JAMB Mathematics Past Questions & Answers - Page 269

1,341.

Calculate the standard deviation of the following data: 7, 8, 9, 10, 11, 12, 13.

A.

2

B.

4

C.

10

D.

11

Correct answer is A

\(\begin{array}{c|c} x & x - x & (x - x)^2\\ \hline 7 & -3 & 9\\8 & -2 & 4 \\9 & -1 & 1\\10 & 0 & 1\\11 & 1 & 1\\ 12 & 2 & 4\\13 & 3 & 9\\ \hline & & 28\end{array}\)

S.D = \(\sqrt{\frac{\sum(x - x)^2}{N}}\)

= \(\sqrt{\frac{\sum d^2}{N}}\)

= \(\sqrt{\frac{28}{7}}\)

= \(\sqrt{4}\)

= 2

1,342.

A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime

A.

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

B.

\(\frac{5}{11}\)

C.

\(\frac{6}{11}\)

D.

\(\frac{8}{11}\)

Correct answer is A

Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors

itself and 1

The prime numbers among the group are 23, 29. Probability of choosing a prime number

= \(\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}\)

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

1,343.

\(\begin{array}{c|c} Class & Frequency\\ \hline 1 - 5 & 2\\6 - 10 & 4\\11 - 15 & 5\\16 - 20 & 2 \\ 21 - 25 & 3\\26 - 30 & 2\\31 - 35 & 1\\36 - 40 & 1 \end{array}\)
Find the median of the observation in the table given

A.

11.5

B.

12.5

C.

14.0

D.

14.5

Correct answer is D

Median = L1 + (\(\frac{Ef}{fm}\)) - fo

\(\frac{\sum f}{2}\)

= \(\frac{20}{2}\)

= 10, L1 = 10.5, fo = 6, fm = 5

Median = 10.5 + \(\frac{(10 - 6)}{5}\)5

= 10.5 + 4

= 14.5

1,344.

The mean of the ages of ten secondary school pupils is 16 but when the age of their teacher is added to it the men becomes 19. Find the age of the teacher

A.

27

B.

35

C.

38

D.

49

Correct answer is D

Average age of 110 students = 16

∴ Total age = 16 x 10 = 160 years

Age of teachers = x, total number of people now = 11

mean age = 19

Total age of new group = 19 x 11 = 209

Age of teachers = x = (209 - 160) = 49 yrs

1,345.

\(\begin{array}{c|c} Weight(s) & 0 -10 & 10 - 20 & 20 - 30 & 40 - 50\\ \hline \text{Number of coconuts} & 10 & 27 & 19 & 6 & 2\end{array}\)
Estimate the mode of the frequency distribution above

A.

13.2g

B.

15.0g

C.

16.8g

D.

17.5g

Correct answer is C

Mode = a + \(\frac{(b - a)(F_m - F_b)}{2F_m - F_a - F_b}\)

= \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\)

= \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\)

= 10 + \(\frac{170}{25}\)

= 10 + 6.8

= 16.8