Build the mean from equal sharing and balance, locate the median and see why extreme values cannot move it, count modes, pool frequency tables, estimate grouped means, and meet the law of large numbers.
Subject: Statistics · 75 slides · applied lesson
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Title
Statistics · §2.5
Mean, median and mode, each built from a picture
Objectives
Concept
\[ u + v + w = w + u + v \]
Any order
Why: totals stay the same
\[ 3 \times u = u + u + u \]
Multiply
Why: copies of one number
\[ u \div v = w \iff w \times v = u \]
Divide back
Why: checks every share
\[ \text{halfway} = (u + v) \div 2;\ \ u, u \to u \]
Halfway
Why: equal steps from both ends
\[ 5,\ 1,\ 3 \to 1,\ 3,\ 5 \]
Sort
Why: smallest value first
\[ u - v < 0 \text{ if } u < v \]
Signs
Why: below gives a negative
\[ \text{rel. freq.} = \text{count} \div \text{total} \]
Relative frequency
Why: each value's share of the data
\[ 2 \div 3 \approx 0.67 \]
Round
Why: ≈ marks two kept decimals
Prediction
Figure (svg): The town's incomes in $ thousands: full scale 0 to 5000 with 49 dots at 30 and one at 5000; a zoom from 0 to 200 with the 49 dots at 30
Predict first
A leaflet reports one 'average income' for this town of 50.
Could that single number mislead?
Correct: Yes: one income can drag it far from the 49
Why: The picture shows 49 of the 50 dots at $30,000 and one far away. A number built by adding everyone's income feels that far dot. The deck builds the tools to measure how much.
Worked example
Figure (svg): The town's incomes in $ thousands: full scale 0 to 5000 with 49 dots at 30 and one at 5000; a zoom from 0 to 200 with the 49 dots at 30
Incomes in $ thousands.
\[ \textcolor{#1f5fbf}{5000} \div \textcolor{#1f5fbf}{30} \approx 166.7 \]
Divide 5000 by 30
Why: measures how extreme one income is
Figure (svg): The town's incomes in $ thousands: full scale 0 to 5000 with 49 dots at 30 and one at 5000, with a striped bar from 0 to 5000 made of incomes of 30 end to end, labelled about 166.7 incomes of 30; a zoom from 0 to 200 with the 49 dots at 30
\[ 49 \times \textcolor{#1f5fbf}{30} = \textcolor{#1f5fbf}{1470} \]
Total the 49 ordinary incomes
Why: weigh them against one
\[ \textcolor{#1f5fbf}{1470} < \textcolor{#1f5fbf}{5000} \]
Compare with the far income
Why: yes: one outweighs all 49
\[ 166.7 \times \textcolor{#1f5fbf}{30} = \textcolor{#1f5fbf}{5001} \approx \textcolor{#1f5fbf}{5000} \]
Check: undo the division
Why: rounding returns about 5000
Needed: one number for the town's centre.
Section
Idea 1 of 4
Concept
Figure (svg): Five columns of blue blocks for the students' movie counts 0, 1, 1, 2 and 6, on a scale from 0 to 6
Goal: one number for a typical student's week.
Discussion prompt
If every movie were pooled and dealt out evenly, how would you find each student's share?
Answer:
Add all the movies, then split that total into five equal stacks.
Worked example
Figure (svg): Five columns of blue blocks for the students' movie counts 0, 1, 1, 2 and 6, on a scale from 0 to 6
\[ \textcolor{#1f5fbf}{0} + \textcolor{#1f5fbf}{6} = \textcolor{#1f5fbf}{6} \]
Add the smallest and largest
Why: halfway needs both ends' sum
\[ \textcolor{#1f5fbf}{6} \div 2 = \textcolor{#6b7280}{3} \]
Halve the sum
Why: tests the ends' midpoint as centre
Figure (svg): The five movie columns 0, 1, 1, 2, 6 with a grey dashed line at height 3, halfway between the smallest and largest
\[ \textcolor{#1f5fbf}{0},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{2} < \textcolor{#6b7280}{3} \]
List the students below 3
Why: tests if 3 is typical
Only the two ends counted.
\[ \textcolor{#6b7280}{3} - \textcolor{#1f5fbf}{0} = \textcolor{#1f5fbf}{6} - \textcolor{#6b7280}{3} \]
Check: equal distance to each end
Why: halfway means equal gaps
Worked example
Figure (svg): Five columns of blue blocks for the students' movie counts 0, 1, 1, 2 and 6, on a scale from 0 to 6
\[ \textcolor{#1f5fbf}{0 + 1 + 1 + 2 + 6} = \textcolor{#1f5fbf}{10} \]
Pool every student's movies
Why: sharing must keep the total
Figure (svg): The five movie columns emptied to dashed outlines; all 10 blocks moved into one pooled row on top
\[ \textcolor{#1f5fbf}{10} \div 5 = \textcolor{#6b7280}{2} \]
Deal the pool to 5 students
Why: evening out shows the typical count
Figure (svg): Five equal columns of 2 blue blocks each, with a grey dashed line at 2 marked 2 each
That equal share is the mean: 2 movies.
