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x 1 2 3 4 5 Total f y + 2 y - 1 2y - 3 y + 4 3y - 4 8y - 2 fx y + 2 2y - 2 6y - 9 4y + 16 15y - 20 28y - 13 Mean = \(\frac{\sum fx}{\sum f}\) \(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\) \(\implies 14(28y - 13) = 43(8y - 2)\) \(392y - 182 = 344y - 86\) \(392y - 344y = -86 + 182 \implies 48y = 96\) \(y = 2\)

x = \(\frac{\sum x}{N}\) = \(\frac{18}{3}\) = 6 \(\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}\) M.D = \(\frac{|x - x|}{N}\) = \(\frac{6}{3}\) = 2

Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h N = Sum of frequencies L = lower class boundary of median class f = sum of all frequencies below L fm = frequency of modal class and h = class width of median class Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5 = 11 + (\(\frac{25 - 21}{20}\))5 = 11 + (\(\frac{(4)}{20}\)) 11 + 1 = 12

\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\) x = \(\frac{\sum fx}{\sum f}\) = \(\frac{24}{8}\) = 3 Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\) = \(\frac{18}{8}\) = \(\frac{9}{4}\)

40 = 20 - x + x + 35 - x 40 = 55 - x x = 55 - 40 = 15 ∴ P(both) \(\frac{15}{40}\) = \(\frac{3}{8}\)

(a) x f fx 15 1 15 25 4 100 35 12 420 45 24 1080 55 18 990 65 8 520 75 3 225 70 3350 \(Mean = \frac{\sum fx}{\sum f} = \frac{3350}{70} = 47.8571\) \(\approxeq 47.9\) (to 3 sig. figs) (b)(i) Mode = 45 (score with the highest frequency) (ii) Range = Highest mark - Lowest mark = 75 - 15 = 60.

(a) 1st bag contains 12 white (W) balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls. P(two black balls from the 1st bag and two black balls from 2nd bag) = \((\frac{8}{20} \times \frac{7}{19}) \times (\frac{15}{25} \times \frac{14}{24})\) = \(\frac{14}{95} \times \frac{7}{20}\) = \(\frac{49}{950}\) (b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag) = \(\frac{12}{20} \times \frac{8}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{12}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{10}{25} \times \frac{15}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{15}{25} \times \frac{10}{24}\) = \(\frac{42}{475} + \frac{42}{475} + \frac{7}{190} + \frac{7}{190}\) = \(\frac{238}{950}\)

(a) Age in years No of people Cum Freq 10 8 8 15 19 27 20 10 37 25 7 44 30 6 50 (b) (c)(i) Median age = 14.5 years (ii) Number of people who are at most 15 years of age = 27 people (iii) Number of people who are between 20 and 25 years = people who are 25 years - people who are 20 years = 44 - 37 = 7 people

15, 20, 25, 26, 28, 29, 33, 37, 38, 39 median = \(\frac{28 + 29}{2}\) = 28.5

Mode = 5 Median = 5 (m, n) = (5, 5)

12, 12, 12, 13, 14, 15 Median = \(\frac{12 + 13}{2}\) = 12.5

Prob(less than 3) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

No of possible outcome = 36 Required outcome (R) = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} n(R) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Prob(obtaining a 2) = \(\frac{30}{200}\) = \(0.15\)

Prob(less than 3) = \(\frac{25 + 30}{200}\) = \(\frac{11}{40}\) = 0.275