WASSCE Further Maths: Statistics and Probability Practice Questions & Answers

In a histogram, the area of each rectangular bar represents the frequency of that class. Therefore, the total area of all the bars combined is directly proportional to the total frequency of the distribution.

Using the law of total probability, $P(C) = P(A \cap C) + P(B \cap C)$P(C)=P(A∩C)+P(B∩C).
$P(C) = (P(A) \times P(C|A)) + (P(B) \times P(C|B)) = (0.6 \times 0.8) + (0.4 \times 0.3) = 0.48 + 0.12 = 0.60$P(C)=(P(A)×P(C∣A))+(P(B)×P(C∣B))=(0.6×0.8)+(0.4×0.3)=0.48+0.12=0.60.

The mean of a grouped frequency distribution is given by the formula $\bar{x} = \frac{\sum fx}{\sum f}$xˉ=∑f∑fx​.
Substituting the given values, $\bar{x} = \frac{280}{40} = 7$xˉ=40280​=7.

A key condition for the binomial distribution is that the probability of success, denoted as $p$p, must remain constant across all trials. Therefore, stating that the probability must vary is incorrect.

Spearman's rank correlation is $r_s = 1 - \frac{6 \sum D^2}{n(n^2 - 1)}$rs​=1−n(n2−1)6∑D2​.
With $n = 5$n=5 and $\sum D^2 = 8$∑D2=8,
$r_s = 1 - \frac{6 \times 8}{5(25 - 1)} = 1 - \frac{48}{120} = 1 - 0.4 = 0.6$rs​=1−5(25−1)6×8​=1−12048​=1−0.4=0.6.

By definition, mutually exclusive events cannot occur at the same time. Therefore, their intersection is an empty set, making $P(A \cap B) = 0$P(A∩B)=0.

Variance is given by $\frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2$∑f∑fx2​−(∑f∑fx​)2.
Mean $= \frac{100}{20} = 5$=20100​=5.
Variance $= \frac{580}{20} - 5^2 = 29 - 25 = 4$=20580​−52=29−25=4.
Standard Deviation $= \sqrt{4} = 2$=4​=2.

An ogive, or cumulative frequency curve, represents the accumulation of frequencies up to a certain point. Therefore, the cumulative frequency is plotted against the corresponding upper class boundary of each class interval.

Let $X$X be the number of heads. $X \sim B(n=4, p=0.5)$X∼B(n=4,p=0.5).
$P(X=2) = \binom{4}{2} (0.5)^2 (0.5)^2 = 6 \times \left(\frac{1}{2}\right)^4 = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8}$P(X=2)=(24​)(0.5)2(0.5)2=6×(21​)4=6×161​=166​=83​.

A downward trend from left to right means as one variable increases, the other decreases. The tight clustering indicates a 'strong' relationship, hence a strong negative correlation.