\[ 5 \times \textcolor{#6b7280}{2} = \textcolor{#1f5fbf}{10} \]
Check: stack the shares back up
Why: five columns rebuild the pool
Worked example
Figure (svg): Five columns of blue blocks for the students' movie counts 0, 1, 1, 2 and 6, on a scale from 0 to 6
x: a value · n: how many · Σ: add them all
\[ \textcolor{#1f5fbf}{{\textstyle\sum} x} = \textcolor{#1f5fbf}{0 + 1 + 1 + 2 + 6} \]
Pool as Σx
Why: a symbol covers any list's total
Figure (svg): The five movie columns emptied to dashed outlines; all 10 blocks moved into one pooled row on top
\[ n = 5 \]
Count as n
Why: the divisor must fit any size
x̄ (x-bar): the mean of a sample
\[ \textcolor{#6b7280}{\bar{x}} = \frac{\textcolor{#1f5fbf}{{\textstyle\sum} x}}{n} \]
Divide Σx by n
Why: the dealing move, for any list
Figure (svg): Five equal columns of 2 blue blocks each, with a grey dashed line at 2 marked 2 each
\[ \textcolor{#1f5fbf}{10} \div 5 = \textcolor{#6b7280}{2} \]
Check with the movies
Why: matches the dealt-out share
Worked example
Figure (svg): Movie counts 0, 1, 1, 2, 6 as dots on a number line from 0 to 6, a grey dashed line at the mean 2
\[ \textcolor{#1f5fbf}{0} - \textcolor{#6b7280}{2},\ \textcolor{#1f5fbf}{1} - \textcolor{#6b7280}{2},\ \textcolor{#1f5fbf}{1} - \textcolor{#6b7280}{2} = \textcolor{#1f5fbf}{-2},\ \textcolor{#1f5fbf}{-1},\ \textcolor{#1f5fbf}{-1} \]
Subtract 2 from 0, 1, 1
Why: balance weighs distances each side
Figure (svg): Movie counts 0, 1, 1, 2, 6 as dots on a number line from 0 to 6, a grey dashed line at the mean 2, and 3 blue arrows from 2 to the values (−2, −1, −1, 0, +4 in order)
\[ \textcolor{#1f5fbf}{2} - \textcolor{#6b7280}{2},\ \textcolor{#1f5fbf}{6} - \textcolor{#6b7280}{2} = \textcolor{#1f5fbf}{0},\ \textcolor{#1f5fbf}{+4} \]
Subtract 2 from 2 and 6
Why: the other side must match
Figure (svg): Movie counts 0, 1, 1, 2, 6 as dots on a number line from 0 to 6, a grey dashed line at the mean 2, and 5 blue arrows from 2 to the values (−2, −1, −1, 0, +4 in order)
\[ \textcolor{#1f5fbf}{-2 - 1 - 1} = \textcolor{#1f5fbf}{-4} \]
Total the left pulls
Why: to weigh against the right side
\[ \textcolor{#1f5fbf}{-4} + \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{0} \]
Add the right pull
Why: zero turn: it balances
Figure (svg): Movie counts 0, 1, 1, 2, 6 as dots on a number line from 0 to 6, a grey dashed line at the mean 2, and 5 blue arrows from 2 to the values (−2, −1, −1, 0, +4 in order); label: pulls total 0
\[ \textcolor{#6b7280}{2} + (\textcolor{#1f5fbf}{-2}) = \textcolor{#1f5fbf}{0},\ \ \textcolor{#6b7280}{2} + \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{6} \]
Check: hop back from 2
Why: arrows land on the data
Worked example
Figure (svg): Dot stacks for the sample 1, 1, 1, 2, 2, 3, 4, 4, 4, 4, 4: three dots at 1, two at 2, one at 3, five at 4, on an axis from 0 to 5
\[ \begin{aligned} &\textcolor{#1f5fbf}{1 + 1 + 1 + 2 + 2 + 3} \\ &{+}\ \textcolor{#1f5fbf}{4 + 4 + 4 + 4 + 4} = \textcolor{#1f5fbf}{30} \end{aligned} \]
Add all eleven values
Why: the pool the mean will share
\[ n = 11 \]
Count the dots
Why: each value gets one share
\[ \textcolor{#6b7280}{\bar{x}} = \textcolor{#1f5fbf}{30} \div 11 \approx \textcolor{#6b7280}{2.73} \]
Divide Σx by n
Why: applies the mean rule just built
Figure (svg): Dot stacks for the sample with a grey line at the mean 2.73, between the stacks at 2 and 3
\[ 11 \times \textcolor{#6b7280}{2.73} = \textcolor{#1f5fbf}{30.03} \approx \textcolor{#1f5fbf}{30} \]
Check: rebuild the pool
Why: near 30 after rounding
Trap
Figure (svg): The five movie counts in cells, 0, 1, 1, 2, 6, with the 0 cell dashed and set aside
\[ \textcolor{#1f5fbf}{1 + 1 + 2 + 6} = \textcolor{#1f5fbf}{10} \]
Drop the 0
Why: it adds no movies
\[ \textcolor{#1f5fbf}{10} \div 4 = \textcolor{#6b7280}{2.5} \]
Divide by 4
Why: counts only watchers
\[ \textcolor{#6b7280}{2.5} \ne \textcolor{#6b7280}{2} \]
Compare with 2
Why: fails: five students share
Figure (svg): The five movie counts in cells, 0, 1, 1, 2, 6, all five kept
\[ \textcolor{#1f5fbf}{0 + 1 + 1 + 2 + 6} = \textcolor{#1f5fbf}{10} \]
Keep the 0
Why: still one student
\[ \textcolor{#1f5fbf}{10} \div 5 = \textcolor{#6b7280}{2} \]
Divide by 5
Why: n counts zeros too
\[ 5 \times \textcolor{#6b7280}{2} = \textcolor{#1f5fbf}{10} \]
Check: multiply back
Why: five shares rebuild the 10
Prediction
Figure (svg): Movie counts 0, 1, 1, 2, 6 as dots on a number line from 0 to 6, a grey dashed line at the mean 2
Predict first
A sixth student, who watched 2 movies, joins these five.
What happens to the mean of 2?
Correct: It stays 2
Why: The pool gains 2 movies and the count gains one share, so each share is unchanged. On the balance picture the newcomer sits on the pivot and pulls neither way.
Worked example
Figure (svg): Movie counts 0, 1, 1, 2, 6 as dots on a number line from 0 to 6, a grey dashed line at the mean 2
\[ \textcolor{#1f5fbf}{10} + \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{12} \]
Add the newcomer's 2 movies
Why: sharing must keep every movie
Figure (svg): Movie counts 0, 1, 1, 2, 6 and a newcomer's 2 as dots on a number line from 0 to 6, a grey dashed line at the mean 2
\[ 5 + 1 = 6 \]
Add one to the count
Why: the newcomer also gets a share
\[ \textcolor{#1f5fbf}{12} \div 6 = \textcolor{#6b7280}{2} \]
Deal the new pool out
Why: the mean is each student's share
Figure (svg): Movie counts 0, 1, 1, 2, 6 and a newcomer's 2 as dots on a number line from 0 to 6, a grey dashed line at the mean 2, and arrows from 2 to each value (−2, −1, −1, 0, +4, 0); label: pulls total 0
\[ 6 \times \textcolor{#6b7280}{2} = \textcolor{#1f5fbf}{12} \]
Check: six shares rebuild the pool
Why: undoes the deal exactly
Section
Idea 2 of 4
Concept
Figure (svg): Cells for the corrected movie counts 0, 1, 1, 2, 26 in order, positions 1 to 5 underneath
The fifth student actually watched 26 movies, not 6.
Discussion prompt
Which way, and how far, will the mean of 2 move?
Answer:
Up, since every share grows; the next slide measures how far.
Worked example
Figure (svg): Dot plot of the corrected movie counts 0, 1, 1, 2, 26 on an axis from 0 to 30
\[ \textcolor{#1f5fbf}{0 + 1 + 1 + 2 + 26} = \textcolor{#1f5fbf}{30} \]
Pool the corrected counts
Why: the mean shares the new total
\[ \textcolor{#1f5fbf}{30} \div 5 = \textcolor{#6b7280}{6} \]
Share among the 5 students
Why: the mean to test
Figure (svg): Dot plot of the corrected movie counts 0, 1, 1, 2, 26 on an axis from 0 to 30, with a grey dashed line at the mean 6, right of four dots
\[ \textcolor{#1f5fbf}{0},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{2} < \textcolor{#6b7280}{6} \]
List the students under the mean
Why: tests if 6 is typical
One far value moved the mean from 2 to 6.
