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Mean x = \(\frac{\sum x}{n}\) = [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1)] \(\div\) 5 = [11 - 1 + 3t - 3t] \(\div\) 5 = 10 \(\div\) 5 = 2

The number with the highest frequency = 4

First arrange the numbers in order of magnitude; 1,2,3,3,4,5,5,5,5,6,7,8,9,9,10 Hence the median = 5

Let E demote the event of obtaining at least a 4 Then n(E) = 16 + 10 + 14 = 40 Hence, prob (E) = \(\frac{n(E)}{n(S)}\) \( = \frac{40}{100}\) \( = \frac{2}{5}\)

at most 2 marks = 5 + 2 + 4 students = 11 students probability(at most 2 marks) = \(\frac{11}{20}\)

\(\begin{array}{c|c} height(x) & frequency(f) & fx \\ \hline 2 & 2 & 4\\ 3 & 4 & 12\\ 4 & 5 & 20 \\ 5 & 3 & 15\\ 6 & 1 & 6\end{array}\) mean\(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{57}{15}\) = 3.8

prob(p) = \(\frac{1}{5}\) prob(Q) = \(\frac{1}{4}\) Prob(neither p) = 1 - \(\frac{1}{5}\) \(\frac{5 - 1}{5} = \frac{4}{5}\) prob(neither Q) = 1 - \(\frac{1}{4}\) \(\frac{4 - 1}{4} = \frac{3}{4}\) prob(neither of them) = \(\frac{4}{5} \times \frac{3}{4} = \frac{12}{20}\) = \(\frac{3}{5}\)

at most 2 marks = 5 + 2 + 4 students = 11 students probability(at most 2 marks) = \(\frac{11}{20}\)

Let A and B denote the events that the man wins contracts X and Y respectively. Then P(A) = 0.5 P(A') = 1 - 0.5 = 0.5 P(B') = 0.3 P(B) = 1 - 0.3 = 0.7 (a) The probability that the man wins both contracts = \(0.5 \times 0.7 = 0.35\). (b) The probability that the man wins exactly one of the contracts is \(P(A) \times P(B') + P(B) \times P(A')\) = \(0.5 \times 0.3 + 0.7 \times 0.5\) = \(0.15 + 0.35\) = \(0.50\) (c) Neither of the contracts (i.e not X, not Y) = \(P(A') \times P(B')\) = \(0.5 \times 0.3\) = \(0.15\)

(a) Scores (x) Frequency (f) \(fx\) 1 2 2 2 5 10 3 13 39 4 11 44 5 9 45 6 10 60 \(\sum f = 50\) \(\sum fx = 200\) \(Mean (\bar{x}) = \frac{\sum fx}{\sum f} = \frac{200}{50} = 4\) \(d = x - 4\) \(|d|\) f \(f|d|\) -3 3 2 6 -2 2 5 10 -1 1 13 13 0 0 11 0 1 1 9 9 2 2 10 20 \(\sum f|d| = 58\) Hence, Mean Deviation = \(\frac{\sum f|d|}{\sum f} = \frac{58}{50}\) = 1.16 (b) Let E denote the event of getting a score of at least 4. \(n(E) = 11 + 9 + 10 = 30\) \(p(E) = \frac{n(E)}{n(S)} = \frac{30}{50}\) = \(0.6\)

\(\sum x\) = (t + 2) + (2t + 4) + (3t + 2) + 2t = 8t N = 4_ ∴ Mean, x = \(\frac{\sum x}{N} = \frac{8t}{4} = 2t\) = 2t

N = \(\sum f\) = 15 Hence the median age is the \(\frac{N + 1}{2}\)th age, i.e. \(\frac{15 + 1}{2}\)th = 8th From the table, the age that falls on the 8th position when arranged in ascending order is 25years

Range = Highest score - Lowest score = 10 - 3 = 7

\((1 \leq x \leq 25)\) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25} Number N of x divisible by both 2 and 3 is 4. n(\(\varepsilon\)) = 25 = \(\frac{N}{n(\varepsilon)}\) = \(\frac{4}{25}\)

n(apples) = 9 n(bananas) = 8 n(oranges) = 7 n(\(\varepsilon\)) = 24 Hence Prob(not apple, nor orange) = Prob(banana) = \(\frac{8}{24}\) = \(\frac{1}{3}\)