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Given from 41 to 56 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56 The nos multiple of 9 are: 45, 54 P(multiple of 9) = \(\frac{2}{16}\) = \(\frac{1}{8}\)

Mode = 12

No expanded working has been attached for this question yet. The verified answer is Option C.

Probability that a student in favor is group A = \(\frac{128}{128 + 32 + 96 + 48}\) = \(\frac{128}{304}\) = \(\frac{8}{19}\)

Probability that the student is against the proposal = \(\frac{32 + 48}{128 + 32 + 96 + 48}\) = \(\frac{80}{304}\) = \(\frac{5}{19}\)

(a) Let S = sample space = integers from 5 to 25 inclusive. F = multiple of 5 ; T = multiple of 3. S = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25} F = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24} P(F or T) = \(\frac{5}{21} + \frac{7}{21} = \frac{12}{21} = \frac{4}{7}\) (b) Let B = blue balls and R = red balls n(B) = 4 ; n(R) = 6. (i) P(both are red) = \(\frac{6}{10} \times \frac{6}{10} = \frac{36}{100}\) = \(\frac{9}{25}\) (ii) P(both are same colour) = P(both are red) or P(both are blue) = \(\frac{9}{25} + (\frac{2}{5} \times \frac{2}{5})\) = \(\frac{9}{25} + \frac{4}{25}\) = \(\frac{13}{25}\)

No expanded working has been attached for this question yet. The verified answer is Option E.

Probability = 5/20 = 1/4

Arrange the numbers in order, 1.6, 2.1, 2.1| 2.4, 2.6| 2.6, 2.6, 3.7 n = median = \(\frac{2.4 + 2.6}{2}\) = 2.5 m = mode = 2.6 ∴ (n, m) = (2.5, 2.6)

(a) (b)(i) The mode is 61kg. (ii) From the total number of students = 12, then the median is the average of the 6th and 7th weight. From the table, the median = 61kg. (c) Mean weight = \(\frac{\sum fx}{\sum f} = \frac{55 \times 2 + 57 \times 1 + 59 \times 2 + 61 \times 4 + 63 \times 3}{12}\) = \(\frac{110 + 57 + 118 + 244 + 189}{12} = \frac{718}{12} = 59.833kg\) \(\approxeq 60kg\)

(a) Sample space for a pair of fair dice = 36 results + 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Total with sum at least 9 = 10 P(sum at least 9) = \(\frac{10}{36} = \frac{5}{18}\) (b) P(civil servant owns a car) = \(\frac{1}{6}\) P(civil servant owns no car) = \(\frac{5}{6}\) (i) P(A and B own a car) = \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\) (ii) P(C and D, only one owns a car) = P(C owns, not D) + P(not C but D owns) = \((\frac{1}{6} \times \frac{5}{6}) + (\frac{5}{6} \times \frac{1}{6})\) = \(\frac{5}{36} + \frac{5}{36}\) = \(\frac{10}{36} = \frac{5}{18}\) (iii) P(of X, Y and Z, only one owns a car) = P(X owns a car, not Y, not Z) + P(not X, Y owns a car, not Z) + P(not X, not Y, Z owns a car) = \(3(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6})\) = \(3 \times \frac{25}{216}\) = \(\frac{25}{72}\)

The mode=4. It has the highest frequency of 7

Score 0 1 2 3 4 5 Total Freq 2 3 4 2 7 2 20 \(fx\) 0 3 8 6 28 10 55 \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{55}{20}\) = \(2.75\)

There are 20 students. Hence, the median is found by the average of the 10th and 11th score. = \(\frac{3 + 3}{2}\) = 3

No. of possible outcome = 40 No. of expected outcome = 16 + 9 = 25 Probability = 25/40 = 5/8