\[ 5 \times \textcolor{#6b7280}{6} = \textcolor{#1f5fbf}{30} \]
Check: rebuild the pool
Why: undoing the division returns 30
Worked example
Figure (svg): Ordered cells 0, 1, 1, 2, 26 with positions 1 to 5 underneath
\[ \textcolor{#1f5fbf}{0},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{2},\ \textcolor{#1f5fbf}{26} \]
Order the counts
Why: a middle needs sorted data
\[ \text{2 values} \mid \text{middle} \mid \text{2 values} \]
Split evenly around one value
Why: locates a centre ignoring distances
Figure (svg): Ordered cells 0, 1, 1, 2, 26 with position 3 highlighted and braces marking 2 left and 2 right
\[ \text{3rd value} = \textcolor{#6b7280}{1} \]
Read the middle cell
Why: the value, not the place, answers
The middle of ordered data is the median.
\[ \textcolor{#1f5fbf}{0},\ \textcolor{#1f5fbf}{1} \le \textcolor{#6b7280}{1} \le \textcolor{#1f5fbf}{2},\ \textcolor{#1f5fbf}{26} \]
Check: two values on each side
Why: a true middle has equal sides
Worked example
Figure (svg): Two ordered lists in cells with positions underneath: n = 5 holds 0, 1, 1, 2, 26; n = 6 holds 0, 1, 1, 2, 4, 26
\[ \text{places } 1, 2, \ldots, n \]
Number the ordered cells
Why: positions can locate a middle
\[ \textcolor{#6b7280}{\text{middle place}} = (1 + n) \div 2 \]
Go halfway from place 1 to n
Why: the halfway rule works on places
Figure (svg): Two ordered lists, n = 5 (0, 1, 1, 2, 26) and n = 6 (0, 1, 1, 2, 4, 26): the n = 5 list has cell 3 outlined in grey with braces counting 2 places on each side; the n = 6 list has a grey line between cells 3 and 4 with braces counting 3 on each side
\[ (1 + 5) \div 2 = \textcolor{#6b7280}{3} \]
Check n = 5 on the cells
Why: cell 3 has 2 each side
Worked example
Figure (svg): Ordered cells 0, 1, 1, 2, 4, 26 with positions 1 to 6
A sixth student watched 4 movies.
\[ 1 + 6 = 7 \]
Add places 1 and 6
Why: the middle place needs both ends
\[ 7 \div 2 = \textcolor{#6b7280}{3.5} \]
Halve 7
Why: halfway takes half the ends' sum
Figure (svg): Ordered cells 0, 1, 1, 2, 4, 26 with a grey line between positions 3 and 4 marked place 3.5
\[ 1, 2, 3 \mid \textcolor{#6b7280}{3.5} \mid 4, 5, 6 \]
Check: count places each side
Why: equal counts make it the middle
Worked example
Figure (svg): Ordered cells 0, 1, 1, 2, 4, 26 with a grey line between positions 3 and 4 marked place 3.5
\( {7 \div 2 = 3.5} \)
\[ \text{3rd} = \textcolor{#1f5fbf}{1},\ \ \text{4th} = \textcolor{#1f5fbf}{2} \]
Read places 3 and 4
Why: no value sits at place 3.5
Figure (svg): Ordered cells 0, 1, 1, 2, 4, 26 with positions 3 and 4 highlighted
\[ \textcolor{#1f5fbf}{1} + \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{3} \]
Add the middle pair
Why: halfway needs their sum
\[ \textcolor{#1f5fbf}{3} \div 2 = \textcolor{#6b7280}{1.5} \]
Halve the sum
Why: even n: median sits between middles
Figure (svg): Ordered cells 0, 1, 1, 2, 4, 26 with positions 3 and 4 highlighted and a grey line between them marked 1.5
\[ \textcolor{#6b7280}{1.5} - \textcolor{#1f5fbf}{1} = \textcolor{#1f5fbf}{2} - \textcolor{#6b7280}{1.5} \]
Check equal distances
Why: 0.5 to each neighbour
Worked example
Figure (svg): A dot plot from 0 to 30 labelled before: 0, 1, 1, 2, 26 with grey lines at median 1 and mean 6
\[ \textcolor{#1f5fbf}{0},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{1},\ \textcolor{#1f5fbf}{2},\ \textcolor{#1f5fbf}{260} \]
Replace 26 with 260
Why: the order is unchanged
Figure (svg): Two dot plots from 0 to 30: before, 0, 1, 1, 2, 26 with grey lines at median 1 and mean 6; after, 0, 1, 1, 2 with a blue arrow off the axis to 260
\[ \text{3rd value} = \textcolor{#6b7280}{1} \]
Read position 3 again
Why: medians depend on order
Figure (svg): The two dot plots; the after plot gains a grey line at median 1
\[ \textcolor{#1f5fbf}{0 + 1 + 1 + 2 + 260} = \textcolor{#1f5fbf}{264} \]
Pool the new counts
Why: the mean feels size
\[ \textcolor{#1f5fbf}{264} \div 5 = \textcolor{#6b7280}{52.8} \]
Share among 5
Why: the extreme value drags it
Figure (svg): The two dot plots; the after plot gains a grey arrow off the axis marked mean 52.8
Values far from the rest: extreme values.
\[ 5 \times \textcolor{#6b7280}{52.8} = \textcolor{#1f5fbf}{264} \]
Check: multiply back
Why: five shares must rebuild 264
Worked example
Figure (svg): Example 2.26's 40 ordered ages in four rows of ten, each row labelled with its first position 1, 11, 21, 31
M names the median.
\[ 1 + 40 = 41 \]
Add places 1 and 40
Why: aims halfway along the ages
\[ 41 \div 2 = 20.5 \]
Halve 41
Why: tells which two ages to read
Figure (svg): Example 2.26's 40 ordered ages in four rows of ten, each row labelled with its first position 1, 11, 21, 31; positions 20 and 21 highlighted, both 24
\[ \text{20th} = \textcolor{#1f5fbf}{24},\ \ \text{21st} = \textcolor{#1f5fbf}{24} \]
Read places 20 and 21
Why: no age sits at place 20.5
\[ M = \textcolor{#6b7280}{24} \]
Take halfway: 24 and 24
Why: equal ages: halfway is that age
\[ 19 \text{ below } \textcolor{#6b7280}{24},\ \ 19 \text{ above} \]
Check each side of 24
Why: equal counts confirm
Trap
\( {1 + 40 = 41}\;\;\Rightarrow\;\;\allowbreak {41 \div 2 = 20.5} \)
\[ M = 20.5 \text{ years} \]
Call it the age
Why: a place, not a value
\[ \text{under } 20.5\text{: } 16;\ \text{over: } 24 \]
Count each side
Why: fails: 16 against 24
Figure (svg): A bar of the 40 ages split at 20.5 years: 16 ages under 20.5 and 24 ages over, both in blue
\[ \text{20th} = \textcolor{#1f5fbf}{24},\ \ \text{21st} = \textcolor{#1f5fbf}{24} \]
Read places 20, 21
Why: a place points to values
\[ \textcolor{#1f5fbf}{24} + \textcolor{#1f5fbf}{24} = \textcolor{#1f5fbf}{48} \]
Add the pair
Why: halfway needs their sum
\[ M = \textcolor{#1f5fbf}{48} \div 2 = \textcolor{#6b7280}{24} \]
Halve 48
Why: completes the halfway rule
\[ 19 \text{ below } \textcolor{#6b7280}{24},\ \ 19 \text{ above} \]
Check each side of 24
Why: equal halves: a middle
Prediction
Figure (svg): Home values on an axis from 250 to 400 ($ thousands): a blue bar of 29 homes at 280, a bar of 30 at 315, and an arrow off the axis to 1 home at 2500
Predict first
Try It 2.27: 60 home values, one of them $2,500,000.
