Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com GCSE: Histograms Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com Objectives: To understand why a histogram is useful for displaying data, and how to both draw and interpret a histogram. Last modified: 5th November 2016

Click to Start Bromanimation Pablo is hosting a party. He counts how many people are between 15 and 20, and 20 and 50. Age (years) Frequency 15 ≤ a < 20 15 20 ≤ a < 50 Why is below graph somewhat unhelpful. How could we fix it? Click to Start Bromanimation 15 The 15 people in the second group are more spread out in age, but this graph seems to suggest that people’s ages are spread out uniformly between 15 and 50. Frequency 10 20 30 40 50 Age

Click to Start Bromanimation Let’s presume that within each age group, the ages are evenly spread. Then there would 3 people of each age in the 15-20 group, and 0.5 people of each age in the 20-50 group. Age (years) Frequency 15 ≤ a < 20 15 20 ≤ a < 50 ? ? Click to Start Bromanimation 3 2 1 The resulting diagram is known as a histogram. The ‘frequency per age’ is known as the ‘frequency density’. In general, given the frequency and class width, we can calculate it using: Frequency Density = Frequency Class Width Frequency Density Estimated Frequency ? 10 20 30 40 50 Age

Bar Charts vs Histograms For discrete data. Frequency given by height of bars. Histograms For continuous data. Data divided into (potentially uneven) intervals. Frequency given by area of bars. ? ? ? ? Frequency Density Frequency 1.0m 1.2m 1.4m 1.6m 1.8m 6 7 8 9 Height Shoe Size

Examples ? ? ? ? ? ? ? ? 1 Weight (w kg) Frequency Frequency Density 40 4 10 < w ≤ 15 6 1.2 15 < w ≤ 35 52 2.6 35 < w ≤ 45 10 1 ? ? Freq ? F.D. Width ? Frequency = 40 ? 2 5 4 3 2 1 Frequency = 15 ? Frequency = 25 ? Frequency Density Frequency = 30 ? 10 20 30 40 50 Height (m)

? ? ? ? ? ? ? 3 The Box of Helpfulness Freq F.D. Width Always start by adding a Frequency Density column 3 Frequency Density 30  30 = 1 ? Bro Tip: For this kind of question, first find a ‘complete’ set of information: i.e. for the first row in this question, we have the frequency and the drawn bar, so can work out the F.D. scale. ? 84 4.2 (using graph) ? 60 ? 6 (using graph) ? 40  20 = 2 ? 18  30 = 0.6 ? 1 2 3 4 5 6 7 8 The Box of Helpfulness We don’t know the scale on the frequency density axis. Can we work it out using the first row of the table? This triangle will help throughout. Freq F.D. Width

Test Your Understanding (on your sheet) 4 3 2 1 5 FD 3.8 4.8 4.2 3.1 0.8 ? Bro Hint: The second row has a ‘complete’ set of information (bar and frequency) ? 31

Quickfire Questions – Determining F.D. scale Work out the scales on the frequency density axis. 4 3 2 1 ? ? 16 12 8 4 ? 2 1 Frequency Density Frequency Density Frequency Density 0 10 20 0 10 20 20 28 36 Height (m) 0≤𝑥<15 Frequency 30 Height (m) 0≤𝑥<5 Frequency 60 Height (m) 20≤𝑥<32 Frequency 6

Questions Provided collection of past GCSE questions. (Answers on next slides)

Question 1 FD 3.2 2 1.2 0.6 0.4 ? 12 6 ? 3.2 2.4 1.6 0.8 4.0

Question 2 FD 0.8 1 1.6 2 1.2 ? ? 2.0 1.5 1.0 0.5

Question 3 FD 3 5 3.6 1.2 ? (b) Work out an estimate for the number of cars with a speed of more than 85 km/h. Note that 85 to 100 is three-quarters of the 80 to 100 interval. Thus we can estimate we’ll have 𝟑 𝟒 of the 24 drivers in this group. 𝟑 𝟒 ×𝟐𝟒=𝟏𝟖 4 3 2 1 5 ?

