Fold a dot plot to test it for symmetry, watch the three centres split when the fold fails, measure how far one value moves the mean and why it leaves the median alone, and read a data set's stretch back out of its mean and median.
Subject: Statistics · 68 slides · applied lesson
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Title
Statistics · §2.6
Read the shape first: it decides where each centre lands
Objectives
Concept
\[ \bar{x} = {\textstyle\sum} x \div n \]
Mean
Why: total Σx shared among n values
\[ \text{place} = (1 + n) \div 2 \]
Median place
Why: counts along the sorted list
\[ M = (u + v) \div 2 \]
Even count
Why: M: halfway between two middles
\[ \text{mode} = \text{commonest value} \]
Mode
Why: answers what happens most often
\[ (u + v) \div 2 \]
Halfway
Why: equal steps from both ends
\[ 3 \times u = u + u + u \]
Multiply
Why: copies of one number
\[ \frac{u + v}{w} = \frac{u}{w} + \frac{v}{w} \]
Share a sum
Why: each part keeps its portion
\[ u - v < 0 \text{ if } u < v \]
Signs
Why: below gives a negative
\[ u \div v = w \Rightarrow w \times v = u \]
Divide back
Why: checks every share
Prediction
Figure (svg): Three dot plots on one scale from 4 to 10, each with a grey dashed line at 7: Group A with 16 marks spread evenly either side, Group B with 10 marks reaching down to 4 and stopping at 8, Group C with 10 marks starting at 6 and reaching up to 10
§2.6's three data sets, read as quiz marks.
Predict first
Every group's commonest mark is 7.
Which group's mean mark is smallest?
Correct: Group B
Why: Group B reaches down to 4 and stops at 8, so its marks sit lower than the other two. A mean adds every mark, so the group whose marks lean low pools the least. This deck measures all three and says how far apart they end up.
Worked example
Figure (svg): Three dot plots on one scale from 4 to 10, each with a grey dashed line at 7: Group A with 16 marks spread evenly either side, Group B with 10 marks reaching down to 4 and stopping at 8, Group C with 10 marks starting at 6 and reaching up to 10
\[ \text{A}:\ 5 \text{ below } \textcolor{#1f5fbf}{7},\ \ 5 \text{ above} \]
Count A each side of 7
Why: shape before arithmetic
Figure (svg): Three dot plots on one scale from 4 to 10, each with a grey dashed line at 7: Group A with 16 marks spread evenly either side, Group B with 10 marks reaching down to 4 and stopping at 8, Group C with 10 marks starting at 6 and reaching up to 10, with Group A's counts printed beside its name
\[ \text{B}:\ 5 \text{ below } \textcolor{#1f5fbf}{7},\ \ 1 \text{ above} \]
Count B the same way
Why: a second group to weigh
Figure (svg): Three dot plots on one scale from 4 to 10, each with a grey dashed line at 7: Group A with 16 marks spread evenly either side, Group B with 10 marks reaching down to 4 and stopping at 8, Group C with 10 marks starting at 6 and reaching up to 10, with Groups A and B's counts printed beside their names
\[ \text{C}:\ 1 \text{ below } \textcolor{#1f5fbf}{7},\ \ 5 \text{ above} \]
Count C the same way
Why: one shape still unweighed
Figure (svg): Three dot plots on one scale from 4 to 10, each with a grey dashed line at 7: Group A with 16 marks spread evenly either side, Group B with 10 marks reaching down to 4 and stopping at 8, Group C with 10 marks starting at 6 and reaching up to 10, with all three groups' counts printed beside their names
\[ 5 = 5,\ \ 5 > 1,\ \ 1 < 5 \]
Compare the six counts
Why: only A answers itself
\[ 16 - 5 - 5 = 6 \]
Check: take A's 10 off 16
Why: the rest all hold 7
Section
Idea 1 of 4
Concept
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10
Sixteen marks out of 10, one dot each.
Discussion prompt
Which single mark would you report as Group A's typical result?
Answer:
All three §2.5 rules answer 7 here. Why do they agree?
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10
f: how many marks share one value.
\[ f = 6 \text{ at } \textcolor{#1f5fbf}{7} \]
Count the tallest stack
Why: commonest mark, quickest guess
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with the six-dot stack at 7 boxed and labelled mode 7
\[ 16 - 6 = 10 \]
Take the 7s off 16
Why: how many the guess ignores
\[ 10 > 6 \]
Compare the two counts
Why: most students sit outside
\[ 6 + 10 = 16 \]
Check: add both counts
Why: nobody was placed twice
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10
\[ 3 \times \textcolor{#1f5fbf}{6} = \textcolor{#1f5fbf}{18} \]
Total the three 6s
Why: counting beats writing copies
\[ 6 \times \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{42} \]
Total the six 7s
Why: the tallest stack weighs most
\[ 3 \times \textcolor{#1f5fbf}{8} = \textcolor{#1f5fbf}{24} \]
Total the three 8s
Why: no repeated mark left out
\[ \textcolor{#1f5fbf}{4 + 5 + 18 + 42 + 24 + 9 + 10} = \textcolor{#1f5fbf}{112} \]
Add the seven part totals
Why: Σx must hold all 16
\[ \textcolor{#1f5fbf}{112} \div 16 = \textcolor{#6b7280}{7} \]
Share the pool among 16
Why: one mark if all were level
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey dashed line at the mean 7
\[ 16 \times \textcolor{#6b7280}{7} = \textcolor{#1f5fbf}{112} \]
Check: rebuild the pool
Why: sixteen shares return Σx
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey dashed line at the mean 7
\[ 1 + 16 = 17 \]
Add the outer places
Why: a median needs ends
\[ 17 \div 2 = 8.5 \]
Halve 17
Why: even n, no single middle
\[ \text{8th} = \textcolor{#1f5fbf}{7},\ \ \text{9th} = \textcolor{#1f5fbf}{7} \]
Read places 8 and 9
Why: these two decide M
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey dashed line at the mean 7 and a grey dotted line at the median 7
\[ M = \textcolor{#6b7280}{7} \]
Take halfway of two 7s
Why: equal middles need nothing
\[ \textcolor{#6b7280}{\bar{x}} = \textcolor{#6b7280}{7},\ \ M = \textcolor{#6b7280}{7},\ \ \text{mode} = \textcolor{#1f5fbf}{7} \]
Set the centres out
Why: three rules, one answer
\[ 5 + 6 = 11 \ge 9 \]
Check: count to place 9
Why: the six 7s reach past
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey dashed fold line at 7
\[ \textcolor{#1f5fbf}{7} - \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{3},\ \ \textcolor{#1f5fbf}{10} - \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{3} \]
Measure 4 and 10
Why: equal gaps make a mirror
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey fold line at 7 and 1 row of blue double arrows below the axis measuring equal gaps from 7
\[ \textcolor{#1f5fbf}{7} - \textcolor{#1f5fbf}{5} = \textcolor{#1f5fbf}{2},\ \ \textcolor{#1f5fbf}{9} - \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{2} \]
Measure 5 and 9
Why: a second pair confirms
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey fold line at 7 and 2 rows of blue double arrows below the axis measuring equal gaps from 7
\[ \textcolor{#1f5fbf}{7} - \textcolor{#1f5fbf}{6} = \textcolor{#1f5fbf}{1},\ \ \textcolor{#1f5fbf}{8} - \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{1} \]
Measure 6 and 8
Why: three answered by three
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey fold line at 7 and 3 rows of blue double arrows below the axis measuring equal gaps from 7
\[ \textcolor{#1f5fbf}{7} - \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{0} \]
Measure the six 7s
Why: on the line, reaching nowhere
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey fold line at 7 and 4 rows of blue double arrows below the axis measuring equal gaps from 7
Mirror images each side: the data is symmetrical.