Which describes a typical home better?
Correct: The median
Why: The $2,500,000 home is an extreme value. It swells the pooled total and so the mean, but in the ordered list it only takes the last place, so the median ignores its size. The next two slides compute both.
Worked example
Figure (svg): A bar of the 60 ordered home values in $ thousands: 29 at 280 in places 1 to 29, 30 at 315 in places 30 to 59, 1 at 2500 in place 60
\[ 1 + 60 = 61 \]
Add places 1 and 60
Why: locates the middle home
\[ 61 \div 2 = 30.5 \]
Halve 61
Why: shows which homes to read
Figure (svg): A bar of the 60 ordered home values in $ thousands: 29 at 280 in places 1 to 29, 30 at 315 in places 30 to 59, 1 at 2500 in place 60; a grey line at place 30.5, inside the 315 block
\[ \text{30th} = \textcolor{#1f5fbf}{315},\ \ \text{31st} = \textcolor{#1f5fbf}{315} \]
Read places 30 and 31
Why: no home sits at place 30.5
\[ M = \textcolor{#6b7280}{315} \]
Take halfway: 315 and 315
Why: equal middles need no averaging
\[ \text{places } 1\text{–}30 \le \textcolor{#6b7280}{315},\ \ 31\text{–}60 \ge \textcolor{#6b7280}{315} \]
Check: read the bar around 315
Why: 30 each side, ties allowed
Worked example
Figure (svg): Home values on an axis from 250 to 400 ($ thousands): a blue bar of 29 homes at 280, a bar of 30 at 315, and an arrow off the axis to 1 home at 2500
\[ 29 \times \textcolor{#1f5fbf}{280},\ 30 \times \textcolor{#1f5fbf}{315} = \textcolor{#1f5fbf}{8120},\ \textcolor{#1f5fbf}{9450} \]
Total each ordinary bar
Why: counted repeated adding
\[ \textcolor{#1f5fbf}{8120 + 9450 + 2500} = \textcolor{#1f5fbf}{20070} \]
Add both bars and 2500
Why: Σx for all 60 homes
\[ \textcolor{#1f5fbf}{20070} \div 60 = \textcolor{#6b7280}{334.5} \]
Share among 60
Why: the mean is one home's cut
Figure (svg): The home-value bars with a grey line at the mean 334.5, right of both bars
\[ \textcolor{#1f5fbf}{315} < \textcolor{#6b7280}{334.5} < \textcolor{#1f5fbf}{2500} \]
Place the mean
Why: tests whether the mean is typical
\[ 60 \times \textcolor{#6b7280}{334.5} = \textcolor{#1f5fbf}{20070} \]
Check: rebuild the pool
Why: undoes the share exactly
Faded example
Figure (svg): Try It 2.26's 39 ordered transplant waits in rows of ten, labelled with first positions 1, 11, 21, 31
Try It 2.26: 39 ordered waits, in months.
Fill in the blanks
position = (1 + 39) ÷ 2 = 20; M = 13 months
Why: 1 + 39 = 40 and 40 ÷ 2 = 20, a whole position, so the median is one wait. Counting along the grid, the 20th wait (the end of the second row) is 13 months.
Worked example
Figure (svg): Try It 2.26's 39 ordered transplant waits in rows of ten, labelled with first positions 1, 11, 21, 31
\[ 1 + 39 = 40 \]
Add places 1 and 39
Why: sets up the middle wait's place
\[ 40 \div 2 = 20 \]
Halve 40
Why: a whole place: read one wait
Figure (svg): Try It 2.26's 39 ordered transplant waits in rows of ten, labelled with first positions 1, 11, 21, 31; position 20 highlighted, value 13
\[ M = \textcolor{#6b7280}{13} \]
Read the 20th cell
Why: odd n: the median is this value
\[ 19 \text{ waits under } \textcolor{#1f5fbf}{13},\ \ 19 \text{ over} \]
Check: count each side of 13
Why: equal halves confirm the 20th
Section
Idea 3 of 4
Concept
Figure (svg): Dot stacks for the sample: three dots at 1, two at 2, one at 3, five at 4, with the counts 3, 2, 1, 5 above the stacks
f: how many times the value x appears.
\[ \begin{array}{c|cccc} x & \textcolor{#1f5fbf}{1} & \textcolor{#1f5fbf}{2} & \textcolor{#1f5fbf}{3} & \textcolor{#1f5fbf}{4} \\ \hline f & 3 & 2 & 1 & 5 \end{array} \]
Discussion prompt
How can the table give the sample's total without writing out all 11 values?
Answer:
Each column repeats one value f times: multiply, then add the columns.
Worked example
Figure (svg): Dot stacks for the sample: three dots at 1, two at 2, one at 3, five at 4, with counts above
\[ \textcolor{#1f5fbf}{1 + 2 + 3 + 4} = \textcolor{#1f5fbf}{10} \]
Add the four listed values
Why: one term per table column
\[ \textcolor{#1f5fbf}{10} \div 4 = \textcolor{#6b7280}{2.5} \]
Divide by the 4 columns
Why: treats columns as single values
\[ \textcolor{#6b7280}{2.5} \ne \textcolor{#6b7280}{2.73} \]
Compare with the true mean
Why: found by adding all 11
Figure (svg): Dot stacks for the sample with counts 3, 2, 1, 5 and a grey line at the mean 2.73
Five 4s counted as one.
