Statistics, permutations and probability
Further Mathematics · WAEC and JAMB · SS2 and SS3
This section closes most papers and is the easiest place to pick up full marks, provided you are organised. Statistics rewards neat tables, while permutations and probability reward one careful decision about whether order matters.
What you need to know
- The mean of a frequency distribution is the sum of fx divided by the sum of f. Build the table with columns for x, f and fx, add the columns, then divide, rather than trying to hold figures in your head.
- The median of grouped data is L + [(n/2 - cfb)/f] x c, where L is the lower class boundary of the median class, cfb is the cumulative frequency before it, f is its frequency and c is the class width. Find the median class first from the cumulative frequency column.
- The mode of grouped data is L + [d1/(d1 + d2)] x c, where d1 is the excess of the modal frequency over the class before and d2 the excess over the class after. The modal class is simply the class with the highest frequency.
- Class boundaries are not class limits. For classes written as 10 to 19 and 20 to 29, the boundary between them is 19.5, and using 19 or 20 instead will shift your median by half a unit.
- Variance is the mean of the squares minus the square of the mean, that is (sum of fx^2)/(sum of f) minus (mean)^2. The standard deviation is the square root of the variance and carries the same unit as the original data.
- Mean deviation is the average of the absolute deviations from the mean, written as (sum of f times |x - mean|)/(sum of f). The absolute value bars matter, because without them the deviations always add to zero.
- A permutation counts arrangements where order matters, and nPr = n!/(n - r)!. A combination counts selections where order does not matter, and nCr = n!/(r!(n - r)!). Ask yourself whether swapping two chosen items gives a different outcome; if it does, use permutations.
- Committees, teams and handshakes are combinations. Positions such as chairman and secretary, seating in a row, and codes are permutations, because the roles or places are distinguishable.
- Arranging n distinct objects in a circle gives (n - 1)! ways, not n!, because rotating the whole circle does not create a new arrangement. Five people around a round table can sit in 4! = 24 ways.
- When letters repeat, divide by the factorial of each repeat count. The word STATISTICS has 10 letters with three S, three T and two I, so the number of distinct arrangements is 10!/(3! x 3! x 2!) = 50400.
- Probability of an event is the number of favourable outcomes divided by the number of equally likely outcomes, always between 0 and 1. The probability that an event does not happen is 1 minus the probability that it does, which is often the fastest route to the answer.
- For the addition rule, P(A or B) = P(A) + P(B) - P(A and B). The subtraction vanishes only when the events are mutually exclusive, that is when they cannot both happen.
- For independent events, P(A and B) = P(A) x P(B). Drawing with replacement keeps events independent; drawing without replacement does not, and then the second probability has a denominator one smaller.
- Conditional probability is P(A given B) = P(A and B)/P(B). Tree diagrams handle these questions well, because each branch carries its own probability and the probabilities along any path are multiplied.
Key terms
- Mean
- The sum of all the values divided by how many values there are, the ordinary average.
- Median
- The middle value when the data is arranged in order, or the value estimated by interpolation for grouped data.
- Standard deviation
- The square root of the variance, measuring how far the values typically spread from the mean.
- Permutation
- An arrangement of objects in which the order of selection matters.
- Combination
- A selection of objects in which the order of selection does not matter.
- Mutually exclusive events
- Events that cannot occur at the same time, so that the probability of both happening is zero.
- Independent events
- Events where the occurrence of one does not change the probability of the other.
Formulae
Mean of grouped data = (sum of fx)/(sum of f)Median = L + [(n/2 - cfb)/f] x cMode = L + [d1/(d1 + d2)] x cVariance = (sum of fx^2)/(sum of f) - (mean)^2Standard deviation = sqrt(variance)Mean deviation = (sum of f|x - mean|)/(sum of f)nPr = n!/(n - r)!nCr = n!/(r!(n - r)!)Circular arrangements of n objects = (n - 1)!Arrangements with repeats = n!/(p! q! r!)P(event) = favourable outcomes / total outcomesP(not A) = 1 - P(A)P(A or B) = P(A) + P(B) - P(A and B)P(A and B) = P(A) x P(B) for independent eventsP(A given B) = P(A and B)/P(B)
Worked examples
The table below shows a distribution: the values of x are 1, 2, 3, 4, 5 with frequencies 3, 5, 7, 3, 2 respectively. Calculate the mean and the standard deviation, correct to two decimal places.
- Total frequency: 3 + 5 + 7 + 3 + 2 = 20.
- Compute fx for each class: 3, 10, 21, 12, 10. Their sum is 56.
- Mean = 56/20 = 2.8.
- Compute fx^2 for each class: 3(1) = 3, 5(4) = 20, 7(9) = 63, 3(16) = 48, 2(25) = 50. Their sum is 184.
- Variance = 184/20 - (2.8)^2 = 9.2 - 7.84 = 1.36.
- Standard deviation = sqrt(1.36) = 1.1662, which is 1.17 to two decimal places.
Answer: Mean = 2.8 and standard deviation = 1.17
A committee of 4 people is to be chosen at random from 5 boys and 4 girls. Find the probability that the committee contains exactly 2 boys and 2 girls.
- Order does not matter in a committee, so use combinations throughout.
- Total number of possible committees: 9C4 = (9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 3024/24 = 126.
- Number of ways to choose 2 boys from 5: 5C2 = (5 x 4)/2 = 10.
- Number of ways to choose 2 girls from 4: 4C2 = (4 x 3)/2 = 6.
- Favourable committees: 10 x 6 = 60.
- Probability = 60/126 = 10/21, which is about 0.476.
Answer: The probability is 10/21, approximately 0.48.
The mistake to avoid
Using permutations where combinations are required. A committee of four from nine is 9C4 = 126, not 9P4 = 3024, and the inflated figure is an immediate zero. In standard deviation work, the other frequent loss is dividing by the number of classes instead of by the total frequency.
In the exam
Draw the full table with columns for x, f, fx and fx^2 and total every column before you compute anything, because the examiner marks the table. In probability, say in one short phrase whether the selection is with or without replacement and whether order matters; that single decision determines the whole answer, and writing it down stops you drifting halfway through.