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Hint: prepare a three-columned table, one for (x), another for the (deviation), and the last for the (squared-deviation). Sum of (x) = 15x, algebraic sum of (deviation) = zero (0) Sum of (squared-deviation) = 10x\(^2\) Variance = \(\frac{∑(squared-deviation)}{\text{n}}\) = \(\frac{10x^2}{5}\) = 2x\(^2\)

Range = Highest - lowest number => 10-4 = 6 Mode is the number with highest occurrence => Mode = 10 Sum = 6 + 10 = 16

If the total number of students represents 100%, then 70% corresponds to the point on the graph where the cumulative frequency reaches 70 From the graph, find where the cumulative frequency line intersects the 70% mark. This typically corresponds to a specific age on the X-axis

The probabilities of the three boys winning are independent and are given as: - Boy A: \( \frac{1}{2} \), - Boy B: \( \frac{1}{3} \), - Boy C: \( \frac{1}{4} \) "One and only one wins" means exactly one boy wins (the other two lose). The three possible cases are: A wins, B loses, C loses \( P = \frac{1}{2} \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} \) B wins, A loses, C loses \( P = \left(1 - \frac{1}{2}\right) \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} \) C wins, A loses, B loses \( P = \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \frac{1}{4} = \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} = \frac{2}{24} \) Add them up: \( \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24} \)

Let the number of people that speak both English and French = x Then (7 - x) + x + (6 - x) = 11 13 - x = 11 \(\implies\) x = 2. \(\therefore\) P(picking a person that speaks both languages) = 2/11

P(man picked at random wears size 9 shoes) = \(\frac{28}{100}\) = \(\frac{7}{25}\)

Marks 3 4 5 6 7 8 Frequency 5 x - 1 x 9 4 1 fx 15 4x - 4 5x 54 28 8 \(\sum fx = 15 + 4x - 4 + 5x + 54 + 28 + 8 = 101 + 9x\) \(\sum f = 5 + x - 1 + x + 9 + 4 + 1 = 18 + 2x\) \(\bar{x} = \frac{\sum fx}{\sum f}\) \(5 = \frac{101 + 9x}{18 + 2x} \implies 101 + 9x = 5(18 + 2x)\) \(101 + 9x = 90 + 10x \implies 101 - 90 = 10x - 9x\) \(11 = x \) (b)(i) Mode = 5. (ii) Median Frequency = 18 + 2(11) = 40. Median position = \(\frac{40}{2} = 20\) 20th position = 5. (c) No of students that scored at least 7 marks = 4 + 1 = 5. Probability of scoring at least 7 marks = \(\frac{5}{40} = \frac{1}{8}\)

Let the four numbers be \(a, b, c, d\). - Mean of the four numbers = 5 → a + b + c + d = 20 - Mean deviation of the four numbers = 3 Mean deviation = \(\frac{|a-5| + |b-5| + |c-5| + |d-5|}{4} = 3\) → Sum of absolute deviations = 12 - Mean deviation of the first three numbers a, b, c = 2 → \(\frac{|a-5| + |b-5| + |c-5|}{3} = 2\) → Sum of absolute deviations for first three = 6 - Therefore, absolute deviation for the fourth number: |d - 5| = 12 - 6 = 6 The fourth number = 6 + 5 = 11.

When the die is rolled, occasion of getting 4 P(obtaining a 4) = \(\frac{1}{6}\) = \(\frac{1}{6}\)

Balls were picked without replacement n( white balls) = 2, n( black balls ) = 3 Total Balls = 2 + 3 = 5 balls Pr( two balls of different colours ) = WB OR BW = \(\frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{2}{4}\) = \(\frac{6}{20} + \frac{6}{20}\) = \(\frac{12}{20}\) = \(\frac{3}{5}\)

mean(x) = \(\frac{\sum x}{N}\) = \(\frac{48}{8}\) = 5.875 re-arranging the numbers; 2, 3, 5, 6, 2, 7, 8, 9 median = \(\frac{6 + 7}{2}\) = \(\frac{1}{2}\) = 6.5 m + 2n = 5.875 + (6.5)2 = 13 + 5.875 = 18.875 = \(\approx\) = 19

\(\begin{array}{c|c} \text{Class intervals} & F & \text{Class boundary}\\ \hline 5 - 7 & 17 & 4.5 - 9.5\\ 10 - 14 & 32 & 9.5 - 14.5\\ 15 - 19 & 25 & 14.5 - 19.5\\ 20 - 24 & 24 & 19.5 - 24.5 \end{array}\) mode = 9.5 + \(\frac{D_1}{D_2 + D_1}\) x C = 9.5 + \(\frac{5(32 - 17)}{2(32) - 17 - 25}\) = 9.5 + \(\frac{75}{22}\) = 12.91 \(\approx\) 12.9

mean (x) = \(\frac{\sum x}{N}\) = k + k + 1 + k + 3 = \(\frac{3k + 3}{3}\) = k + 1 \(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline k & -1 & 1 \\ k + 1 & 0 & 0\\ k + 2 & 1 \\ \hline & & 2\end{array}\) Variance (52) = \(\frac{\sum (x - x)^2}{N}\) = \(\frac{2}{3}\)

Number of red balls = 16, Number of blue balls = 20 Let x represent the No of white balls to be added ∴ Total number of balls = 36 + x Pr( red ball) = \(\frac{ 16}{ 36 + x } = \frac{2}{5}\) (remember that Pr( red ball ) = \frac{2}{5}\)) cross multiplying 2(36 + x) = 80 2x = 80 - 72 2x = 8 x = \(\frac{8}{2}\) = 4

To find the probability that either team X or team Y wins the gold, we can use the formula for the probability of the union of two events: P(x or y) = p(x) + p(y) - p(x n y) But p(x n y ) = 0 p(x) = \(\frac{2}{5}\) p(y) = \(\frac{1}{3}\) p(x or y) = p(x ∪ y) = p(x) + p(y) = \(\frac{2}{5}\) + \(\frac{1}{3}\) = \(\frac{11}{5}\)