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Mark(χ) f fχ 1 2 2 2 2 4 3 8 24 4 4 16 5 4 20 ∑f = 20 ∑fχ = 66 \(Mean = \frac{\sum fx}{\sum f}\) = \(\frac{66}{20}\) = 3.3

pr. (More than one pet) = \(\frac{\text{No. of students with > 1 pet}}{\text{total no. of students}}\) = \(\frac{5 + 10 + 3}{30}\) = \(\frac{18}{30}\) = \(\frac{3}{5}\)

mean age = \(\frac{\text{sum of ages}}{\text{no. of men}}\) 50 = \9\frac{sum}{R}\) sum = 50R.....(1) Sum of ages of the men that left = 55 + 63 = 188 remaining sum = 50R - 118 remaining no. of men = R - 2 now mean age = 50 - 1 = 49 years 49 = \(\frac{50R - 118}{R - 2}\) 49(R - 2) = 50R - 118 49R - 50R = -188 - 98 -R = -20 R = 20

mean \(\leq\) 5; \(\frac{2 + 5 + 2x + 7}{4}\) \(\leq\) 5 = \(\frac{14 + 2x}{4} \leq 5\) = 14 + 2x \(\leq\) 5 x 4 14 + 2x \(\leq\) 20 ; 2x \(\leq\) 20 - 14 2x \(\leq\) 20 - 14 2x \(\leq\) 6 x \(\leq\) \(\frac{6}{2}\) x \(\leq\) 3

(a) Class Intervals freq cum freq class boundaries 20 - 29 10 10 19.5 - 29.5 30 - 39 18 28 29.5 - 39.5 40 - 49 22 50 39.5 - 49.5 50 - 59 25 75 49.5 - 59.5 60 - 69 16 91 59.5 - 69.5 70 - 79 9 100 69.5 - 79.5 (b) (c) (i) Median (\(Q_{2}\)) = 49.5 (ii) Semi- Interquartile range = \(\frac{1}{2} (Q_{3} - Q_{1})\) \(Q_{1} = 38.5\); \(Q_{3} = 59.5\). Semi- Interquartile range = \(\frac{1}{2} (59.5 - 38.5)\) = \(\frac{1}{2} (21)\) = 10.5. (iii) Pr(\(\geq 60kg = \frac{25}{100} = \frac{1}{4}\)

x f fx 5 3 15 8 2 16 6 4 24 k i k ∑f=10 ∑fx=55+k Mean = \(\frac{∑fx}{∑f}\) 5.7 = \(\frac{55 + k}{10}\) 57 = 55 + k k = 57 - 55 k = 2

x ∴x x-\(\overline{X}\) /x-\(\overline{X}\)/ x 1 -1 1 2x 2 0 0 x+1 2 0 0 3x 3 1 1 ∑/x-\(\overline{X}\)/ = 2 Mean(\(\overline{X}\)) = \(\frac{ x + 2x + x + 1 + 3x}{4}\) = 2 2 = \(\frac{7x + 1}{4}\) 8 = 7x + 1 7x = 8 - 1 = 7 x = \(\frac{7}{7}\) = 1 Mean deviation = \(\frac{∑(| x - \overline{X}|)}{n}\) M.D = \(\frac{2}{4}\) = \(\frac{1}{2}\) = 0.5

To find the number of ways to seat 9 people in 3 chairs, we use the permutation formula: \(P(n, r) = \frac{n!}{(n - r)!}\) where \( n \) is the total number of people and \( r \) is the number of chairs. Here, \( n = 9 \) and \( r = 3 \): \(P(9, 3) = \frac{9!}{(9 - 3)!} = \frac{9!}{6!}\) Calculating this, we have: \(P(9, 3) = 9 \times 8 \times 7 = 504\) Thus, the number of ways to seat 9 people in 3 chairs is 504 ways

\(\frac{x + 50 + 70}{3} = 55\) x + 120 = 3(55) x + 120 = 165 = 165 - 120 x = 45%

22 35 41 63 74 82 Median = \(\frac{41 + 63}{2} = \frac{104}{2}\) = 52

Total no of balls = 30 Let x = no. of red balls Pr(red) = \(\frac{x}{30}\) Pr(black) = \(\frac{3}{10} = \frac{9}{30}\) Pr(white) = \(\frac{2}{5} = \frac{12}{30}\) No. of black balls = 9 No. of white balls = 12 9 = 12 + x = 30 x = 30 - 21 x = 9 No. of red balls = 9 OR Let w = no of white balls B = no of black balls R = no of red balls Pr(black balls) = \(\frac{3}{10}\) Pr(white balls) = \(\frac{2}{5}\) But Pr(white balls) = \(\frac{2}{5}\) = \(\frac{W}{30}\) cross multiply W = no of white balls = 12 Pr(black balls) = \(\frac{3}{10}\) = \(\frac{B}{30}\) cross multiply B = no of black balls = 9 Total number of balls in the basket = W + B + R = 30 ( given) = 12 + 9 + R = 30 R = 30 - 12 - 9 = 9 Therefore, the no of red balls = 9.

\(\frac{\text{education}}{\text{graduates}} = \frac{108}{360} \times 4320\) = 1296 Pr(Education) = \(\frac{1296}{4320} = \frac{1}{10}\) pr(not education) = 1 - \(\frac{3}{10}\) = \(\frac{10 - 3}{10}\) = \(\frac{7}{10}\)

\(Mean = \frac{t+2t-1+t-2+2t-1+4t+2t+2}{6}\) Mean = \(\frac{12t-2}{6}\) =\(\frac{2(6t-1)}{6}\) =\(\frac{6t-1}{3}\) =\(\frac{6t}{3}- \frac{1}{3}\) =2t-\(\frac{1}{3}\)

4, 4, 1, 0, 4, 4, 2, 0 Arrange in ascending order: 0, 0, 1, 1, 2, 4, 4, 4, 4 Position of median = N+1 / 2 = 9+1 / 2 = 5th Median = 2

X = 20/4 = 5 mean deviation = 8/4 = 2