\[ 5 + 6 + 5 = 16 \]
Check: add the three counts
Why: none lacks a partner
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey dashed fold line at 7
\( {7 - 4 = 3}\;\;\Rightarrow\;\;\allowbreak {7 - 5 = 2}\;\;\Rightarrow\;\;\allowbreak {7 - 6 = 1}\;\;\Rightarrow\;\;\allowbreak {7 - 7 = 0} \)
\[ \textcolor{#1f5fbf}{4 + 10} = \textcolor{#1f5fbf}{14},\ \ \textcolor{#1f5fbf}{5 + 9} = \textcolor{#1f5fbf}{14} \]
Add the outer pairs
Why: the fold sets a total
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey fold line at 7 and 2 rows of blue double arrows below the axis measuring equal gaps from 7
\[ \textcolor{#1f5fbf}{6 + 8} = \textcolor{#1f5fbf}{14},\ \ \textcolor{#1f5fbf}{7 + 7} = \textcolor{#1f5fbf}{14} \]
Add two more pairs
Why: tests the pattern nearer 7
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey fold line at 7 and 4 rows of blue double arrows below the axis measuring equal gaps from 7
\[ \textcolor{#1f5fbf}{14} \div 2 = \textcolor{#6b7280}{7} \]
Halve one pair's total
Why: each averages the fold
\[ 8 \times \textcolor{#1f5fbf}{14} = \textcolor{#1f5fbf}{112} \]
Check: rebuild Σx
Why: the stacks pooled 112 too
Worked example
Figure (svg): A scale from 8 to 16 with a grey fold line at m, a blue dot one gap below it and a blue dot the same gap above it, both gaps arrowed and labelled 2
\[ (\textcolor{#6b7280}{m} - \textcolor{#1f5fbf}{d}) + (\textcolor{#6b7280}{m} + \textcolor{#1f5fbf}{d}) = \textcolor{#6b7280}{m} + \textcolor{#6b7280}{m} - \textcolor{#1f5fbf}{d} + \textcolor{#1f5fbf}{d} \]
Remove the brackets
Why: gaps can now meet
\[ -\textcolor{#1f5fbf}{d} + \textcolor{#1f5fbf}{d} = 0 \]
Add the two gaps
Why: they cancel out
\[ \textcolor{#6b7280}{m} + \textcolor{#6b7280}{m} = \textcolor{#6b7280}{2m} \]
Add the two folds
Why: the gaps left nothing behind
\[ {\textstyle{\textstyle\sum}} x = k \times \textcolor{#6b7280}{2m} \]
Total k pairs
Why: the pool is just its pairs
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey fold line at 7 and 4 rows of blue double arrows below the axis measuring equal gaps from 7
\[ \textcolor{#6b7280}{2}k\textcolor{#6b7280}{m} \div 2k = \textcolor{#6b7280}{m} \]
Share among 2k marks
Why: k pairs hold all
\[ k = 8,\ \textcolor{#6b7280}{m} = \textcolor{#6b7280}{7}:\ \textcolor{#1f5fbf}{112} \div 16 = \textcolor{#6b7280}{7} \]
Check on Group A
Why: the fold returns
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey dashed fold line at 7
\[ \textcolor{#1f5fbf}{4},\ \textcolor{#1f5fbf}{5},\ \textcolor{#1f5fbf}{6},\ \textcolor{#1f5fbf}{6},\ \textcolor{#1f5fbf}{6} \to 5 \]
Count the marks under 7
Why: a middle needs side counts
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with the counts 5 below 7 and 5 above printed under the axis
\[ \textcolor{#1f5fbf}{8},\ \textcolor{#1f5fbf}{8},\ \textcolor{#1f5fbf}{8},\ \textcolor{#1f5fbf}{9},\ \textcolor{#1f5fbf}{10} \to 5 \]
Count the marks over 7
Why: each partner lies opposite
\[ 5 = 5 \]
Compare the two side counts
Why: pairing forces them to match
\[ \text{places } 6 \text{ to } 11 = \textcolor{#1f5fbf}{7} \]
Name the places the 7s fill
Why: 5 marks sit either side
\[ M = \textcolor{#6b7280}{7} \]
Read places 8 and 9
Why: both middles lie inside
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, with a grey dotted median line at 7 and the counts 5 and 5 under the axis
\[ (1 + 16) \div 2 = 8.5 \]
Check: work out the middle place
Why: 8.5 falls inside that block
Worked example
Figure (svg): Dot plot of the made-up marks 4, 4, 5, 6, 6 on a scale from 3 to 7: two dots at 4, one at 5, two at 6
Made-up marks 4, 4, 5, 6, 6 — five students.
\[ \textcolor{#1f5fbf}{4 + 4 + 5 + 6 + 6} = \textcolor{#1f5fbf}{25} \]
Add the five made-up marks
Why: a fresh set to test
\[ \textcolor{#1f5fbf}{25} \div 5 = \textcolor{#6b7280}{5} \]
Share among 5
Why: the mirror rule predicted this
Figure (svg): Dot plot of the made-up marks 4, 4, 5, 6, 6 on a scale from 3 to 7: two dots at 4, one at 5, two at 6, with a grey dashed line at the mean 5
\[ \text{modes} = \textcolor{#1f5fbf}{4},\ \textcolor{#1f5fbf}{6} \]
Find the two tallest stacks
Why: a tie empties the centre
Figure (svg): Dot plot of the made-up marks 4, 4, 5, 6, 6 on a scale from 3 to 7: two dots at 4, one at 5, two at 6, with a grey mean line at 5 and the two-dot stacks at 4 and 6 boxed as modes
\[ \textcolor{#1f5fbf}{4} \ne \textcolor{#6b7280}{5},\ \ \textcolor{#1f5fbf}{6} \ne \textcolor{#6b7280}{5} \]
Compare both modes with 5
Why: neither peak marks the balance
\[ \textcolor{#6b7280}{5} - \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{1},\ \ \textcolor{#1f5fbf}{6} - \textcolor{#6b7280}{5} = \textcolor{#1f5fbf}{1} \]
Check the fold at 5
Why: the mirror holds, modes differ
Worked example
Figure (svg): Dot plot of Maris' ten letter counts from 0 to 10: one dot at 2, two at 3, three at 4, three at 6, one at 8
Example 2.31: Maris' ten letter counts.
\[ \textcolor{#1f5fbf}{5} - \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{3},\ \ \textcolor{#1f5fbf}{8} - \textcolor{#1f5fbf}{5} = \textcolor{#1f5fbf}{3} \]
Measure 2 and 8
Why: the ends decide fastest
Figure (svg): Dot plot of Maris' ten letter counts from 0 to 10: one dot at 2, two at 3, three at 4, three at 6, one at 8, with a grey fold line at 5 and blue arrows of 3 to the counts 2 and 8
\[ \textcolor{#1f5fbf}{5} - \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{1},\ \ \textcolor{#1f5fbf}{6} - \textcolor{#1f5fbf}{5} = \textcolor{#1f5fbf}{1} \]
Measure 4 and 6
Why: three counts answer three
Figure (svg): Dot plot of Maris' ten letter counts from 0 to 10: one dot at 2, two at 3, three at 4, three at 6, one at 8, with arrows of 3 and of 1 drawn under the axis
\[ \textcolor{#1f5fbf}{5} - \textcolor{#1f5fbf}{3} = \textcolor{#1f5fbf}{2},\ \ \textcolor{#1f5fbf}{7}: \text{none} \]
Seek partners for the 3s
Why: a fold demands 7s
Figure (svg): Dot plot of Maris' ten letter counts from 0 to 10: one dot at 2, two at 3, three at 4, three at 6, one at 8, with a third row of arrows showing a gap of 2 below 5 and the word none above 7
\[ 2 \ne 0 \]
Compare the two tallies
Why: no mirror, no fold
\[ 8 + 2 = 10 \]
Check: add both tallies
Why: all ten counts placed
Trap
\[ \textcolor{#1f5fbf}{1 + 2 + 6 + 9 + 12} = \textcolor{#1f5fbf}{30} \]
Add the values
Why: a pool to share
\[ \textcolor{#1f5fbf}{30} \div 5 = \textcolor{#6b7280}{6} \]
Share among 5
Why: this list's mean
\[ \text{3rd} = \textcolor{#1f5fbf}{6} \]
Read place 3
Why: odd n, one middle
\[ \textcolor{#6b7280}{\bar{x}} = M \Rightarrow \text{folds} \]
Claim a mirror
Why: no gap measured
Figure (svg): The same made-up dot plot with a grey dashed fold line at 6, the dots reaching 5 below it and 6 above
\[ \textcolor{#1f5fbf}{6} - \textcolor{#1f5fbf}{1} = \textcolor{#1f5fbf}{5},\ \ \textcolor{#1f5fbf}{12} - \textcolor{#1f5fbf}{6} = \textcolor{#1f5fbf}{6} \]
Measure both ends
Why: gaps must match
\[ \textcolor{#1f5fbf}{5} \ne \textcolor{#1f5fbf}{6} \]
Compare the gaps
Why: low end nearer
\[ \textcolor{#1f5fbf}{6} - \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{4} \ne \textcolor{#1f5fbf}{3} = \textcolor{#1f5fbf}{9} - \textcolor{#1f5fbf}{6} \]
Check inner pair
Why: no mirror here
Prediction
Figure (svg): Dot plot of the made-up marks 10, 11, 12, 13, 14 on a scale from 9 to 15, with a grey dashed fold line at 12
Predict first
A sixth mark joins the made-up set 10, 11, 12, 13, 14.