Question 4 4 3 2 1 FD 1 3.5 3 1.5 0.5 ? 30 ? ? 60

Question 5 ? 24 ? 30

Summary So Far ? ? ? ? Purpose: Area: Histograms allow us to display continuous data grouped into (potentially non-fixed) intervals. The idea is that they reflect the ‘concentration’ of things within each range of values. Area: The area of a bar is equal to the frequency*. * Actually it’s only proportional to it, but you don’t need to worry about that till A Level. ? ? Working out the F.D. scale: If the frequency is known and the bar height is known, we can work out the scale using the formula on the left. ? Frequency Density Formula: Frequency Density is ‘frequency per unit value’, i.e: ? Freq F.D. Width

Harder Histogram Questions We previously saw that the area gives us frequency. If we don’t have the frequency density axis, we could use the idea that area and frequency are in proportion to each other. One strategy: Since area represents frequency, find out how many people each square is worth. ? 130cm-135cm  20 squares ∴ Each square worth: 10 20 =0.5 children 110-130cm  70 squares 0.5×70=35 children If each square is 0.5 children, we need 12 squares. Click to show?

Harder Histogram Questions However, there’s nothing stopping you using the same approach as before: working out the frequency density axis from a ‘complete’ set of information (i.e. where we have the frequency AND the bar). 3 2 ? 1 𝐹𝑟𝑒𝑞 𝐷𝑒𝑛𝑠𝑖𝑡𝑦= 10 5 =2 So finding area between 110cm and 130cm (as area = frequency) 𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑠= 10×1 + 5×2 + 5×3 =35

Test Your Understanding (use either method) (If using square count method, perhaps count big squares rather than little squares.) 18 big squares → 9 students So 1 square = 0.5 students Total squares = 30 ∴ 15 students OR: F.D. = 𝟗 𝟏𝟓 =𝟎.𝟔 Then use areas of bars. ? 24 (big) squares required (We can have 4 rows of 5 to get first 20 squares, but have to split the last 4 squares across 5, thus use 4/5 of a square for each square at top) This is quite a clumsy method, so the frequency density approach would probably be easier here. b ?

Proportion Histogram Questions Sometimes you have to find the proportion of people/things/animals within some range of values. Key Point: Since area is frequency, the proportion of the area is the same as the proportion of the frequency. e.g. If half the area is above 18m, then half the people are above 18m. 8 7 6 5 4 3 2 1 Total area =50 ? What proportion of people had a height: Between 10 and 14m: 𝟏𝟔 𝟓𝟎 = 𝟖 𝟐𝟓 Between 14 and 18m: 𝟏𝟖 𝟓𝟎 = 𝟗 𝟐𝟓 ? Frequency Density ? Bro Tip: If the frequency density scale is missing, you can set it to what you like for this kind of question. 10 14 18 22 26 Height (m)

Test Your Understanding Bro Tip: I recommend writing the area of each bar on the bar itself, as you’ll likely have to use some areas twice. Solution: Total apples: (40 x 0.12) + (20 x 0.36) + (20 x 0.7) + (20 x 0.56) + (40 x 0.18) = 44.4 Apples in range 140-160g: (20 x 0.36) + (20 x 0.7) + (20 x 0.56) = 32.4 Proportion = 32.4 44.4 = 𝟐𝟕 𝟑𝟕 𝒐𝒓 𝟎.𝟕𝟑 ?

Questions Provided collection of past GCSE questions.

Solutions – Question 1 a If we use 1,2,3,… on frequency density scale: Total area =10+10+8+35+20+6+6+6 =101 Area between 25 and 40 mins: 6+6+6=18 Percentage: 18 101 ×100=𝟏𝟕.𝟖% ? b 𝟖 𝟏𝟎𝟏 of area is between 10-15 mins. 𝟏𝟔÷𝟖 ×𝟏𝟎𝟏=𝟐𝟎𝟐 ?

Solutions – Question 2 10 20 30 40 50 60 Answer: 431 1425 ×285=86.2 ?

Solutions – Question 3 ? 14 𝐹𝐷= 2 1 =2 ∴ Total batteries 12 The histogram shows information about the lifetime of some batteries. ? 2 4 6 8 10 12 14 𝐹𝐷= 2 1 =2 ∴ Total batteries =2+ 0.5×6 + 0.5×12 + 0.5×8 + 1×4 =19 batteries Two of the batteries had a lifetime of between 1.5 and 2.5 years. Find the total number of batteries.

Solutions – Question 4 FD 16 4 2.4 4.8 16 12 8 4 Frequency Density 40 60 56 32 ? B1 for Frequency density label or appropriate units B2 for 4 correct histogram bars sq (B1 for 2 bars correct)

Solutions – Question 5 ? 0.05 0.04 0.03 0.02 0.01 8 6 ?