Notes · Unit 11
Basic Statistics
Fourteen short slides on mean, median and mode, weighted and grouped mean, histograms, probability and expected frequency.
Press Next (or the right arrow key) to move on. On most slides, each press reveals one more idea.
Mean, median, modeWeighted meanGrouped dataProbabilityExpected frequency
1 · Averages
Mean, median and mode
Mean. sum of all values ÷ number of values. For 2, 4, 6, 8, 10 the sum is 30 and n = 5, so the mean is 6.
Median. The middle value of the data arranged in order. For an odd number n of values it is the (n + 1)/2-th value, so with 5 values it is the 3rd.
Mode. The value that occurs most often. Data may have two modes (2, 2, 5, 5, 9) or none when every value occurs once.
Mean, median and mode are measures of central tendency. The range measures spread, so it is not an average.
2 · Finding the median
Order first, then find the middle
Step 1. Arrange the values in order. The median is defined for ordered data, so skipping this step gives a wrong answer.
Odd number of values. Take the single middle value.
Even number of values. Take the mean of the two middle values. For 3, 8, 5, 10: ordered it is 3, 5, 8, 10, so the median is (5 + 8)/2 = 6.5.
For 2, 4, 6, 10 the middle values are 4 and 6, so the median is (4 + 6)/2 = 5.
3 · Choosing an average
Which one tells the truth?
The mean uses every value, so one extreme value pulls it a lot. The median is affected much less.
Marks 40, 45, 50, 55, 100: the mean is 290 ÷ 5 = 58, pulled up by 100. The median is 50, which represents a typical student better.
Houses at about Rs 5 million and one palace at Rs 500 million: the median describes a typical house.
A shoe shop deciding which size to stock most should use the mode, because it is the most frequently sold size.
4 · Working with the mean
Sum, shifts and scaling
Sum from the mean. sum = mean × n. If the mean of 10 values is 35.5, the sum is 355, not 35.5 + 10.
Missing value. The mean of 8 values is 25, so the total is 200. Seven values add to 162, so the eighth is 38.
Add 5 to every value: the mean and the median both go up by 5. Multiply every value by 3: the mean becomes 3 times as large.
Add a value equal to the mean: the mean does not change. With 9 values of mean 20, adding 20 gives (180 + 20)/10 = 20.
5 · Weighted mean
When values are not equally important
Weighted mean = Σwx ÷ Σw
Multiply each value x by its weight w, add, then divide by the total of the weights.
Scores 50, 70, 90 with weights 2, 3, 5: Σwx = 100 + 210 + 450 = 760 and Σw = 10, so the weighted mean is 76.
A final exam is 70% and a quiz 30%, so use weights 70 and 30. The simple mean of the two scores would be unfair.
Combined mean. 20 students with mean 60 and 30 students with mean 70 give (20 × 60 + 30 × 70)/50 = 66, not 65.
6 · Grouped data
Class marks and the mean
Class mark = (lower limit + upper limit) ÷ 2. For 20 – 30 it is 25. It represents the class because the exact values are unknown.
Class boundaries lie halfway across the gap between classes. For 30 – 39 they are 29.5 and 39.5.
Mean = Σfx ÷ Σf
Classes 10–20, 20–30, 30–40, 40–50 with frequencies 4, 6, 8, 2. Class marks: 15, 25, 35, 45. Σfx = 60 + 150 + 280 + 90 = 580 and Σf = 20, so the mean is 29.
Values 2, 4, 6 with frequencies 3, 5, 2: Σfx = 6 + 20 + 12 = 38 and Σf = 10, so the mean is 3.8.
7 · Grouped mode and median
Mode and median formulas
Mode = l + (fm − f1)/(2fm − f1 − f2) × h, where fm is the modal class frequency, f1 the frequency before it and f2 the frequency after it.
Same classes as before: the modal class is 30–40 with l = 30, h = 10, fm = 8, f1 = 6, f2 = 2. Mode = 30 + (8 − 6)/(16 − 6 − 2) × 10 = 32.5.
Median = l + h/f × (n/2 − C), where l is the lower boundary of the median class, h its size, f its frequency, n = Σf and C the cumulative frequency before the median class.
The median class is the class that contains the n/2-th value.
8 · Graphs
Histogram and frequency polygon
A histogram is a chart of adjacent rectangular bars, one for each class interval. The bars touch because the class boundaries join one class to the next.
A frequency polygon plots class marks on the x-axis and frequencies on the y-axis, then joins the points.
Do not plot class limits or boundaries for the polygon. The points are class marks.
9 · Probability
Always between 0 and 1
P(E) = favourable outcomes ÷ total outcomes
Probability lies from 0 to 1, ends included. An impossible event has probability 0 and a certain event has probability 1. So 1.2 or 1.5 can never be a probability.
Complement. E′ and E together make the whole sample space, so P(E) + P(E′) = 1. P(E) = 0.6 and P(E′) = 0.6 cannot both be true.
A bag has 5 red, 3 green and 2 white balls. P(green) = 3/10, so P(not green) = 1 − 3/10 = 7/10.
Letter chosen from ALLAH: L occurs 2 times among 5 letters, so P(L) = 2/5.
10 · Experimental probability
Relative and expected frequency
Relative frequency = times the event occurred ÷ total trials. It is also called experimental probability. Heads 45 times in 120 tosses gives 45/120 = 3/8.
Expected frequency = number of trials × probability. It is a prediction, not a guarantee.
A bulb is defective with probability 0.02: in 5000 bulbs expect 5000 × 0.02 = 100. Rain probability 0.2 over 150 days gives 30 rainy days.
With more trials, relative frequency comes closer to the theoretical probability: 0.168 from 5000 rolls is a better estimate than 0.20 from 50.
After 7 heads in 10 tosses, the next toss is still equally likely to be heads or tails. Earlier results do not change it.
11 · Common mistakes
Five traps to avoid
Finding the median without ordering the data.
Adding instead of multiplying: sum = mean × n.
Wrong total in probability. With 3 red and 7 blue balls, P(red) = 3/10, not 3/7.
Using favourable ÷ total when outcomes are not equally likely. "It may rain or not" does not make P(rain) = 1/2.
Using the simple mean when weights differ.
12 · Quick check
Try these, then tap to check
13 · Your turn
Practise Unit 11
Three timed MCQ tests with an explanation after every answer.
Practice questions are independently written and are not official FBISE papers.