JAMB Mathematics Past Questions & Answers - Page 271

1,351.

In a survey, it was observed that 20 students read newspapers and 35 read novels. If 40 of the students read either newspapers or novels, what is the probability of the students who read both newspapers and novels?

A.

\(\frac{1}{2}\)

B.

\(\frac{2}{3}\)

C.

\(\frac{3}{8}\)

D.

\(\frac{3}{11}\)

Correct answer is C

40 = 20 - x + x + 35 - x

40 = 55 - x

x = 55 - 40

= 15

∴ P(both) \(\frac{15}{40}\)

= \(\frac{3}{8}\)

1,352.

\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & 2 & 1 & 2 & 1 & 2\end{array}\)
Find the variance of the frequency distribution above

A.

\(\frac{3}{2}\)

B.

\(\frac{9}{4}\)

C.

\(\frac{5}{2}\)

D.

3

Correct answer is B

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

x = \(\frac{\sum fx}{\sum f}\)

= \(\frac{24}{8}\)

= 3

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

= \(\frac{18}{8}\)

= \(\frac{9}{4}\)

1,353.

\(\begin{array}{c|c} \text{Class Interval} & 1 - 5 & 6 - 10 & 11 - 15 & 16 - 20 & 21 - 25 \\ \hline Frequency & 6 & 15 & 20 & 7 & 2\end{array}\)
Estimate the median of the frequency distribution above

A.

10\(\frac{1}{2}\)

B.

11\(\frac{1}{2}\)

C.

12

D.

13

Correct answer is C

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

1,354.

Find the mean deviation of the set of numbers 4, 5, 9

A.

zero

B.

2

C.

5

D.

6

Correct answer is B

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

1,355.

\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & y + 2 & y - 2 & 2y - 3 & y + 4 & 3y - 4\end{array}\)
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y

A.

1

B.

2

C.

3

D.

4

Correct answer is B

x 1 2 3 4 5 Total
f y + 2 y - 1 2y - 3 y + 4 3y - 4 8y - 2
fx y + 2 2y - 2 6y - 9 4y + 16 15y - 20 28y - 13

Mean = \(\frac{\sum fx}{\sum f}\)

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

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

\(392y - 182 = 344y - 86\)

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

\(y = 2\)