Which new mark keeps the fold at 12?
Correct: Another 12
Why: A mark on the fold has a gap of 0, so it needs no partner. Each of the other three arrives with a gap that nothing answers: a second 10 would leave two marks at gap 2 below and only one above, and 15 or 9 would need a partner at 9 or 15 that the set does not have.
Worked example
Figure (svg): Dot plot of the made-up marks 10, 11, 12, 13, 14 on a scale from 9 to 15, with a grey dashed fold line at 12
\[ \textcolor{#1f5fbf}{12} - \textcolor{#1f5fbf}{10} = \textcolor{#1f5fbf}{2},\ \ \textcolor{#1f5fbf}{14} - \textcolor{#1f5fbf}{12} = \textcolor{#1f5fbf}{2} \]
Measure the ends from 12
Why: the set folds already
\[ \textcolor{#1f5fbf}{12} - \textcolor{#1f5fbf}{11} = \textcolor{#1f5fbf}{1},\ \ \textcolor{#1f5fbf}{13} - \textcolor{#1f5fbf}{12} = \textcolor{#1f5fbf}{1} \]
Measure the inner pair
Why: both gaps already have partners
\[ \textcolor{#1f5fbf}{12} - \textcolor{#1f5fbf}{12} = \textcolor{#1f5fbf}{0} \]
Measure the newcomer
Why: a gap of nothing asks for none
Figure (svg): Dot plot of the made-up marks 10, 11, 12, 12, 13, 14 on a scale from 9 to 15, with a grey dashed fold line at 12, and two dots stacked at 12
\[ \textcolor{#1f5fbf}{10 + 11 + 12 + 12 + 13 + 14} = \textcolor{#1f5fbf}{72} \]
Total the six marks
Why: a number to test the rule
\[ \textcolor{#1f5fbf}{72} \div 6 = \textcolor{#6b7280}{12} \]
Check: share the new pool
Why: the mean stays on the fold
Section
Idea 2 of 4
Concept
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8
Ten marks out of 10, one dot each.
Discussion prompt
Group B has a mark at 4 and none at 10. What should that do to its three centres?
Answer:
The fold fails and the centres split. The next slides measure the split.
Worked example
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dashed fold line at 7
\[ \textcolor{#1f5fbf}{6}:\ 3,\ \ \textcolor{#1f5fbf}{8}:\ 1 \]
Count at 6 and at 8
Why: a mirror needs equal stacks
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey fold line at 7 and blue arrows of 1 to the marks at 6 and 8
\[ 3 \ne 1 \]
Compare the two stacks
Why: one cannot answer three
\[ \textcolor{#1f5fbf}{5} \to \textcolor{#1f5fbf}{9}?,\ \ \textcolor{#1f5fbf}{4} \to \textcolor{#1f5fbf}{10}? \]
Seek partners for 5 and 4
Why: the fold needs marks up there
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey fold line at 7 and three rows of arrows, the lower two labelled none on the high side
\[ \textcolor{#1f5fbf}{9},\ \textcolor{#1f5fbf}{10}:\ \text{no marks} \]
Read the empty high end
Why: two low marks stay unanswered
\[ 5 \text{ below } \textcolor{#1f5fbf}{7},\ \ 1 \text{ above} \]
Check: count each side of 7
Why: sides this uneven cannot mirror
Worked example
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8
\[ 3 \times \textcolor{#1f5fbf}{6} = \textcolor{#1f5fbf}{18} \]
Total the three 6s
Why: counting shortens the line
\[ 4 \times \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{28} \]
Total the four 7s
Why: the tallest stack dominates
\[ \textcolor{#1f5fbf}{4 + 5 + 18 + 28 + 8} = \textcolor{#1f5fbf}{63} \]
Add the five part totals
Why: Σx must hold all 10
\[ \textcolor{#1f5fbf}{63} \div 10 = \textcolor{#6b7280}{6.3} \]
Share the pool among 10
Why: an equal cut for a lopsided class
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dashed line at the mean 6.3
\[ 10 \times \textcolor{#6b7280}{6.3} = \textcolor{#1f5fbf}{63} \]
Check: rebuild the pool
Why: ten shares return Σx
Worked example
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dashed line at the mean 6.3
\[ 1 + 10 = 11 \]
Add the outer places
Why: the middle needs both ends
\[ 11 \div 2 = 5.5 \]
Halve 11
Why: no mark sits there
\[ \text{5th} = \textcolor{#1f5fbf}{6},\ \ \text{6th} = \textcolor{#1f5fbf}{7} \]
Read places 5 and 6
Why: the middles differ here
\[ \textcolor{#1f5fbf}{6} + \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{13} \]
Add the middle pair
Why: halfway needs their sum
\[ \textcolor{#1f5fbf}{13} \div 2 = \textcolor{#6b7280}{6.5} \]
Halve 13
Why: M lands between two marks
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dashed mean line at 6.3 and a grey dotted median line at 6.5
\[ 5 \text{ marks} \le \textcolor{#1f5fbf}{6},\ \ 5 \ge \textcolor{#1f5fbf}{7} \]
Check: count each side of 6.5
Why: five places either side
Worked example
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dashed mean line at 6.3 and a grey dotted median line at 6.5
\[ \textcolor{#1f5fbf}{7}:\ 4 \text{ marks} \]
Count the tallest stack
Why: the mode is commonest
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with mean and median lines and the four-dot stack at 7 boxed as the mode
\[ \textcolor{#1f5fbf}{6}:\ 3,\ \ \textcolor{#1f5fbf}{4},\ \textcolor{#1f5fbf}{5},\ \textcolor{#1f5fbf}{8}:\ 1 \]
Count the rest
Why: only taller wins
\[ \text{mode} = \textcolor{#1f5fbf}{7} \]
Take the mark with four
Why: a mark, not a count
\[ \textcolor{#6b7280}{6.3} < \textcolor{#6b7280}{6.5} < \textcolor{#1f5fbf}{7} \]
Order the three centres
Why: each rule stopped elsewhere
Marks stretching down: skewed left.
Tail: the long thin side.
\[ \textcolor{#1f5fbf}{7} - \textcolor{#6b7280}{6.3} = \textcolor{#1f5fbf}{0.7},\ \ \textcolor{#1f5fbf}{7} - \textcolor{#6b7280}{6.5} = \textcolor{#1f5fbf}{0.5} \]
Check from the mode
Why: the mean fell further
Worked example
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6
\[ 4 \times \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{28} \]
Total the four 7s
Why: the tallest stack carries most
\[ 3 \times \textcolor{#1f5fbf}{8} = \textcolor{#1f5fbf}{24} \]
Total the three 8s
Why: the second repeat owes its share
\[ \textcolor{#1f5fbf}{6 + 28 + 24 + 9 + 10} = \textcolor{#1f5fbf}{77} \]
Add the five part totals
Why: Σx counts all 10 marks
\[ \textcolor{#1f5fbf}{77} \div 10 = \textcolor{#6b7280}{7.7} \]
Share the pool among 10
Why: an equal cut for a class leaning high
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6, with a grey dashed line at the mean 7.7
\[ 10 \times \textcolor{#6b7280}{7.7} = \textcolor{#1f5fbf}{77} \]
Check: rebuild the pool
Why: ten equal cuts rebuild 77
Faded example
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6, with a grey dashed line at the mean 7.7
Group C's ten marks, already in order.
Fill in the blanks
5th mark = 7; 6th mark = 8; M = 7.5
Why: 1 + 10 = 11 and 11 ÷ 2 = 5.5, so places 5 and 6 decide M. Place 5 holds 7 and place 6 holds 8, and (7 + 8) ÷ 2 = 7.5, a value no student actually scored.
