JAMB Mathematics Past Questions & Answers - Page 271

40 = 20 - x + x + 35 - x

40 = 55 - x

x = 55 - 40

= 15

∴ P(both) $\frac{15}{40}$

= $\frac{3}{8}$

$\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}$

x = $\frac{\sum fx}{\sum f}$

= $\frac{24}{8}$

= 3

Variance (6²) = $\frac{\sum f(\bar{x} - x)^2}{\sum f}$

= $\frac{18}{8}$

= $\frac{9}{4}$

Median = L + [$\frac{\frac{N}{2} - f}{fm}$]h

N = Sum of frequencies

L = lower class boundary of median class

f = sum of all frequencies below L

fm = frequency of modal class and

h = class width of median class

Median = 11 + [$\frac{\frac{50}{2} - 21}{20}$]5

= 11 + ($\frac{25 - 21}{20}$)5

= 11 + ($\frac{(4)}{20}$)

11 + 1 = 12

x = $\frac{\sum x}{N}$

= $\frac{18}{3}$

= 6

$\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}$

M.D = $\frac{|x - x|}{N}$

= $\frac{6}{3}$

= 2

Mean = $\frac{\sum fx}{\sum f}$

$\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}$

$\implies 14(28y - 13) = 43(8y - 2)$

$392y - 182 = 344y - 86$

$392y - 344y = -86 + 182 \implies 48y = 96$

$y = 2$