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JAMB Mathematics Past Questions & Answers - Page 279

1,391.

Let p be a probability function on set S, where S = (a₁, a₂, a₃, a₄). Find P(a₁) if P(a₂) = $\frac{1}{3}$ , p(a₃) = $\frac{1}{6}$ and p(a₄) = $\frac{1}{5}$

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{3}{10}$

View answer & explanation

Correct answer is D

No explanation has been provided for this answer.

1,392.

$\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}$
The median of the distribution above is

4.0

3.4

3.2

3.0

View answer & explanation

Correct answer is B

Median = $\frac{a + b}{fm}$ ( $\frac{1}{2} \sum f - CF_b$ )

= 2.95 + $\frac{0.5}{15}$ (2.-7)

= 2.95 + $\frac{0.5}{15}$ x 13

= 2.95 + 0. 43

= 3.38

= 3.4

1,393.

3.2

3.3

3.7

4.2

View answer & explanation

Correct answer is B

Mode = a + (b - a)(f_m - F_b)

2F_m - F_a - F_b

= 3.0 + $\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10}$

= 3 + $\frac{(6.4)(11)}{30 - 14}$

= 3 + $\frac{4.4}{16}$

= 3 + 0.275

= 3.275

= 3.3cm

1,394.

The variance of the scores 1, 2, 3, 4, 5 is

1, 2

1, 4

2.0

3.0

View answer & explanation

Correct answer is C

$x$	1	2	3	4	5	sum =15
$x - \bar{x}$	-2	-1	0	1	2
$(x - \bar{x})^{2}$	4	1	0	1	4	10

Mean = $\frac{15}{5} = 3$

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

= $\frac{10}{5}$

= $2.0$

1,395.

$\begin{array}{c|c} class& 1 - 3 & 4 - 6 & 7 - 9\\ \hline Frequency & 5 & 8 & 5\end{array}$
Find the standard deviation of the data using the table above

$\sqrt{6}$

$\frac{5}{3}$

$\sqrt{5}$

View answer & explanation

Correct answer is D

$\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 2 & 10 & -3 & 9 & 45\\ 4 - 6 & 8 & 5 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 8 & 40 & 3 & 9 & 45 \\ \hline & 18 & & 90 & & & 90 \end{array}$

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

= $\frac{90}{18}$

= 5

S.D = $\frac{\sum f(x - x)^2}{\sum f}$

= $\frac{90}{18}$

= $\sqrt{5}$