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\(\frac{10 + 8 + 5 + 11 + 12 + 9 + 6 + 3 + 15 + 23}{10}\) = \(\frac{102}{10}\) = 10.2

(3 x 0) + (1 x 2) + (5 x 2) + (5 x 3) + (5 x 4) + (10 x 5) + (4 x 6) + (3 x 7) + (1 x 8) + (2 x 9) \(\div\) 40 = \(\frac{168}{40}\) = 4.2

Number of students in the club is 10 + 24 + 8 + 5 + 3 = 50

Median = \(\frac{N}{2}\)th age = \(\frac{50}{2}\)th age = 25th age = 14 years

\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\) From the table above, event space, n(E) = 6 sample space, n(S) = 36 Hence, probability sum of scores is at least 10, is; \(\frac{n(E)}{n(S)}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

n(red balls) = 5 n(blue balls) = 4 n(\(\iff\)) = 9 Hence, prob (R1, B2) = \(\frac{5}{9} \times \frac{4}{9}\) = \(\frac{20}{81}\)

Lower quartile, Q1 = \(\frac{1}{4}\)Nth term = \(\frac{1}{4}\) x 600th term = 150th term = 15.5 + 3 + 0.5 = 19

In probability, or = + Prob( A or B or Both ) = \(\frac{1}{3} + \frac{1}{5} + \frac{1}{15}\) = \(\frac{5+ 3 + 1}{15}\) = \(\frac{9}{15}\) = \(\frac{3}{5}\)

Prob.(A outcome) = \(\frac{1}{6}\), Prob.(B outcome) = ¼ P(A ∪ B) = 1/12, Prob.(A ∩ B) Prob.(A ∩ B) = Prob.(A) + Prob.(B) − Prob.(A ∪ B) 1/6 + 1/4 − 1/12 = (2 + 3 − 1) ÷ 12 = 4/12 = 1/3 Prob.(A ∩ B) = 1/3

To calculate the mean of grouped data, - First step to determine the midpoint (x) of each interval or class. 0 - 4 ►2 5 - 9 ► 7 10 - 14 ►12 These midpoints must then be multiplied by the frequencies of the corresponding classes: 2 X 2 = 4 1 X 7 = 7 2 X 12 = 24 Mean = ( 24 + 7 + 4) \div ( 2 + 1 + 2 ) : Mean = 7

Pr(Ebou did not hit the target) = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\) = Pr(E'), Pr(Omar did not hit the target) = 1 - \(\frac{4}{5}\) = \(\frac{1}{5}\) = Pr(O') Hence the probability that only Kebba will hit the target = Pr(K) x Pr(E') x Pr(O') = \(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\) = \(\frac{1}{30}\)

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

mean = \(\bar{x}\) = \(\frac{\sum x}{n}\) = \(\frac{5 + 3 + 0 + 7 + 2 + 1}{6}\) \(\frac{18}{6}\) = 3 \(\begin{array}{c|c} x & x - \bar{x} & |x - \bar{x}|\\ \hline 5 & -2 & 2 \\ \hline 3 & 0 & 0 \\ \hline 0 & -3 & 3 \\ \hline 7 & 4 & 4 \\ \hline 2 & -1 & 1 \\ \hline 1 & -2 & 2 \\ \hline & & & \sum|x - \bar{x}| \\ \hline & & & 12 \end{array}\) hence, MD = \(\frac{\sum|x - \bar{x}}{n}|\) \(\frac{12}{6}\) = 2

Marks (%) frequency (f) Cumulative frequency Upper class boundaries 0 - 9 7 7 9.5 10 - 19 11 18 19.5 20 - 29 17 35 29.5 30 - 39 20 55 39.5 40 - 49 29 84 49.5 50 - 59 34 118 59.5 60 - 69 30 148 69.5 70 - 79 25 173 79.5 80 - 89 21 194 89.5 90 - 99 6 200 99.5 (b) (i) If 5% of the candidates passed the examination, then (100 - 5)% = 95% passed with a mark \(\leq\) the lowest mark for distinction. \(\text{95% of 200} = \frac{95}{100} \times 200 \) = 190 candidates. From the ogive, 190 corresponds to 79.5 + 8 = 87.5 marks (to one decimal place) (ii) From the ogive, the number of candidates who scored at most 45% is 69. Hence, the probability of selecting a candidate who scored at most 45% = \(\frac{69}{200} = 0.345 \approxeq 0.3\) (1 d.p)

y = x2 - 2x - 3, Then \(\frac{\delta y}{\delta x} = 2x - 2\) But at minimum point,\(\frac{\delta y}{\delta x} = 0\), Which means 2x - 2 = 0 2x = 2 x = 1. Hence the minimum value of y = x2 - 2x - 3 is; ymin = (1)2 - 2(1) - 3 ymin = 1 - 2 - 3 ymin = -4