Worked example
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6, with a grey dashed line at the mean 7.7
\[ 1 + 10 = 11 \]
Add the outer places
Why: ten marks, so both ends again
\[ 11 \div 2 = 5.5 \]
Halve 11
Why: places 5 and 6 decide
\[ \text{5th} = \textcolor{#1f5fbf}{7},\ \ \text{6th} = \textcolor{#1f5fbf}{8} \]
Read places 5 and 6
Why: the middles disagree here
\[ \textcolor{#1f5fbf}{7} + \textcolor{#1f5fbf}{8} = \textcolor{#1f5fbf}{15} \]
Add the pair
Why: the halving starts here
\[ \textcolor{#1f5fbf}{15} \div 2 = \textcolor{#6b7280}{7.5} \]
Halve 15
Why: M falls between marks
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6, with a grey dashed mean line at 7.7 and a grey dotted median line at 7.5
\[ \textcolor{#6b7280}{7.5} - \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{0.5},\ \ \textcolor{#1f5fbf}{8} - \textcolor{#6b7280}{7.5} = \textcolor{#1f5fbf}{0.5} \]
Check: step to each middle mark
Why: equal steps confirm halfway
Worked example
Figure (svg): Two dot plots on one scale from 4 to 10: Group B with a grey dashed mean line at 6.3 and a grey dotted median line at 6.5, and Group C with a mean line at 7.7 and a median line at 7.5
\[ \textcolor{#1f5fbf}{7}:\ 4 \text{ marks} \]
Count Group C's tallest stack
Why: its mode, for the ordering
Figure (svg): Two dot plots on one scale from 4 to 10: Group B with a grey dashed mean line at 6.3 and a grey dotted median line at 6.5, and Group C with a mean line at 7.7 and a median line at 7.5, and Group C's four-dot stack at 7 boxed as its mode
\[ \textcolor{#1f5fbf}{7} < \textcolor{#6b7280}{7.5} < \textcolor{#6b7280}{7.7} \]
Order Group C's centres
Why: Group B ran the other way
\[ \textcolor{#6b7280}{6.3} - \textcolor{#6b7280}{6.5} = \textcolor{#1f5fbf}{-0.2} \]
Take B's mean less its median
Why: the sign names the pulled side
\[ \textcolor{#6b7280}{7.7} - \textcolor{#6b7280}{7.5} = \textcolor{#1f5fbf}{+0.2} \]
Take C's mean less its median
Why: same size, the other side
\[ \textcolor{#6b7280}{6.5} + (\textcolor{#1f5fbf}{-0.2}) = \textcolor{#6b7280}{6.3},\ \ \textcolor{#6b7280}{7.5} + \textcolor{#1f5fbf}{+0.2} = \textcolor{#6b7280}{7.7} \]
Check: add each gap to its median
Why: both land back on their means
Trap
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with the four-dot stack at 7 boxed as the mode
\[ \textcolor{#1f5fbf}{7}:\ 4 \text{ marks} \]
Find the tall stack
Why: eyes go there
\[ \text{peak high} \Rightarrow \text{right} \]
Judge by the peak
Why: thin side rules
\[ \textcolor{#6b7280}{6.3} < \textcolor{#6b7280}{6.5} \]
Compare the centres
Why: 6.3 is lower
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dashed line at the mean 6.3 sitting left of the median 6.5
\[ \textcolor{#1f5fbf}{4},\ \textcolor{#1f5fbf}{5}:\ 1;\ \ \textcolor{#1f5fbf}{9},\ \textcolor{#1f5fbf}{10}:\ \text{none} \]
Read both ends
Why: one thin, one empty
\[ \text{skewed left} \]
Name the side
Why: the tail decides
\[ \textcolor{#6b7280}{6.3} < \textcolor{#6b7280}{6.5} < \textcolor{#1f5fbf}{7} \]
Check the order
Why: the mean is lowest
Prediction
Figure (svg): Dot plot of nine made-up waiting times on a scale from 0 to 20 minutes: two dots at 2, three at 3, two at 4, one at 5 and one far out at 18
Predict first
Made-up waiting times in minutes: 2, 2, 3, 3, 3, 4, 4, 5, 18.
Which way do these times stretch?
Correct: To the high side, towards 18
Why: Eight of the nine waits sit between 2 and 5, and the ninth is 18, far above them, with nothing below 2 to answer it. The long thin side is the high side. The mean is not needed for this: the shape is visible in the plot, and the mean only measures how far the stretch reaches.
Worked example
Figure (svg): Dot plot of nine made-up waiting times on a scale from 0 to 20 minutes: two dots at 2, three at 3, two at 4, one at 5 and one far out at 18
\[ \textcolor{#1f5fbf}{5} - \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{3} \]
Measure the eight-wait pack
Why: they crowd three minutes
\[ \textcolor{#1f5fbf}{18} - \textcolor{#1f5fbf}{5} = \textcolor{#1f5fbf}{13} \]
Measure the reach up
Why: thin side, far longer
\[ \textcolor{#1f5fbf}{2 + 2 + 3 + 3 + 3 + 4 + 4 + 5 + 18} = \textcolor{#1f5fbf}{44} \]
Total all nine waits
Why: a mean sizes the stretch
\[ \textcolor{#1f5fbf}{44} \div 9 \approx \textcolor{#6b7280}{4.9} \]
Share among 9
Why: one wait lifts the cut
Figure (svg): Dot plot of nine made-up waiting times on a scale from 0 to 20 minutes: two dots at 2, three at 3, two at 4, one at 5 and one far out at 18, with a grey dashed line at the mean 4.9
\[ \text{5th of 9} = \textcolor{#1f5fbf}{3} \]
Read the middle wait
Why: odd n gives one
\[ \textcolor{#6b7280}{4.9} > \textcolor{#1f5fbf}{3} \]
Check: mean against median
Why: the mean climbed the tail
Section
Idea 3 of 4
Concept
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, redrawn on a scale from 4 to 14 with the top mark still at 10
Idea 1 found Group A: Σx = 112, x̄ = 7, M = 7.
Discussion prompt
Which of Group A's three centres move when its 10 becomes 14?
Answer:
Only the mean moves, and by far less than 4.
