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Median = 0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5 Median mark = \(\frac{3+3}{2} = \frac{6}{2} = 3\)

Prob (2 or 3) = prob(2) + prob(3) Prob (mark 2) \(=\frac{7}{22}\) + Prob(mark 3) = \(\frac{5}{22}\) Prob(2 or 3) \(=\frac{7}{22}+\frac{5}{22}=\frac{12}{22}=\frac{6}{11}\)

Prob(RW or WR) = Prob(RW) + Prob(WR) \(Prob(R) = \frac{3}{5}:Prob(W) = \frac{2}{5}\\ Prob(RW\hspace{1mm}or\hspace{1mm}WR)=\frac{3}{5}\times\frac{2}{5}+\frac{2}{5}\times \frac{3}{5}\\ \frac{6}{25}+\frac{6}{25}=\frac{12}{25}\)

Given the first term \( a = 1.2 \) and common difference \( d = 2 \), the first five terms of the AP are: \(a_1 = 1.2\) \(a_2 = 1.2 + 2 = 3.2\) \(a_3 = 1.2 + 4 = 5.2 \) \(a_4 = 1.2 + 6 = 7.2 \) \(a_5 = 1.2 + 8 = 9.2\) The sum of the first five terms is: \(\text{Sum} = 1.2 + 3.2 + 5.2 + 7.2 + 9.2 = 26\) The mean is calculated as: \(\text{Mean} = \frac{26}{5} = 5.2\) Thus, the mean of the first five terms is \(5.2\).

The mean of 2, 3, 3, 3, 4,4, and 3 = \(\frac{ 2 + 3 + 3 + 3 + 2+ 4 + 4 +3}{ 8} = \frac{24}{8}\) = 3 The mode of 2, 3, 3, 3, 4, 4, and 3 = 3 Hence, the mean of 2,3,3,3,4,4, and 3 divided by its mode = \(\frac{3}{3}\) = 1

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

\(Mean = \frac{30+56+31+55+43+44}{6}\) = \(\frac{259}{6}\) = 43.167 Median = 30,31,43,44,55,56 = \(\frac{43+44}{2}\) =\( \frac{87}{2}\) = 43.5 Median-mean = 43.5 - 43.17 = 0.33

Mean = \(\frac{(0 \times 1) + (1 \times 2) + (2 \times 2) + (3 \times 1) + (4 \times 9)}{1 + 2 + 2 + 1 + 9}\) = \(\frac{(0 + 2 + 4 + 3 + 36)}{15}\) = \(\frac{45}{15}\) =3

Black balls = 5 Red balls = 3 P(B) = \(\frac{5}{8}\), P(R) = \(\frac{3}{8}\) P( a Black and a Red) = black ball and red ball or red ball and black ball i.e BR + RB = (\(\frac{5}{8}\)) * (\(\frac{3}{7}\)) + (\(\frac{3}{8}\)) * (\(\frac{5}{7}\)) = (\(\frac{15}{56}\)) + (\(\frac{15}{56}\)) = \(\frac{30}{56}\) = \(\frac{15}{28}\)

From the given table, the most common mark is 3 with a frequency of 18

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

Member that are neither doctors nor lawyers = 60-(15+9)=36 Probability (Not doctors ad not lawyers) \(=\frac{36}{60}\\ =\frac{6}{10}=\frac{3}{5}\)

The two dice produce 36 equally likely outcomes in total. The sums divisible by 3 are exactly 3, 6, 9, and 12. The complete list of favorable pairs \((d_1, d_2)\) is: Sum = 3 (2 ways): (1,2), (2,1) Sum = 6 (5 ways): (1,5), (2,4), (3,3), (4,2), (5,1) Sum = 9 (4 ways): (3,6), (4,5), (5,4), (6,3) Sum = 12 (1 way): (6,6) Total favourable outcomes: 2 + 5 + 4 + 1 = 12. Therefore, the probability is \(\frac{12}{36} = \frac{1}{3}.\)

\(\begin{array}{c|c} \text{Age in years} & 10 & 20 & 30 & 40 & 50 \\\hline \text{frequency} & 3 & 4 & 5 & 6 & 2\end{array}\) \(\sum f \) = number of people involved in the pollution \(\sum f\) = 3 + 4 + 5 + 6 + 2 = 20

x 0 1 2 3 4 5 7 2 7 7 2 3 9 fx 0 7 14 6 12 45 (a)(i) Median = \(\frac{1}{2} (N + 1)\) = \(\frac{1}{2}(30 + 1) = 15.5\) = 15th and 16th position = \(\frac{2 + 2}{2} = 2\) (ii) Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{84}{30}\) = \(2.8 \approxeq 3\) (nearest whole no). (b)