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(a) Class Interval Tally Freq 45 - 49 |||| | 6 50 - 54 |||| |||| 9 55 - 59 |||| |||| 10 60 - 64 |||| |||| || 12 65 - 69 |||| || 7 70 - 74 |||| | 6 (b) Class Interval Mid-value (x) x - 62 \((x - 62)^{2}\) \(f\) \(f(x - 62)\) \(f(x - 62)^{2}\) 45 - 49 47 -15 225 6 -90 1350 50 - 54 52 -10 100 9 -90 900 55 - 59 57 -5 25 10 -50 250 60 - 64 62 0 0 12 0 0 65 - 69 67 5 25 7 35 175 70 - 74 72 10 100 6 60 600 \(\sum\) 50 -135 3275 \(Mean (\bar{x}) = A + \frac{\sum f(x - A)}{\sum f}\) = \(62 + \frac{-135}{50}\) = \(62 - 2.7 = 59.3\) Standard deviation = \(\sqrt{\frac{\sum f(x - A)}{\sum f}}\) = \(\sqrt{\frac{3275}{50}}\) = \(\sqrt{65.5}\) = \(8.093 \approxeq 8.1\) (to 1 decimal place)

(b) |AC| = 7cm ; |QB| = 2.9cm.

(a) Distance along arc PQ = \(\frac{\theta}{360^\circ} \times 2 \pi r\) where \(r = R \cos \theta\) \(|PQ| = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 6400 \times \cos 40^\circ\) \(|PQ| = \frac{1}{6} \times 2 \times \frac{22}{7} \times 6400 \times 0.7660\) = \(5134.74 km \approxeq 5135km\) (to the nearest whole number) (b) Difference in latitude between Q and T = 40^\circ - 28^\circ = 12^\circ \(\therefore |QT| \text{along the line of longitude} = \frac{12}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\frac{1}{30} \times 2 \times \frac{22}{7} \times 6400 = 1340.95km \approxeq 1341km\) (c) Average speed = \(\frac{\text{Total distance covered}}{\text{Total time taken}}\) = \(\frac{5134.74 + 1340.95}{12} = 539.64km/hr\) \(\approxeq 540km/hr\) (to 3 significant figures)

< QSR = 90^\circ (angle in a semi-circle) \(\therefore\) < SQR = 180^\circ - (90^\circ + 62^\circ) = 28^\circ

\(\frac{0.009}{0.012}\) =\(\frac{0.009}{0.012}\) = \(\frac{9}{12}\) = 0.75 =\(7.5 \times 10^{-1}\)

XZ = 40 x sin30o = 20m

P = 2x (angle in the same segment); x = p/2

Area of ABDE = b x h 12.2 x 4 = 48.8cm 2

From the graph; when sin x = cos x x = 45º

L = 14cm, r = 7cm; \(\theta\)= ? L = \(\frac{\theta}{360}\) x 2\pir 14= \(\frac{\theta\}{360}\) x 2\pi(7) \(\theta = \frac{360 \times 14}{44}\) =114.55^\circ

tan\(\theta\) = 9/20 = 0.45 \(\theta\) = tan\(^{-1}\) 0.45 = 24.2^\circ the bearing of y from x = 180 0 - 0 = 180^\circ - 24.2^\circ the bearing = 155.8^\circ \(\approxeq\) 156^\circ

In \(\Delta DOB\), let < DOB = \(\alpha\) In \(\Delta DOB\), \(5^2 = 3^2 + s^2\) \(s^2 = 25 - 9 = 16\) \(s = 4cm\) \(\sin \alpha = \frac{4}{5}\) \(\alpha = \frac{< AOB}{2}\) Length of chord = \(2r \sin (\frac{\theta}{2})\) |OB| = r = 5cm L = \(2(5)(\frac{4}{5})\) = 8 cm

Total population = 1400 + 550 + 200 + 150 + 400 = 2700 grown up girls = 200 = \(\frac {200}{2700}\) x \(\frac {360}{1}\) =80/3 = 26.7^\circ

If the data is arranged either in ascending or descending order, the 15th and 16th scores are in the middle, which is 5 and 5, respectively median = \(\frac{5 + 5}{2}\) = 5 since n is even i.e, 30

< PBQ = 180^\circ - 75^\circ (opp angles of a cyclic quadrilateral) = 105^\circ