Statistics, Year 10: sampling, histograms and cumulative frequency

The population (in statistics)

HintThe same word is used for everyone who lives in a country.

The whyA sample is a smaller group chosen from it. Results from the sample are used to estimate what is true of the whole group.

A random sample

HintNames drawn from a hat produce one.

The whyGiving everyone the same chance avoids favouring any one type of member, so the sample is more likely to represent the whole population.

Those who do not eat school dinners

HintThink about who would never be standing there.

The whyA biased sample over-represents some parts of the population. Pupils who dislike the dinners and bring packed lunches have no chance of being asked, so the results will look more favourable than the truth.

144

HintAssume the whole school behaves like the sample, and scale up.

The why12 out of 50 is 24%, and 24% of 600 = 144. Equivalently, the school is 12 times the size of the sample, and 12 × 12 = 144. It is an estimate: another sample would give a slightly different figure.

The sample of 200 — larger samples vary less

HintOne unusual person changes a small group's result far more than a big group's.

The whyEvery sample differs a little from the population by chance. In a large sample those chance differences tend to even out, so its results are closer to the truth; no sample can be certain.

Frequency density

HintIt is the quantity plotted up the vertical axis, and its name suggests how tightly the data are packed.

The whyDividing by the class width makes classes of different widths comparable: it gives the frequency per unit of width.

1.5

HintFind how wide the class is, then share the frequency across that width.

The whyClass width = 30 − 20 = 10. Frequency density = 15 ÷ 10 = 1.5.

area

HintIt combines how wide the bar is with how tall it is.

The whyBar area = class width × frequency density = frequency. So to read a frequency from a histogram, multiply the width of the bar by its height.

Wide classes would look too important

HintA wider bar of a given height covers more of the page.

The whyA class that is twice as wide collects more data just by being wide. If its bar is also drawn to the full frequency, the eye sees an area far bigger than its share of the data. Using frequency density makes area match frequency.

30

HintOnly half the width of the bar is wanted.

The whyFrequency = width × frequency density. For the part from 40 to 50 the width is 10, so the estimate is 10 × 3 = 30. The whole bar holds 20 × 3 = 60 values. It is an estimate because the values are assumed to be spread evenly across the class.

Cumulative frequency

HintThe first word of the name means "building up as it goes".

The whyEach value is the total number of data values up to the top of that class. The final cumulative frequency equals the total number of values.

13

HintInclude every class that ends at or below that time.

The whyAdd the frequencies of all the classes up to 20: 4 + 9 = 13. Thirteen journeys took 20 minutes or less.

The upper class boundary

HintThe running total only becomes true once the whole group has been counted.

The whyThe cumulative frequency for the class 10 < t ≤ 20 counts every value up to 20, so it is plotted at t = 20. Plotting it anywhere else shifts the whole curve sideways.

40

HintThe median has an equal number of values on each side of it.

The whyThe median is the middle value, so read across from half the total: 80 ÷ 2 = 40. Go across to the curve and then down to the height axis.

15 pupils

HintThe graph tells you how many had finished by then; the question asks about the rest.

The why45 pupils took 30 minutes or less. The others took longer: 60 − 45 = 15.

155

HintOne point has to stand for the whole class, so it goes in the middle.

The whyEach point of a frequency polygon is plotted at the midpoint of its class: (150 + 160) ÷ 2 = 155. The points are then joined with straight lines.

Class B's pupils are generally taller

HintLook at where each peak sits along the horizontal axis.

The whyA polygon that sits further to the right shows larger values. If the two polygons also have a similar shape, the heights are spread out in a similar way and only the most common height group differs.

Both can be drawn on the same axes

HintLines can overlap without hiding each other; solid blocks cannot.

The whyWith the two polygons on one diagram, differences in position and spread can be seen at a glance, class by class.

20 < m ≤ 30

HintThe highest point marks the centre of the group that holds the most parcels.

The whyPoints are plotted at class midpoints, so a point above m = 25 belongs to the class 20 < m ≤ 30. The modal class is the one with the greatest frequency.

gaps

HintThe scale along the bottom runs on without a break, so one class ends exactly where the next begins.

The whyContinuous data can take any value, so the classes join up along a number line and the bars touch. Bars for separate categories, such as favourite colours, are drawn apart.