin{align} & \text{Rainbow Farm produce higher quality eggs as a whole since the median mass of eggs} \\ & \text{produced is higher.} \\ \\ & \text{Rainbow Farm produce a larger proportion of Grade 1 and Grade 2 eggs since the lower} \\ & \text{quartile mass of eggs produced is higher and graded Grade 2.} \end{align}