WAEC Further Mathematics Past Questions & Answers - Page 59

291.

Marks 2 3 4 5 6 7 8
No of students 5 7 9 6 3 6 4

The table above shows the distribution of marks by some candidates in a test. If a student is selected at random, what is the probability that she scored at least 6 marks?

A.

\(\frac{3}{40}\)

B.

\(\frac{1}{4}\)

C.

\(\frac{13}{40}\)

D.

\(\frac{27}{40}\)

Correct answer is C

No explanation has been provided for this answer.

292.

Marks 2 3 4 5 6 7 8
No of students 5 7 9 6 3 6 4

The table above shows the distribution of marks by some candidates in a test. Find, correct to one decimal place, the mean of the distribution.

A.

5.5

B.

5.3

C.

5.2

D.

4.7

Correct answer is D

Marks(\(x\) 2 3 4 5 6 7 8 Total
Frequency \(f\) 5 7 9 6 3 6 4 40
\(fx\) 10 21 36 30 18 42 32 189

Mean \(\bar{x} = \frac{\sum fx}{\sum f} = \frac{189}{40}\)

= \(4.725 \approxeq 4.7\)

293.

Marks 2 3 4 5 6 7 8
No of students 5 7 9 6 3 6 4

The table above shows the distribution of marks by some candidates in a test. What is the median score?

A.

3.5

B.

4.0

C.

4.5

D.

5.0

Correct answer is B

No explanation has been provided for this answer.

294.

Two forces \(F_{1} = (10N, 020°)\) and \(F_{2} = (7N, 200°)\) act on a particle. Find the resultant force.

A.

(3 N, 020°)

B.

(3 N, 200°)

C.

(17 N, 020°)

D.

(17 N, 200°)

Correct answer is A

No explanation has been provided for this answer.

295.

The probability of Jide, Atu and Obu solving a given problem are \(\frac{1}{12}\), \(\frac{1}{6}\) and \(\frac{1}{8}\) respectively. Calculate the probability that only one solves the problem.

A.

\(\frac{1}{576}\)

B.

\(\frac{55}{576}\)

C.

\(\frac{77}{576}\)

D.

\(\frac{167}{576}\)

Correct answer is D

\(P(Jide) = \frac{1}{12}; P(\text{not Jide}) = \frac{11}{12}\)

\(P(Atu) = \frac{1}{6}; P(\text{not Atu}) = \frac{5}{6}\)

\(P(Obu) = \frac{1}{8}; P(\text{not Obu}) = \frac{7}{8}\)

\(P(\text{only one of them}) = P(\text{Jide not Atu not Obu}) + P(\text{Atu not Jide not Obu}) + P(\text{Obu not Jide not Atu})\)

= \((\frac{1}{12} \times \frac{5}{6} \times \frac{7}{8}) + (\frac{1}{6} \times \frac{11}{12} \times \frac{7}{8}) + (\frac{1}{8} \times \frac{11}{12} \times \frac{5}{6})\)

= \(\frac{35}{576} + \frac{77}{576} + \frac{55}{576}\)

= \(\frac{167}{576}\)