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\(\int_0^1 4x - 6^3\sqrt x^2\) dx =\(\int_0^1 4x - 6x^{2/3}\) dx =\((2x^2 - \frac{8x^{5/3}}{5})_0^1\) =\((2(1)^2 - \frac{18(1)^{5/3}}{5}) - (2(0)^2 - \frac{18(0)^{5/3}}{5}\) = - \(\frac{8}{5}\) - 0 - \(\frac{8}{5}\)

Class Interval Class Mark Frequency (f) fx 5 - 6 5.5 29 5.5 x 29 = 159.5 7 - 8 7.5 40 7.5 x 40 = 300 9 - 10 9.5 38 9.5 x 38 = 361 \(\sum f = 107\) \(\sum fx = 820.5\) Mean = \(\frac{\sum fx}{\sum f} = \frac {820.5}{107}\) = 7.7 (1 d.p)

The first five prime numbers are 2, 3, 5, 7 and 11 M.D = \(\frac {\sum (x - x̄)}{n}\) n = 5 \(x̄ = \frac {\sum x}{n} = \frac {2 + 3 + 5 + 7 + 11}{5} = \frac {28}{5} = 5.6\) M.D = \(\frac {(2 - 5.6) + (3 - 5.6) + (5 - 5.6) + (7 - 5.6) + (11 - 5.6)}{5}\) = \(\frac {(-3.6) + (-2.6) + (-0.6) + (1.4) + (5.4)}{5} = \frac {13.6}{5}\) Take the absolute value of all numbers in bracket = \(\frac {3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac {13.6}{5}\) \(\therefore M.D = 2.72\)

\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) th = 54th value i.e the median age is in the interval "7 - 8" (29 + 25 = 54) Median = \(1_m + (\frac {^{\sum f} /_2 - f_b}{f_m})\)c \(l_m\)=lower class boundary of the median age = 6.5 \(\sum f\) = 107 \(f_b\) = sum of all frequencies before the median age = 29 \(f_m\)=frequency of the median age = 40 c = class width = 8.5 - 6.5 = 2 Median = 6.5 + \((\frac {^{107}/_2 - 29}{40})\) x 2 = 6.5 + \((\frac {53.5 - 29}{40})\) x 2 = 6.5 + \((\frac {24.5}{40})\) x 2 = 6.5 + 1.225 \(\therefore\) median age = 7.725

Number of students who scored atleast 25% = 5 + 3 + 8 = 16

Total number of students = total frequency = 3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53 Since the failed mark was 4, Total number of students that passed = 12 + 7 + 5 + 10 = 34 Pr( passed) = \(\frac{34}{53}\) = 0. 64

Number of students that scored at least 3 marks = 9 + 1 + 12 + 7 + 5 + 10 = 44

x̄ = \(\frac{ 14 + 15 + 16 +17 + 18 + 19}{6} = \frac{99}{6}\) = 16.5 M . D = \(\frac{ (14 - 16.5) + (15 - 16.5) + (16 - 16.5) + (17 - 16.5) + (18 - 16. 5) + (19 - 16.5) }{ 6}\) M.D = \(\frac{(-2.5) + (-1.5) + (-0.5) + (0.5) + (1.5) + (2.5)}{6}\) Taking the absolute value of the deviations M.D = \(\frac{2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5}{6}\) M.D = \(\frac{9}{6}\) = 1.5

a. X F FX 3 4 12 4 6 24 5 4 20 6 5 30 7 6 42 8 2 16 Σf =27 Σfx = 144 Median position = \(\frac{∑F + 1}{2}\) (since ∑F is odd so we added 1) \(\frac{27 + 1}{2} = \frac{28}{2}\) = 14th position. counting down the frequency column, the median position occurs at 4. Therefore, the median height is 5m. b i. mean(x̄) = \(\frac{Σfx}{Σf}\) = \(\frac{144}{27}\) ∴ x̄ = 5.3m ( to 1 d.p) ii. x f fx \(fx^2\) 3 4 12 (3 x 3)4 =36 4 6 24 (4 x 4)6 =96 5 4 20 ( 5 x 5)4 =100 6 5 30 (6 x 6)5 =180 7 6 42 (7 x 7)6 =294 8 2 16 (8 x 8)2 =128 Σf =27 Σfx = 144 Σ\(fx^2\) = 834 Standard deviation = \(\sqrt{vaiance} = \sqrt{\frac{Σfx^2}{Σf} - (\frac{Σfx}{Σf})^2}\) S D = \(\sqrt{\frac{834}{27} - (\frac{144}{27})^2}\) S.D = \(\sqrt{30.888 - (5.333)^2}\) = \(\sqrt{30.888 - 28.441}\) = \(\sqrt{2.447}\) = 1.5738 ≈ 1.6 ( to 1 d.p)

Average performance rating of Team B = \(\frac{7+9+1+9+6}{5} = \frac{32}{5}\) = 6.4 Average performance rating of Team A = \(\frac{5+3+6+10+2}{5} = \frac{26}{5}\) = 5.2 ∴ The difference in the average performance ratings between Team B and Team A = 6.4 - 5.2 = 1.2

Total number of students that scored at most 50 marks = 100 + 80 + 60 + 40 + 80 = 360

SYLLABUS has 8 letters, 2S's and 2L's \(\frac{8!}{2! 2!}\)

If the majority are women implies that: 3 Women and 2 Men Or 4 Women and 1 Man = \(^4C_3 \times^6C_2+^4C_4\times^6C_1\) = \(4 \times 15 + 1 \times 6\) = 60 + 6 = 66

Sample space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Total number of possible outcomes = 2 x 2 x 2 = 8 favourable number of outcomes = 7 ∴Pr(at least one head) = \(\frac {7}{8}\)

Let the number of white balls be n The number of red balls = 8 Now, the probability of drawing a white ball = \(\frac {n}{n + 8}\) The probability of drawing a red ball = \(\frac {8}{n + 8}\) Since Pr(White ball) = \(\frac{1}{2}\) x Pr(Red ball) \(\therefore \frac {n}{n + 8} = \frac {1}{2}\times \frac {8}{n + 8}\) = \(\frac {n}{n + 8} = \frac {4}{n + 8}\) \(\therefore n = 4\) Number of white balls = 4 The possible number of outcomes = n + 8 = 4 + 8 = 12 Pr(Red ball and White ball) = Pr(Red ball) x Pr(White ball) Pr(Red ball) = \(\frac {8}{12}\) Pr(Red ball) = \(\frac {4}{11}\) Pr(Red ball and white ball) = \(\frac {8}{12} \times \frac {4}{11}\) \(\therefore\)Pr(Red ball and white ball) = \(\frac {8}{33}\)