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14
\[ \textcolor{#1f5fbf}{14} - \textcolor{#1f5fbf}{10} = \textcolor{#1f5fbf}{4} \]
Measure how far it travels
Why: only this value changed
\[ \textcolor{#6b7280}{7} + \textcolor{#1f5fbf}{4} = \textcolor{#6b7280}{11} \]
Add the whole move
Why: the quickest guess, worth testing
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with the top mark moved out to 14 and a grey dashed line at 11, right of every other mark
\[ 16 \times \textcolor{#6b7280}{11} = \textcolor{#1f5fbf}{176} \]
Rebuild the pool from 11
Why: a mean must give Σx back
\[ \textcolor{#1f5fbf}{176} - \textcolor{#1f5fbf}{112} = \textcolor{#1f5fbf}{64} \]
Subtract the old pool
Why: the guess invented marks
\[ \textcolor{#1f5fbf}{64} \ne \textcolor{#1f5fbf}{4} \]
Compare with the real move
Why: sixteen times too much
\[ \textcolor{#1f5fbf}{112} + \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{116} \ne \textcolor{#1f5fbf}{176} \]
Check: pool the re-marked data
Why: the true total is 60 short
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14
\[ \textcolor{#1f5fbf}{112} + \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{116} \]
Add the move to the pool
Why: no other mark changed
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with the top mark moved from 10 out to 14, a hollow ring left at 10 and a blue arrow labelled moves 4
\[ \textcolor{#1f5fbf}{116} \div 16 = \textcolor{#6b7280}{7.25} \]
Share the new pool
Why: the mean rule is unchanged
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with the top mark at 14 and a grey dashed line at the mean 7.25
\[ \textcolor{#6b7280}{7.25} - \textcolor{#6b7280}{7} = \textcolor{#1f5fbf}{0.25} \]
Subtract the old mean
Why: how far the balance travelled
\[ \textcolor{#1f5fbf}{4} \div 16 = \textcolor{#1f5fbf}{0.25} \]
Share the move itself
Why: the same number, shorter road
\[ 16 \times \textcolor{#1f5fbf}{0.25} = \textcolor{#1f5fbf}{4} \]
Check: rebuild the move
Why: sixteen shares make 4
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with the top mark moved out to 14, a hollow ring at 10 and a grey line at the mean 7.25
c: one value's move · n: how many values
\[ {\textstyle{\textstyle\sum}} x + \textcolor{#1f5fbf}{c} \]
Add the move
Why: no other value moved
\[ ({\textstyle{\textstyle\sum}} x + \textcolor{#1f5fbf}{c}) \div n \]
Share the new pool
Why: the rule fits any list
\[ {\textstyle{\textstyle\sum}} x \div n + \textcolor{#1f5fbf}{c} \div n \]
Split the share
Why: each part keeps its own
\[ \textcolor{#6b7280}{\bar{x}} + \textcolor{#1f5fbf}{c} \div n \]
Name the first part
Why: it is the old mean
\[ \text{shift} = \textcolor{#1f5fbf}{c} \div n \]
Subtract the old mean
Why: the rest is its travel
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with the top mark at 14, a grey dashed line at the new mean 7.25 and a grey dotted line at the old mean 7
\[ \textcolor{#1f5fbf}{c} = \textcolor{#1f5fbf}{4},\ n = 16:\ \textcolor{#1f5fbf}{4} \div 16 = \textcolor{#1f5fbf}{0.25} \]
Check on Group A
Why: it matches the long road
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with the top mark moved out to 14 and a grey dashed mean line at 7.25
\[ \textcolor{#1f5fbf}{14} > \textcolor{#1f5fbf}{10} > \textcolor{#1f5fbf}{9} \]
Compare the moved mark
Why: it was largest and still is
\[ 1 + 16 = 17 \]
Add the outer places
Why: the count has not changed
\[ 17 \div 2 = 8.5 \]
Halve 17
Why: places 8 and 9 decide M
\[ \text{8th} = \textcolor{#1f5fbf}{7},\ \ \text{9th} = \textcolor{#1f5fbf}{7} \]
Read places 8 and 9
Why: both sit far from the move
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with the top mark at 14, a grey dashed mean line at 7.25 and a grey dotted median line at 7
\[ M = \textcolor{#6b7280}{7} \]
Take halfway between two 7s
Why: the median has not stirred
\[ 5 \text{ below } \textcolor{#1f5fbf}{7},\ \ 5 \text{ above} \]
Check: count each side again
Why: the split at 7 survived
Worked example
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with a grey dashed line at the mean 7
\[ \textcolor{#1f5fbf}{8} \div 16 = \textcolor{#1f5fbf}{0.5} \]
Move the mark twice as far
Why: tests what distance controls
Figure (svg): Dot plot of Group A's 16 marks on a scale from 4 to 10: one dot at 4, one at 5, three at 6, six at 7, three at 8, one at 9, one at 10, on a scale from 4 to 14 with grey lines at 7.25 and 7.5 labelled with their shifts
\[ \textcolor{#1f5fbf}{4} \div 32 = \textcolor{#1f5fbf}{0.125} \]
Share the same move among 32
Why: tests what class size controls
\[ \textcolor{#1f5fbf}{0} \div 16 = \textcolor{#1f5fbf}{0} \]
Leave the mark alone
Why: a rule must hold at its edge
\[ \textcolor{#1f5fbf}{0.5} > \textcolor{#1f5fbf}{0.25} > \textcolor{#1f5fbf}{0.125} \]
Rank the three shifts
Why: long moves in small classes count
The mean always travels towards the moved mark.
\[ 16 \times \textcolor{#1f5fbf}{0.5} = \textcolor{#1f5fbf}{8} \]
Check: rebuild the longest move
Why: sixteen shares add to 8
Worked example
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6, on a scale from 6 to 20
Group C pooled to 77, mean 7.7.
\[ \textcolor{#1f5fbf}{20} - \textcolor{#1f5fbf}{10} = \textcolor{#1f5fbf}{10} \]
Measure the mark's move
Why: the rule needs a distance
\[ \textcolor{#1f5fbf}{10} \div 10 = \textcolor{#1f5fbf}{1} \]
Share the move among 10
Why: n is the group size
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6, on a scale from 6 to 20 with the top mark moved from 10 out to 20, a hollow ring at 10 and a blue arrow labelled moves 10
\[ \textcolor{#6b7280}{7.7} + \textcolor{#1f5fbf}{1} = \textcolor{#6b7280}{8.7} \]
Add the shift on
Why: the rule's predicted balance
Figure (svg): Dot plot of Group C's 10 marks from 4 to 10: one dot at 6, four at 7, three at 8, one at 9, one at 10, nothing below 6, on a scale from 6 to 20 with the top mark at 20 and a grey dashed line at the mean 8.7
\[ \textcolor{#1f5fbf}{77} + \textcolor{#1f5fbf}{10} = \textcolor{#1f5fbf}{87} \]
Pool the re-marked data
Why: an independent road there
\[ \textcolor{#1f5fbf}{87} \div 10 = \textcolor{#6b7280}{8.7} \]
Check: share the pool
Why: the long road agrees
Trap
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dotted median line at 6.5
\[ \textcolor{#1f5fbf}{4} \to \textcolor{#1f5fbf}{9} \]
Re-mark B's 4 as 9
Why: nine marks hold
\[ M = \textcolor{#6b7280}{6.5} \]
Keep the old median
Why: a mark crossed
\[ \text{5th} = \textcolor{#1f5fbf}{7},\ \ \text{6th} = \textcolor{#1f5fbf}{7} \]
Read both middles
Why: they are new
Figure (svg): Dot plot of the re-marked Group B, 5, 6, 6, 6, 7, 7, 7, 7, 8, 9, on a scale from 4 to 10, with a grey dotted median line at 7
\[ \textcolor{#1f5fbf}{5},\ \textcolor{#1f5fbf}{6},\ \textcolor{#1f5fbf}{6},\ \textcolor{#1f5fbf}{6},\ \textcolor{#1f5fbf}{7},\ \textcolor{#1f5fbf}{7},\ \textcolor{#1f5fbf}{7},\ \textcolor{#1f5fbf}{7},\ \textcolor{#1f5fbf}{8},\ \textcolor{#1f5fbf}{9} \]
Sort the new ten
Why: order rules M
\[ M = \textcolor{#6b7280}{7} \]
Halve two 7s
Why: the new middles agree
\[ \textcolor{#6b7280}{7} - \textcolor{#6b7280}{6.5} = \textcolor{#1f5fbf}{0.5} \]
Check M's move
Why: crossings shift it
Prediction
Figure (svg): A bar of the club's 50 ordered places with a grey dashed line at place 25.5 and a blue arrow leaving the top place labelled 60 to 80
Predict first
One member's age is corrected from 60 to 80 years.
What happens to the club's two centres?
Correct: The mean rises 0.4; the median stays 31
Why: The correction moves one age by 80 − 60 = 20, and the shift rule shares that among all 50 members: 20 ÷ 50 = 0.4, so the mean becomes 34.4. Both 60 and 80 sit above the middle of the ordered ages, so the members at places 25 and 26 keep their ages and the median stays 31.
Worked example
Figure (svg): A bar of the club's 50 ordered places with a grey dashed line at place 25.5 and a blue arrow leaving the top place labelled 60 to 80
\[ \textcolor{#1f5fbf}{80} - \textcolor{#1f5fbf}{60} = \textcolor{#1f5fbf}{20} \]
Measure the correction
Why: a distance drives the rule
\[ \textcolor{#1f5fbf}{20} \div 50 = \textcolor{#1f5fbf}{0.4} \]
Share it among 50
Why: everyone holds part
\[ \textcolor{#6b7280}{34} + \textcolor{#1f5fbf}{0.4} = \textcolor{#6b7280}{34.4} \]
Add the shift on
Why: the club's new balance
\[ 1 + 50 = 51 \]
Add the outer places
Why: the median rule wants ends
\[ 51 \div 2 = 25.5 \]
Halve 51
Why: places 25 and 26 decide
Figure (svg): A bar of the club's 50 ordered places split at place 25.5, with a grey line there and a blue arrow at the top place labelled 60 to 80
\[ \textcolor{#1f5fbf}{60} > \textcolor{#6b7280}{31},\ \ \textcolor{#1f5fbf}{80} > \textcolor{#6b7280}{31} \]
Check: both ages exceed 31
Why: no middle place changed hands
Faded example
Figure (svg): A bar of 25 ordered places with a grey dashed line at place 13 and a blue arrow at the top place labelled up 5
Made up: 25 scores, mean 62. One score rises 5.
Fill in the blanks
shift = 5 ÷ 25 = 0.2; new mean = 62.2
Why: The shift rule shares the 5-point rise among all 25 scores: 5 ÷ 25 = 0.2, so the mean climbs from 62 to 62.2. The median only moves if that score crossed the middle place, which is place 13.