\[ 4 \times \textcolor{#6b7280}{2.5} = \textcolor{#1f5fbf}{10} \]
Check the attempt's arithmetic
Why: 4 listed values, not all 11
Worked example
Figure (svg): Dot stacks for the sample: three dots at 1, two at 2, one at 3, five at 4, with counts above
\[ 3 \times \textcolor{#1f5fbf}{1},\ 2 \times \textcolor{#1f5fbf}{2},\ 1 \times \textcolor{#1f5fbf}{3},\ 5 \times \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{3},\ \textcolor{#1f5fbf}{4},\ \textcolor{#1f5fbf}{3},\ \textcolor{#1f5fbf}{20} \]
Total each stack
Why: a stack repeats one value
Figure (svg): Dot stacks for the sample with stack totals 3, 4, 3, 20 above the first 4 stacks
\[ \textcolor{#1f5fbf}{3 + 4 + 3 + 20} = \textcolor{#1f5fbf}{30} \]
Add the stack totals
Why: the mean needs the sample's total
\[ 3 + 2 + 1 + 5 = 11 \]
Add the frequencies
Why: each value once: n
\[ \textcolor{#1f5fbf}{30} \div 11 \approx \textcolor{#6b7280}{2.73} \]
Share the pool among 11
Why: the list's mean rule still applies
Figure (svg): Dot stacks for the sample with stack totals 3, 4, 3, 20 above the first 4 stacks and a grey line at the mean 2.73
\[ 11 \times \textcolor{#6b7280}{2.73} \approx \textcolor{#1f5fbf}{30} \]
Check: rebuild the pool
Why: shares times n return it
Worked example
Figure (svg): Dot stacks for the sample: three dots at 1, two at 2, one at 3, five at 4, with counts above
\[ f \times \textcolor{#1f5fbf}{x} \]
Total one stack
Why: multiplying replaces repeated adding
Figure (svg): Dot stacks for the sample with stack totals 3 above the first 1 stacks
\[ \textcolor{#1f5fbf}{{\textstyle\sum} f x} \]
Add every stack's total
Why: gives the mean's numerator
Figure (svg): Dot stacks for the sample with stack totals 3, 4, 3, 20 above the first 4 stacks
\[ n = {\textstyle\sum} f \]
Add the frequencies
Why: the mean divides by n
\[ \textcolor{#6b7280}{\bar{x}} = \frac{\textcolor{#1f5fbf}{{\textstyle\sum} f x}}{{\textstyle\sum} f} \]
Share the pool among n
Why: the same share as a list
Figure (svg): Dot stacks for the sample with stack totals 3, 4, 3, 20 above the first 4 stacks and a grey line at the mean 2.73
\[ \frac{\textcolor{#1f5fbf}{30}}{11} \approx \textcolor{#6b7280}{2.73} \]
Check with the sample
Why: matches the eleven-value list
Concept
Figure (svg): Dot stacks for the 11 cup orders: three at size 1, two at 2, one at 3, five at 4, with a grey line at the mean 2.73 where no cup size sits
The §2.5 sample, read as 11 cup orders by size.
Discussion prompt
The mean order is size 2.73. Does it tell the café which size to stock extra?
Answer:
No: no cup is size 2.73. The café needs the size ordered most often.
Worked example
Figure (svg): Dot stacks for the sample: three dots at 1, two at 2, one at 3, five at 4, with counts above
\[ 5 > 3 > 2 > 1 \]
Rank the frequencies
Why: the mode is the commonest value
\[ f = 5 \text{ at } \textcolor{#1f5fbf}{x = 4} \]
Find the value with f = 5
Why: modes are values, not counts
Figure (svg): Dot stacks for the sample with counts above and the tallest stack, five dots at 4, boxed
\[ \text{mode} = \textcolor{#1f5fbf}{4},\ \ \textcolor{#6b7280}{\bar{x}} \approx \textcolor{#6b7280}{2.73} \]
Set the mode beside the mean
Why: different questions, different centres
\[ 1, 1, 1, 2, 2, 3, \textcolor{#1f5fbf}{4, 4, 4, 4, 4} \]
Check: count the 4s in the list
Why: five of them, the tallest stack
Worked example
Figure (svg): Example 2.28's 20 exam scores as one row of dots per score: one dot each at 50, 53, 76, 78, 81, 83, 90, 93; two at 59 and 63; five at 72; three at 84
\[ \textcolor{#1f5fbf}{59},\ \textcolor{#1f5fbf}{63}:\ f = 2 \text{ each} \]
Count the 59s and 63s
Why: only repeats can win
Figure (svg): Example 2.28's 20 exam scores as one row of dots per score: one dot each at 50, 53, 76, 78, 81, 83, 90, 93; two at 59 and 63; five at 72; three at 84; counts printed for 59, 63
\[ \textcolor{#1f5fbf}{72}: f = 5,\ \ \textcolor{#1f5fbf}{84}: f = 3 \]
Count the 72s and 84s
Why: a bigger repeat could overtake
Figure (svg): Example 2.28's 20 exam scores as one row of dots per score: one dot each at 50, 53, 76, 78, 81, 83, 90, 93; two at 59 and 63; five at 72; three at 84; counts printed for 59, 63, 72, 84
\[ 5 > 3 > 2 \]
Rank the repeat counts
Why: the largest f marks the mode
\[ \text{mode} = \textcolor{#1f5fbf}{72} \]
Take the score with f = 5
Why: a score, not a count
Figure (svg): Example 2.28's 20 exam scores as one row of dots per score: one dot each at 50, 53, 76, 78, 81, 83, 90, 93; two at 59 and 63; five at 72; three at 84; counts printed for 59, 63, 72, 84; the row for 72 boxed
\[ \text{other rows: } f \le 3 < 5 \]
Check: scan every dot row
Why: no rival reaches 5
Trap
Figure (svg): Example 2.29's five scores as tallies on one line: 430 with two dots, 480 with two dots, 495 with one dot
\[ \textcolor{#1f5fbf}{430}: 2,\ \ \textcolor{#1f5fbf}{480}: 2,\ \ \textcolor{#1f5fbf}{495}: 1 \]
Count each score
Why: tallies come first
\[ \text{largest } f = 2 \]
Find the largest count
Why: a mode needs the top frequency
\[ \text{mode} = 2 \]
Call 2 the mode
Why: fails: nobody scored 2
Figure (svg): The same tallies with the two-dot groups 430 and 480 boxed
\[ f = 2:\ \textcolor{#1f5fbf}{430},\ \textcolor{#1f5fbf}{480} \]
Take scores with f = 2
Why: a mode is a data value
\[ \text{modes} = \textcolor{#1f5fbf}{430},\ \textcolor{#1f5fbf}{480} \]
Keep both tied values
Why: two modes: bimodal data
\[ \textcolor{#1f5fbf}{495}: 1 < 2 \]
Check the unboxed 495
Why: it misses the tie
Prediction
Predict first
Favourite colours: red, red, red, green, green, yellow, purple, black, blue.
Which centre can this data have?
Correct: Only the mode
Why: Colours cannot be added, so there is no pool to share; they have no natural order, so there is no middle. They can still be counted, and red is the most frequent.