Worked example
Figure (svg): A bar of 25 ordered places with a grey dashed line at place 13 and a blue arrow at the top place labelled up 5
\[ \textcolor{#1f5fbf}{5} \div 25 = \textcolor{#1f5fbf}{0.2} \]
Share the rise among 25
Why: the rule divides by n
\[ \textcolor{#6b7280}{62} + \textcolor{#1f5fbf}{0.2} = \textcolor{#6b7280}{62.2} \]
Add the shift to the mean
Why: the new equal cut
\[ 25 \times \textcolor{#6b7280}{62} = \textcolor{#1f5fbf}{1550} \]
Rebuild the old pool
Why: the long way, as a cross-check
\[ \textcolor{#1f5fbf}{1550} + \textcolor{#1f5fbf}{5} = \textcolor{#1f5fbf}{1555} \]
Add the rise to that pool
Why: only one score changed
\[ \textcolor{#1f5fbf}{1555} \div 25 = \textcolor{#6b7280}{62.2} \]
Check: share the new pool
Why: the long road lands too
Section
Idea 4 of 4
Concept
Figure (svg): Two dot plots on one scale from 0 to 10: Terry's counts with a tall stack at 3 and single counts out at 7 and 9, and Davis' counts with a tall stack at 3 and nothing above 4
Example 2.31: ten letter counts each, one dot per word.
Discussion prompt
Whose counts stretch to the high side, Terry's or Davis'?
Answer:
Terry's: 7 and 9 hang far above the rest. Can numbers alone say so?
Worked example
Figure (svg): Dot plot of Terry's ten letter counts from 0 to 10: one dot at 1, two at 2, four at 3, one at 4, one at 7, one at 9
\[ 4 \times \textcolor{#1f5fbf}{3} = \textcolor{#1f5fbf}{12} \]
Total Terry's four 3s
Why: counting shortens the pool
\[ \textcolor{#1f5fbf}{1 + 2 + 2 + 12 + 4 + 7 + 9} = \textcolor{#1f5fbf}{37} \]
Add the seven part totals
Why: Σx must hold ten counts
\[ \textcolor{#1f5fbf}{37} \div 10 = \textcolor{#6b7280}{3.7} \]
Share the pool among 10
Why: Terry's mean, for comparison
Figure (svg): Dot plot of Terry's ten letter counts from 0 to 10: one dot at 1, two at 2, four at 3, one at 4, one at 7, one at 9, with a grey dashed line at the mean 3.7
\[ \textcolor{#1f5fbf}{3}:\ 4 \text{ counts} \]
Find Terry's tallest stack
Why: a peak is the obvious reference
Figure (svg): Dot plot of Terry's ten letter counts from 0 to 10: one dot at 1, two at 2, four at 3, one at 4, one at 7, one at 9, with a grey mean line at 3.7 and the four-dot stack at 3 boxed as the mode
\[ \textcolor{#6b7280}{3.7} > \textcolor{#1f5fbf}{3} \]
Compare the mean with it
Why: the mean settled above the crowd
\[ 10 \times \textcolor{#6b7280}{3.7} = \textcolor{#1f5fbf}{37} \]
Check: rebuild Terry's pool
Why: ten cuts give the pool back
Worked example
Figure (svg): Dot plot of Terry's ten letter counts from 0 to 10: one dot at 1, two at 2, four at 3, one at 4, one at 7, one at 9, with a grey dashed line at the mean 3.7
\[ 1 + 10 = 11 \]
Add the outer places
Why: a place before a value
\[ 11 \div 2 = 5.5 \]
Halve 11
Why: 5.5 sits between two counts
\[ \text{5th} = \textcolor{#1f5fbf}{3},\ \ \text{6th} = \textcolor{#1f5fbf}{3} \]
Read the two middles
Why: both land in the 3s
\[ M = \textcolor{#6b7280}{3} \]
Take halfway of two 3s
Why: no averaging needed
\[ \textcolor{#6b7280}{3.7} > \textcolor{#6b7280}{3} \]
Compare the two
Why: the mean drifted high
Figure (svg): Dot plot of Terry's ten letter counts from 0 to 10: one dot at 1, two at 2, four at 3, one at 4, one at 7, one at 9, with a grey dashed mean line at 3.7 and a grey dotted median line at 3
\[ \text{places } 1\text{–}5 \le \textcolor{#1f5fbf}{3},\ \ 6\text{–}10 \ge \textcolor{#1f5fbf}{3} \]
Check: read the places
Why: five each side
Worked example
Figure (svg): Dot plot of Davis' ten letter counts from 0 to 10: two dots at 1, one at 2, five at 3, two at 4
\[ 2 \times \textcolor{#1f5fbf}{1} = \textcolor{#1f5fbf}{2} \]
Total Davis' two 1s
Why: his shortest words repeat
\[ 5 \times \textcolor{#1f5fbf}{3} = \textcolor{#1f5fbf}{15} \]
Total his five 3s
Why: the tallest stack weighs most
\[ 2 \times \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{8} \]
Total his two 4s
Why: no repeat left out
\[ \textcolor{#1f5fbf}{2 + 2 + 15 + 8} = \textcolor{#1f5fbf}{27} \]
Add the four part totals
Why: Σx must hold ten counts
\[ \textcolor{#1f5fbf}{27} \div 10 = \textcolor{#6b7280}{2.7} \]
Share the pool among 10
Why: Davis' mean, beside Terry's
Figure (svg): Dot plot of Davis' ten letter counts from 0 to 10: two dots at 1, one at 2, five at 3, two at 4, with a grey dashed line at the mean 2.7
\[ 10 \times \textcolor{#6b7280}{2.7} = \textcolor{#1f5fbf}{27} \]
Check: rebuild Davis' pool
Why: the cut times ten is Σx
Worked example
Figure (svg): Dot plot of Davis' ten letter counts from 0 to 10: two dots at 1, one at 2, five at 3, two at 4, with a grey dashed line at the mean 2.7
Ten counts again: middle place 5.5.
\[ \text{5th} = \textcolor{#1f5fbf}{3},\ \ \text{6th} = \textcolor{#1f5fbf}{3} \]
Read Davis' middles
Why: the same place as Terry
\[ M = \textcolor{#6b7280}{3} \]
Take halfway between two 3s
Why: his median is a count
\[ \textcolor{#6b7280}{2.7} < \textcolor{#6b7280}{3} \]
Compare his mean with it
Why: this mean drifted low
Figure (svg): Dot plot of Davis' ten letter counts from 0 to 10: two dots at 1, one at 2, five at 3, two at 4, with a grey dashed mean line at 2.7 and a grey dotted median line at 3
\[ \textcolor{#6b7280}{3} - \textcolor{#6b7280}{2.7} = \textcolor{#1f5fbf}{0.3},\ \ \textcolor{#6b7280}{3.7} - \textcolor{#6b7280}{3} = \textcolor{#1f5fbf}{0.7} \]
Measure both from 3
Why: Terry's strayed further
\[ \text{places } 4 \text{ to } 8 = \textcolor{#1f5fbf}{3} \]
Check: name the 3s' places
Why: 5 and 6 sit inside
Worked example
Figure (svg): Two dot plots on one scale from 0 to 10: Terry's counts with a tall stack at 3 and single counts out at 7 and 9, and Davis' counts with a tall stack at 3 and nothing above 4, each with a grey dashed mean line and a grey dotted median line
\[ \textcolor{#6b7280}{3.7} - \textcolor{#6b7280}{3} = \textcolor{#1f5fbf}{+0.7} \]
Subtract Terry's median
Why: a signed gap carries direction
\[ \textcolor{#6b7280}{2.7} - \textcolor{#6b7280}{3} = \textcolor{#1f5fbf}{-0.3} \]
Subtract Davis' median
Why: his sign comes out reversed
\[ \textcolor{#1f5fbf}{+0.7}:\ \text{high side} \]
Read Terry's sign
Why: his 7 and 9 stretch upward
\[ \textcolor{#1f5fbf}{-0.3}:\ \text{low side} \]
Read Davis' sign
Why: his two 1s stretch downward
In both samples the mean fell on the stretched side.