Worked example
Figure (svg): The nine favourite colours as one row of dots per colour: red 3, green 2, yellow, purple, black and blue 1 each
\[ \textcolor{#1f5fbf}{\text{red}}: 3,\ \ \textcolor{#1f5fbf}{\text{green}}: 2 \]
Count the repeated colours
Why: tallies need no arithmetic
Figure (svg): The nine favourite colours as one row of dots per colour: red 3, green 2, yellow, purple, black and blue 1 each; counts printed for red, green
\[ \text{the other four}: 1 \text{ each} \]
Count the other colours
Why: completes the tally for ranking
Figure (svg): The nine favourite colours as one row of dots per colour: red 3, green 2, yellow, purple, black and blue 1 each; counts printed for red, green, yellow, purple, black, blue
\[ 3 > 2 > 1 \]
Rank the counts
Why: the most-chosen colour is the mode
\[ \text{mode} = \textcolor{#1f5fbf}{\text{red}} \]
Take the colour counted 3 times
Why: categories can have modes
Figure (svg): The nine favourite colours as one row of dots per colour: red 3, green 2, yellow, purple, black and blue 1 each; counts printed for red, green, yellow, purple, black, blue; the red row boxed
\[ \text{longest row} = \textcolor{#1f5fbf}{\text{red}},\ 3 \text{ dots} \]
Check: read the longest dot row
Why: no other colour has 3
Faded example
Figure (svg): Table 2.25's 30 students as dot stacks on an axis from 0 to 4 movies, counts not yet shown
\[ \begin{aligned} \text{movies}&:\ \textcolor{#1f5fbf}{0, 1, 2, 3, 4} \\ \text{shares}&:\ 5/30,\ 15/30,\ 6/30,\ 3/30,\ 1/30 \end{aligned} \]
\[ f = 5,\ 15,\ 6,\ 3,\ 1 \]
Multiply each share by 30
Why: share times total undoes ÷ total
Figure (svg): Table 2.25's 30 students as dot stacks on an axis from 0 to 4 movies: 5 at 0, 15 at 1, 6 at 2, 3 at 3, 1 at 4
Fill in the blanks
Σfx = 40; x̄ ≈ 1.33 movies
Why: The stack totals f × x are 0, 15, 12, 9, 4, which add to 40, and 40 ÷ 30 ≈ 1.33 movies.
Worked example
Figure (svg): Table 2.25's 30 students as dot stacks on an axis from 0 to 4 movies: 5 at 0, 15 at 1, 6 at 2, 3 at 3, 1 at 4
\( {f = 5,\ 15,\ 6,\ 3,\ 1} \)
\[ 5 \times \textcolor{#1f5fbf}{0},\ 15 \times \textcolor{#1f5fbf}{1},\ 6 \times \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{0},\ \textcolor{#1f5fbf}{15},\ \textcolor{#1f5fbf}{12} \]
Total stacks 0–2
Why: f × x skips adding copies
Figure (svg): Table 2.25's 30 students as dot stacks on an axis from 0 to 4 movies: 5 at 0, 15 at 1, 6 at 2, 3 at 3, 1 at 4; stack totals 0, 15, 12 above
\[ 3 \times \textcolor{#1f5fbf}{3},\ 1 \times \textcolor{#1f5fbf}{4} = \textcolor{#1f5fbf}{9},\ \textcolor{#1f5fbf}{4} \]
Total stacks 3 and 4
Why: no column's movies left out
Figure (svg): Table 2.25's 30 students as dot stacks on an axis from 0 to 4 movies: 5 at 0, 15 at 1, 6 at 2, 3 at 3, 1 at 4; stack totals 0, 15, 12, 9, 4 above
\[ \textcolor{#1f5fbf}{0 + 15 + 12 + 9 + 4} = \textcolor{#1f5fbf}{40} \]
Add the stack totals
Why: Σfx is the mean's numerator
\[ \textcolor{#1f5fbf}{40} \div 30 \approx \textcolor{#6b7280}{1.33} \]
Share among 30
Why: Σf is the class size
Figure (svg): Table 2.25's 30 students as dot stacks on an axis from 0 to 4 movies: 5 at 0, 15 at 1, 6 at 2, 3 at 3, 1 at 4; stack totals 0, 15, 12, 9, 4 above; grey line at the mean 1.33
\[ 30 \times \textcolor{#6b7280}{1.33} = \textcolor{#1f5fbf}{39.9} \approx \textcolor{#1f5fbf}{40} \]
Check: multiply back
Why: near 40, rounding
Concept
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2
The club rolled the die 20 times, shown in rows of five.
Discussion prompt
Should the club judge the die from the first five rolls, or wait for all twenty?
Answer:
Compare both averages with a fair die's mean: the next slides compute all three.
Worked example
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2
μ (mu): a population's mean. Here: faces 1 to 6.
\[ \textcolor{#1f5fbf}{1 + 2 + 3 + 4 + 5 + 6} = \textcolor{#1f5fbf}{21} \]
Pool the six faces
Why: a fair die favours no face
\[ \textcolor{#6b7280}{\mu} = \textcolor{#1f5fbf}{21} \div 6 = \textcolor{#6b7280}{3.5} \]
Share among the 6 faces
Why: the target samples aim at
\[ \textcolor{#1f5fbf}{6 + 5 + 1 + 6 + 4} = \textcolor{#1f5fbf}{22} \]
Pool the first five rolls
Why: a small sample to judge by
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2; row totals shown for 1 rows; row 1 highlighted
\[ \textcolor{#6b7280}{\bar{x}} = \textcolor{#1f5fbf}{22} \div 5 = \textcolor{#6b7280}{4.4} \]
Share among n = 5
Why: this sample's estimate of μ
\[ 5 \times \textcolor{#6b7280}{4.4} = \textcolor{#1f5fbf}{22} \]
Check: multiply back
Why: returns the sample total
Worked example
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2; row totals shown for 1 rows; row 1 highlighted
\( {\text{rolls 1–5: } 22} \)
\[ \textcolor{#1f5fbf}{2 + 3 + 5 + 4 + 2} = \textcolor{#1f5fbf}{16} \]
Add row 2
Why: sample 2's x̄ needs it
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2; row totals shown for 2 rows; row 2 highlighted
\[ \textcolor{#1f5fbf}{3 + 1 + 6 + 2 + 5} = \textcolor{#1f5fbf}{17} \]
Add row 3
Why: shows how x̄ varies
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2; row totals shown for 3 rows; row 3 highlighted
\[ \textcolor{#1f5fbf}{4 + 3 + 1 + 6 + 2} = \textcolor{#1f5fbf}{16} \]
Add row 4
Why: sample 4's x̄ needs it
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2; row totals shown for 4 rows; row 4 highlighted
\[ \textcolor{#1f5fbf}{22 + 16 + 17 + 16} = \textcolor{#1f5fbf}{71} \]
Add the row totals
Why: one pool for 20 rolls
Figure (svg): Twenty die rolls in rows of five: 6 5 1 6 4 / 2 3 5 4 2 / 3 1 6 2 5 / 4 3 1 6 2; row totals shown for 4 rows
\[ \textcolor{#1f5fbf}{16 + 16 + 17 + 22} = \textcolor{#1f5fbf}{71} \]
Check: add in reverse
Why: any order, same total
Worked example
Figure (svg): Plot of the mean of the rolls against rolls so far: 4.4 after 5 rolls, with a grey dashed line at μ 3.5
\( {\mu = 3.5},\quad\allowbreak \allowbreak {n = 5:\ \bar{x} = 4.4},\quad\allowbreak \allowbreak {\textstyle{\textstyle\sum} x = 71} \)
\[ \textcolor{#1f5fbf}{71} \div 20 = \textcolor{#6b7280}{3.55} \]
Share among 20
Why: the longer run's x̄
Figure (svg): Plot of the mean of the rolls against rolls so far: 4.4 after 5 rolls, 3.55 after 20, with a grey dashed line at μ 3.5
\[ \textcolor{#6b7280}{4.4} - \textcolor{#6b7280}{3.5},\ \ \textcolor{#6b7280}{3.55} - \textcolor{#6b7280}{3.5} = \textcolor{#1f5fbf}{0.9},\ \ \textcolor{#1f5fbf}{0.05} \]
Subtract μ from each
Why: size the two errors
\[ \textcolor{#1f5fbf}{0.05} < \textcolor{#1f5fbf}{0.9} \]
Compare misses
Why: does a bigger sample help?