\[ \textcolor{#6b7280}{3} + \textcolor{#1f5fbf}{0.7} = \textcolor{#6b7280}{3.7},\ \ \textcolor{#6b7280}{3} - \textcolor{#1f5fbf}{0.3} = \textcolor{#6b7280}{2.7} \]
Check: step from 3 by each gap
Why: both steps land on their means
Trap
\[ \textcolor{#6b7280}{4.6} - \textcolor{#6b7280}{4} = \textcolor{#1f5fbf}{+0.6} \]
Subtract the median
Why: the sign names the pulled side
\[ \textcolor{#1f5fbf}{+0.6} \Rightarrow \text{stretches high} \]
Read the sign
Why: the plot disagrees
\[ \textcolor{#1f5fbf}{8} - \textcolor{#1f5fbf}{6} = \textcolor{#1f5fbf}{2},\ \ \textcolor{#1f5fbf}{4} - \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{2} \]
Measure both ends
Why: the reaches agree
Figure (svg): Dot plot of Maris' ten letter counts from 0 to 10: one dot at 2, two at 3, three at 4, three at 6, one at 8, with a grey dashed fold line at 5 and two counts at 3 facing an empty 7
\[ \textcolor{#1f5fbf}{5} - \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{3},\ \ \textcolor{#1f5fbf}{8} - \textcolor{#1f5fbf}{5} = \textcolor{#1f5fbf}{3} \]
Measure the ends
Why: test the fold
\[ \textcolor{#1f5fbf}{3}:\ 2 \text{ counts},\ \ \textcolor{#1f5fbf}{7}:\ \text{none} \]
Count 3 against 7
Why: weight lies low
\[ \textcolor{#6b7280}{4.6} > \textcolor{#6b7280}{4} \]
Check the centres
Why: a tendency only
Worked example
Figure (svg): Dot plot of Group B's 10 marks from 4 to 10: one dot at 4, one at 5, three at 6, four at 7, one at 8, nothing above 8, with a grey dashed mean line at 6.3 and a grey dotted median line at 6.5
\[ \textcolor{#1f5fbf}{20} - \textcolor{#1f5fbf}{8} = \textcolor{#1f5fbf}{12} \]
Push the top mark to 20
Why: tests which centre notices
\[ \textcolor{#1f5fbf}{12} \div 10 = \textcolor{#1f5fbf}{1.2} \]
Share the push among 10
Why: the shift rule gives the travel
\[ \textcolor{#6b7280}{6.3} + \textcolor{#1f5fbf}{1.2} = \textcolor{#6b7280}{7.5} \]
Add the shift to the mean
Why: one mark carried it past 7
Figure (svg): Dot plot of Group B with its top mark pushed out to 20, on a scale from 4 to 20, with a grey dashed mean line at 7.5 and a grey dotted median line at 6.5
\[ M = \textcolor{#6b7280}{6.5} \]
Read the median again
Why: no mark changed places
Report the median when data stretches one way.
\[ \textcolor{#6b7280}{7.5} - \textcolor{#6b7280}{6.3} = \textcolor{#1f5fbf}{1.2} \]
Check: measure the mean's travel
Why: it matches the shared push
Prediction
Figure (svg): A bar of 200 ordered places split in half at place 100.5, the lower half labelled 2 items or fewer and the upper half 2 items or more
Predict first
A shop reports three centres for 200 orders: mode 1 item, median 2 items, mean 5.4 items.
Which statement do those three numbers support?
Correct: At least 100 orders are below the mean
Why: A median of 2 means at least 100 of the 200 orders are 2 items or fewer, and 2 is below 5.4. It is the median, not the mean, that splits the count in half; the commonest order is the mode, 1 item; and a mean sitting far above the median is the mark of a one-sided stretch, not an even spread.
Worked example
Figure (svg): A bar of the shop's 200 orders, in size order, not yet divided
\[ 1 + 200 = 201 \]
Add the outer places
Why: 200 orders have a middle too
\[ 201 \div 2 = 100.5 \]
Halve 201
Why: places 100 and 101 decide
\[ \text{places } 1\text{–}100 \le \textcolor{#1f5fbf}{2} \]
Read the places below
Why: a median of 2 caps them
Figure (svg): A bar of 200 ordered places split at place 100.5, the lower half labelled 2 items or fewer, with a grey line at the middle place
\[ \textcolor{#1f5fbf}{2} < \textcolor{#6b7280}{5.4} \]
Compare the cap with 5.4
Why: all 100 sit under it too
\[ 200 - 100 = 100 \]
Check: count what is left
Why: at most 100 beat the mean
Faded example
Figure (svg): A stem-and-leaf plot of the presidents' ages: stem 4 with leaves 6 and 9, stem 5 with leaves 3, 6, 7, 7, 7 and 8, stem 6 with twelve leaves 0, 0, 3, 3, 4, 4, 5, 6, 7, 7, 7 and 8
Try It 2.31: stem = tens, leaf = ones, 20 ages.
Fill in the blanks
middle places = 10 and 11; M = 61.5 years
Why: 1 + 20 = 21 and 21 ÷ 2 = 10.5, so places 10 and 11 decide M. The first two stems hold 8 ages, so place 9 is the first leaf of stem 6; place 10 is 60 and place 11 is 63, and (60 + 63) ÷ 2 = 61.5.
Worked example
Figure (svg): A stem-and-leaf plot of the presidents' ages: stem 4 with leaves 6 and 9, stem 5 with leaves 3, 6, 7, 7, 7 and 8, stem 6 with twelve leaves 0, 0, 3, 3, 4, 4, 5, 6, 7, 7, 7 and 8
\[ 1 + 20 = 21 \]
Add the outer places
Why: the stem plot is sorted
\[ 21 \div 2 = 10.5 \]
Halve 21
Why: places 10 and 11 decide
\[ 2 + 6 = 8 \]
Add the stems' leaves
Why: tells where sixties begin
\[ \text{10th} = \textcolor{#1f5fbf}{60},\ \ \text{11th} = \textcolor{#1f5fbf}{63} \]
Read places 10 and 11
Why: the sixties open here
Figure (svg): A stem-and-leaf plot of the presidents' ages: stem 4 with leaves 6 and 9, stem 5 with leaves 3, 6, 7, 7, 7 and 8, stem 6 with twelve leaves 0, 0, 3, 3, 4, 4, 5, 6, 7, 7, 7 and 8; the leaves at places 10 and 11 boxed
\[ \textcolor{#1f5fbf}{60} + \textcolor{#1f5fbf}{63} = \textcolor{#1f5fbf}{123} \]
Add the pair
Why: a sum comes before halving
\[ \textcolor{#1f5fbf}{123} \div 2 = \textcolor{#6b7280}{61.5} \]
Check: halve that total
Why: 61.5 sits 1.5 either way
Worked example
Figure (svg): Dot plot of 20 presidents' ages at death on a scale from 45 to 70: single dots at 46, 49, 53, 56 and 58, three at 57, two at 60, two at 63, two at 64, one each at 65, 66 and 68, three at 67
\[ \textcolor{#1f5fbf}{46 + 49} = \textcolor{#1f5fbf}{95} \]
Total the forties row
Why: one stem at a time
\[ 3 \times \textcolor{#1f5fbf}{57} = \textcolor{#1f5fbf}{171} \]
Total the three 57s
Why: three equal ages multiply once
\[ \textcolor{#1f5fbf}{53 + 56 + 58} = \textcolor{#1f5fbf}{167} \]
Add the single fifties
Why: three ages appear once
\[ \textcolor{#1f5fbf}{171} + \textcolor{#1f5fbf}{167} = \textcolor{#1f5fbf}{338} \]
Add the fifties parts
Why: that stem's own pool
\[ \textcolor{#1f5fbf}{95} + \textcolor{#1f5fbf}{338} = \textcolor{#1f5fbf}{433} \]
Check: add the two stems
Why: eight ages, each counted once
Worked example
Figure (svg): Dot plot of 20 presidents' ages at death on a scale from 45 to 70: single dots at 46, 49, 53, 56 and 58, three at 57, two at 60, two at 63, two at 64, one each at 65, 66 and 68, three at 67
\( {95 + 338 = 433} \)
\[ 2 \times \textcolor{#1f5fbf}{60},\ 2 \times \textcolor{#1f5fbf}{63},\ 2 \times \textcolor{#1f5fbf}{64} = \textcolor{#1f5fbf}{120},\ \textcolor{#1f5fbf}{126},\ \textcolor{#1f5fbf}{128} \]