Law of large numbers: bigger n, x̄ likely nearer μ.
\[ \textcolor{#6b7280}{3.5} + \textcolor{#1f5fbf}{0.9} = \textcolor{#6b7280}{4.4},\ \ \textcolor{#6b7280}{3.5} + \textcolor{#1f5fbf}{0.05} = \textcolor{#6b7280}{3.55} \]
Check: add misses to μ
Why: lands on both means
Verdict: no sign of unfairness.
Worked example
Figure (svg): Two dot plots on axes from 1 to 6: the 20 single rolls spread from 1 to 6; the means-of-five row still empty
\( {\text{row totals } 22,\ 16,\ 17,\ 16},\quad\allowbreak \allowbreak {\text{row 1: } \bar{x} = 4.4} \)
\[ \textcolor{#1f5fbf}{16} \div 5,\ \textcolor{#1f5fbf}{17} \div 5,\ \textcolor{#1f5fbf}{16} \div 5 = \textcolor{#6b7280}{3.2},\ \textcolor{#6b7280}{3.4},\ \textcolor{#6b7280}{3.2} \]
Share each later row among 5
Why: shows whether samples agree
Figure (svg): Two dot plots on axes from 1 to 6: the 20 single rolls spread from 1 to 6; the four means of five, 4.4, 3.2, 3.4, 3.2, bunched near the grey line at μ 3.5
\[ \textcolor{#6b7280}{4.4},\ \textcolor{#6b7280}{3.2},\ \textcolor{#6b7280}{3.4},\ \textcolor{#6b7280}{3.2} \ne \textcolor{#6b7280}{\mu} \]
Compare each x̄ with μ = 3.5
Why: tests whether any estimate is exact
\[ \textcolor{#6b7280}{3.2} \times 5 = \textcolor{#1f5fbf}{16},\ \ \textcolor{#6b7280}{3.4} \times 5 = \textcolor{#1f5fbf}{17} \]
Check: rebuild two row totals
Why: undoes each share
Concept
Figure (svg): Two dot plots on axes from 1 to 6: the 20 single rolls spread from 1 to 6; the four means of five, 4.4, 3.2, 3.4, 3.2, bunched near the grey line at μ 3.5
Discussion prompt
Which spreads more: single rolls, or the four means of five rolls?
Answer:
Single rolls: 1 to 6. Means: 3.2 to 4.4. Each mean is a statistic.
A statistic's values over very many samples: its sampling distribution.
Section
Idea 4 of 4
Concept
Figure (svg): Example 2.30's eight grade classes from 50 to 98.5 drawn as dashed boxes on a grade axis from 50 to 100, with counts 1, 0, 4, 4, 2, 3, 4, 1 above them
Example 2.30: last test's grades, in eight classes from 50 to 98.5.
Discussion prompt
What single number could stand in for the 4 grades in class 62.5–68.5?
Answer:
A point inside the class; the next slide tests the ends and the middle.
Worked example
Figure (svg): One class from 62.5 to 68.5 on a grade axis from 60 to 70 holding 4 unknown grades
m: a class's midpoint, halfway between its limits.
\[ \textcolor{#1f5fbf}{68.5} - \textcolor{#1f5fbf}{62.5} = \textcolor{#1f5fbf}{6} \]
Find the class width
Why: worst miss for a class end
Figure (svg): One class from 62.5 to 68.5 on a grade axis from 60 to 70 holding 4 unknown grades, a blue arrow across it marked width 6
\[ \textcolor{#1f5fbf}{62.5} + \textcolor{#1f5fbf}{68.5} = \textcolor{#1f5fbf}{131} \]
Add the class limits
Why: a midpoint needs their sum
\[ \textcolor{#1f5fbf}{131} \div 2 = \textcolor{#1f5fbf}{65.5} \]
Halve the sum
Why: the halfway rule gives m
Figure (svg): One class from 62.5 to 68.5 on a grade axis from 60 to 70 holding 4 unknown grades, a blue arrow across it marked width 6, its midpoint 65.5 marked m
\[ \textcolor{#1f5fbf}{68.5} - \textcolor{#1f5fbf}{65.5} = \textcolor{#1f5fbf}{3} \]
Measure m to the top limit
Why: a grade's largest miss from m
Figure (svg): One class from 62.5 to 68.5 on a grade axis from 60 to 70 holding 4 unknown grades, a blue arrow across it marked width 6, its midpoint 65.5 marked m, arrows of 3 from the midpoint to each end
\[ \textcolor{#1f5fbf}{65.5} - \textcolor{#1f5fbf}{62.5} = \textcolor{#1f5fbf}{3} \]
Check the lower side
Why: equal reach both ways
Prediction
Figure (svg): One class from 62.5 to 68.5 on a grade axis from 60 to 70 holding 4 unknown grades, its midpoint 65.5 marked m
Predict first
The estimate totals this class's 4 grades as 4 × 65.5.
When is that the real total?
Correct: Only when the grades average to 65.5
Why: 4 × 65.5 is the total of four grades whose equal share is 65.5. A list with a different average has a different total, however close its grades sit to m; unequal grades can still average to 65.5.