Total the three pairs
Why: counting each repeat
\[ 3 \times \textcolor{#1f5fbf}{67} = \textcolor{#1f5fbf}{201} \]
Total the three 67s
Why: the commonest age weighs most
\[ \textcolor{#1f5fbf}{65 + 66 + 68} = \textcolor{#1f5fbf}{199} \]
Add the single sixties
Why: no repeat here to shorten
\[ \textcolor{#1f5fbf}{120 + 126 + 128 + 201 + 199} = \textcolor{#1f5fbf}{774} \]
Add the five parts
Why: this stem's pool
\[ \textcolor{#1f5fbf}{433} + \textcolor{#1f5fbf}{774} = \textcolor{#1f5fbf}{1207} \]
Check: add both stems
Why: all 20 ages once
Worked example
Figure (svg): Dot plot of 20 presidents' ages at death on a scale from 45 to 70: single dots at 46, 49, 53, 56 and 58, three at 57, two at 60, two at 63, two at 64, one each at 65, 66 and 68, three at 67
\[ \textcolor{#1f5fbf}{1207} \div 20 = \textcolor{#6b7280}{60.35} \]
Share the pool among 20
Why: one age if all were level
Figure (svg): Dot plot of 20 presidents' ages at death on a scale from 45 to 70: single dots at 46, 49, 53, 56 and 58, three at 57, two at 60, two at 63, two at 64, one each at 65, 66 and 68, three at 67, with a grey dashed line at the mean 60.35
\[ \textcolor{#6b7280}{60.35} - \textcolor{#6b7280}{61.5} = \textcolor{#1f5fbf}{-1.15} \]
Subtract the median
Why: the sign names a side
Figure (svg): Dot plot of 20 presidents' ages at death on a scale from 45 to 70: single dots at 46, 49, 53, 56 and 58, three at 57, two at 60, two at 63, two at 64, one each at 65, 66 and 68, three at 67, with a grey dashed mean line at 60.35 and a grey dotted median line at 61.5
\[ \textcolor{#1f5fbf}{-1.15}:\ \text{low side} \]
Read the sign
Why: the stretch runs towards 46
\[ \textcolor{#1f5fbf}{57} - \textcolor{#1f5fbf}{46} = \textcolor{#1f5fbf}{11},\ \ \textcolor{#1f5fbf}{68} - \textcolor{#1f5fbf}{57} = \textcolor{#1f5fbf}{11} \]
Measure each end from 57
Why: equal reach, unequal filling
\[ 2 \text{ ages below } \textcolor{#1f5fbf}{53},\ \ 6 \text{ above } \textcolor{#1f5fbf}{64} \]
Check: count both thin ends
Why: two ages hold the low one
Pattern
\[ \text{shift} = \textcolor{#1f5fbf}{c} \div n,\qquad \textcolor{#6b7280}{\bar{x}} - M \text{ names the stretched side} \]
Check
Figure (svg): A bar of 20 ordered places with a grey line at place 10.5 and a blue arrow leaving the top place labelled up c
Check your understanding
One student's score in a class of 20 is re-marked upward. How far must it rise to lift the class mean from 5 to 5.4?
Answer: A
Why: The shift rule says the mean travels c ÷ n. Here the travel is 5.4 − 5 = 0.4 and n = 20, so c ÷ 20 = 0.4 and c = 0.4 × 20 = 8 marks.
Worked example
Figure (svg): A bar of 20 ordered places with a grey line at place 10.5 and a blue arrow at the top place labelled up 8
\[ \textcolor{#6b7280}{5.4} - \textcolor{#6b7280}{5} = \textcolor{#1f5fbf}{0.4} \]
Measure the mean's required travel
Why: the shift rule's known side
\[ \textcolor{#1f5fbf}{0.4} \times 20 = \textcolor{#1f5fbf}{8} \]
Undo the sharing by 20
Why: multiplying back gives the move
\[ 20 \times \textcolor{#6b7280}{5} = \textcolor{#1f5fbf}{100} \]
Rebuild the old pool
Why: a second route to the answer
\[ \textcolor{#1f5fbf}{100} + \textcolor{#1f5fbf}{8} = \textcolor{#1f5fbf}{108} \]
Add the 8-mark rise
Why: only one score changed
\[ \textcolor{#1f5fbf}{108} \div 20 = \textcolor{#6b7280}{5.4} \]
Check: share the new pool
Why: the mean lands on target
Check
Figure (svg): Four dot plots on one scale from 1 to 10, one per made-up list: 1, 6, 7, 8, 9; 1, 2, 3, 4, 10; 3, 4, 5, 6, 7; and 2, 5, 5, 5, 8
Check your understanding
Which of these made-up lists of five values has its mean below its median?
Answer: A
Why: Its total is 31, so the mean is 6.2, while its third value is 7. The lone 1 stretches the low side and drags the mean under the median.
Worked example
Figure (svg): Four dot plots on one scale from 1 to 10, one per made-up list: 1, 6, 7, 8, 9; 1, 2, 3, 4, 10; 3, 4, 5, 6, 7; and 2, 5, 5, 5, 8, each with a grey dashed mean line at 6.2, 4, 5 and 5
\[ \textcolor{#1f5fbf}{31},\ \textcolor{#1f5fbf}{20},\ \textcolor{#1f5fbf}{25},\ \textcolor{#1f5fbf}{25} \]
Total each list in turn
Why: a mean starts from its pool
\[ \textcolor{#1f5fbf}{31} \div 5,\ \textcolor{#1f5fbf}{20} \div 5,\ \textcolor{#1f5fbf}{25} \div 5 = \textcolor{#6b7280}{6.2},\ \textcolor{#6b7280}{4},\ \textcolor{#6b7280}{5} \]
Share each pool among 5
Why: three shares cover four
\[ \text{3rd values} = \textcolor{#1f5fbf}{7},\ \textcolor{#1f5fbf}{3},\ \textcolor{#1f5fbf}{5},\ \textcolor{#1f5fbf}{5} \]
Read place 3 of each
Why: odd n puts M there
\[ \textcolor{#6b7280}{6.2} < \textcolor{#1f5fbf}{7};\ \ \textcolor{#6b7280}{4} > \textcolor{#1f5fbf}{3};\ \ \textcolor{#6b7280}{5} = \textcolor{#1f5fbf}{5} \]
Compare each with its M
Why: only the first falls short
\[ \textcolor{#1f5fbf}{9} - \textcolor{#1f5fbf}{7} = \textcolor{#1f5fbf}{2},\ \ \textcolor{#1f5fbf}{7} - \textcolor{#1f5fbf}{1} = \textcolor{#1f5fbf}{6} \]
Check: measure each side
Why: the low reach is longer
Worked example
Figure (svg): Three dot plots on one scale from 4 to 10, each with a grey dashed line at 7: Group A with 16 marks spread evenly either side, Group B with 10 marks reaching down to 4 and stopping at 8, Group C with 10 marks starting at 6 and reaching up to 10
\[ \textcolor{#6b7280}{6.3} < \textcolor{#6b7280}{7} < \textcolor{#6b7280}{7.7} \]
Order the three means
Why: the opening question
Figure (svg): Three dot plots on one scale from 4 to 10, each with a grey dashed line at 7: Group A with 16 marks spread evenly either side, Group B with 10 marks reaching down to 4 and stopping at 8, Group C with 10 marks starting at 6 and reaching up to 10, each labelled with its mean: 7, 6.3 and 7.7
\[ \textcolor{#6b7280}{7} - \textcolor{#6b7280}{6.3} = \textcolor{#1f5fbf}{0.7},\ \ \textcolor{#6b7280}{7.7} - \textcolor{#6b7280}{7} = \textcolor{#1f5fbf}{0.7} \]
Measure both from A
Why: equal misses from the fold
\[ 14 - \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{10},\ \ 14 - \textcolor{#1f5fbf}{8} = \textcolor{#1f5fbf}{6} \]
Reflect B's ends through 7
Why: they land on C's
\[ \text{B stretches down, C up} \]
Name what that shows
Why: one set mirrors the other
\[ 14 - \textcolor{#6b7280}{6.3} = \textcolor{#6b7280}{7.7} \]
Check: reflect B's mean
Why: Group C's value returns
Recap
OpenStax Introductory Statistics 2e, §2.6 Skewness and the Mean, Median, and Mode §2.6, pp. 104-106 — Figures 2.16-2.21, Example 2.31 and Try It 2.31 trace back here
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