Worked example
Figure (svg): Class 62.5–68.5 on a grade axis from 62 to 69 with its midpoint m 65.5 as a grey dashed line and a blue dot; rows for list A and list B, still empty
\[ \textcolor{#1f5fbf}{63 + 63 + 64 + 64} = \textcolor{#1f5fbf}{254} \]
Total list A
Why: a real total to test
Figure (svg): Class 62.5–68.5 on a grade axis with m 65.5 as a grey dashed line and a blue dot; list A's dots at 63, 63, 64, 64
\[ 4 \times \textcolor{#1f5fbf}{65.5} = \textcolor{#1f5fbf}{262} \]
Total the midpoint stand-ins
Why: the total the estimate uses
\[ \textcolor{#1f5fbf}{254} \ne \textcolor{#1f5fbf}{262} \]
Compare list A with 262
Why: so 'always' fails
\[ \textcolor{#1f5fbf}{63 + 64 + 67 + 68} = \textcolor{#1f5fbf}{262} \]
Total list B
Why: unequal grades, yet a match
Figure (svg): The same class with list A's dots and list B's dots at 63, 64, 67, 68, two each side of m
\[ \textcolor{#1f5fbf}{262} \div 4 = \textcolor{#6b7280}{65.5} \]
Check: share list B's total
Why: its average is exactly m
Worked example
Figure (svg): Example 2.30's eight grade classes from 50 to 98.5 drawn as dashed boxes on a grade axis from 50 to 100, with counts 1, 0, 4, 4, 2, 3, 4, 1 above them
\[ f \times \textcolor{#1f5fbf}{m} \]
Stack f grades at the midpoint
Why: stand-ins for unknown grades
Figure (svg): Example 2.30's classes as boxes with counts; the class 62.5–68.5 highlighted and its 4 grades drawn as dots stacked at its midpoint
\[ \textcolor{#1f5fbf}{{\textstyle\sum} f m} \]
Add every class's stack total
Why: estimates the pool of grades
Figure (svg): Example 2.30's classes as boxes with counts, each class's f dots stacked at its midpoint
\[ n = {\textstyle\sum} f \]
Add the class frequencies
Why: each student counted once
\[ \textcolor{#6b7280}{\bar{x}} \approx \frac{\textcolor{#1f5fbf}{{\textstyle\sum} f m}}{{\textstyle\sum} f} \]
Share the estimated pool
Why: ≈: exact when stand-in and real totals match
\[ m = \textcolor{#1f5fbf}{x}:\ \frac{\textcolor{#1f5fbf}{{\textstyle\sum} f x}}{{\textstyle\sum} f} \]
Check: shrink classes to points
Why: gives the table mean
Worked example
Figure (svg): Example 2.30's table: classes 50–56.5 to 92.5–98.5 with f = 1, 0, 4, 4, 2, 3, 4, 1
\[ \begin{array}{lcl} \textcolor{#1f5fbf}{50 + 56.5} = \textcolor{#1f5fbf}{106.5} & & \textcolor{#1f5fbf}{74.5 + 80.5} = \textcolor{#1f5fbf}{155} \\ \textcolor{#1f5fbf}{56.5 + 62.5} = \textcolor{#1f5fbf}{119} & & \textcolor{#1f5fbf}{80.5 + 86.5} = \textcolor{#1f5fbf}{167} \\ \textcolor{#1f5fbf}{62.5 + 68.5} = \textcolor{#1f5fbf}{131} & & \textcolor{#1f5fbf}{86.5 + 92.5} = \textcolor{#1f5fbf}{179} \\ \textcolor{#1f5fbf}{68.5 + 74.5} = \textcolor{#1f5fbf}{143} & & \textcolor{#1f5fbf}{92.5 + 98.5} = \textcolor{#1f5fbf}{191} \end{array} \]
Add each class's two limits
Why: a midpoint starts from the ends' sum
\[ \textcolor{#1f5fbf}{191} - \textcolor{#1f5fbf}{98.5} = \textcolor{#1f5fbf}{92.5} \]
Check: undo class 8's sum
Why: returns its lower limit
Worked example
Figure (svg): Example 2.30's table: classes 50–56.5 to 92.5–98.5 with f = 1, 0, 4, 4, 2, 3, 4, 1
\( {\text{sums: } 106.5},\ \allowbreak \allowbreak {119},\ \allowbreak \allowbreak {131},\ \allowbreak \allowbreak {143},\ \allowbreak \allowbreak {155},\ \allowbreak \allowbreak {167},\ \allowbreak \allowbreak {179},\ \allowbreak \allowbreak {191} \)
\[ \begin{array}{lcl} \textcolor{#1f5fbf}{106.5} \div 2 = \textcolor{#1f5fbf}{53.25} & & \textcolor{#1f5fbf}{155} \div 2 = \textcolor{#1f5fbf}{77.5} \\ \textcolor{#1f5fbf}{119} \div 2 = \textcolor{#1f5fbf}{59.5} & & \textcolor{#1f5fbf}{167} \div 2 = \textcolor{#1f5fbf}{83.5} \\ \textcolor{#1f5fbf}{131} \div 2 = \textcolor{#1f5fbf}{65.5} & & \textcolor{#1f5fbf}{179} \div 2 = \textcolor{#1f5fbf}{89.5} \\ \textcolor{#1f5fbf}{143} \div 2 = \textcolor{#1f5fbf}{71.5} & & \textcolor{#1f5fbf}{191} \div 2 = \textcolor{#1f5fbf}{95.5} \end{array} \]
Halve each sum
Why: gives each class its stand-in
Figure (svg): Example 2.30's table: classes 50–56.5 to 92.5–98.5 with f = 1, 0, 4, 4, 2, 3, 4, 1; midpoints m filled for 8 rows
\[ \textcolor{#1f5fbf}{53.25} - \textcolor{#1f5fbf}{50},\ \textcolor{#1f5fbf}{56.5} - \textcolor{#1f5fbf}{53.25} = \textcolor{#1f5fbf}{3.25},\ \textcolor{#1f5fbf}{3.25} \]
Check the first class
Why: m sits midway between its limits
Worked example
Figure (svg): Example 2.30's table: classes 50–56.5 to 92.5–98.5 with f = 1, 0, 4, 4, 2, 3, 4, 1; midpoints m filled for 8 rows
\( {m:\ 53.25},\ \allowbreak \allowbreak {59.5},\ \allowbreak \allowbreak {65.5},\ \allowbreak \allowbreak {71.5},\ \allowbreak \allowbreak {77.5},\ \allowbreak \allowbreak {83.5},\ \allowbreak \allowbreak {89.5},\ \allowbreak \allowbreak {95.5} \)
\[ \begin{array}{lcl} 1 \times \textcolor{#1f5fbf}{53.25} = \textcolor{#1f5fbf}{53.25} & & 2 \times \textcolor{#1f5fbf}{77.5} = \textcolor{#1f5fbf}{155} \\ 0 \times \textcolor{#1f5fbf}{59.5} = \textcolor{#1f5fbf}{0} & & 3 \times \textcolor{#1f5fbf}{83.5} = \textcolor{#1f5fbf}{250.5} \\ 4 \times \textcolor{#1f5fbf}{65.5} = \textcolor{#1f5fbf}{262} & & 4 \times \textcolor{#1f5fbf}{89.5} = \textcolor{#1f5fbf}{358} \\ 4 \times \textcolor{#1f5fbf}{71.5} = \textcolor{#1f5fbf}{286} & & 1 \times \textcolor{#1f5fbf}{95.5} = \textcolor{#1f5fbf}{95.5} \end{array} \]
Multiply each m by its f
Why: every grade counted at its midpoint
Figure (svg): Example 2.30's table: classes 50–56.5 to 92.5–98.5 with f = 1, 0, 4, 4, 2, 3, 4, 1; midpoints m filled for 8 rows; f × m filled for 8 rows
\[ \textcolor{#1f5fbf}{89.5 + 89.5} = \textcolor{#1f5fbf}{179},\ \ \textcolor{#1f5fbf}{179 + 179} = \textcolor{#1f5fbf}{358} \]
Check 4 × 89.5 by doubling twice
Why: four copies rebuild 358
Worked example
Figure (svg): Example 2.30's classes as boxes with counts, each class's f dots stacked at its midpoint
\( {f \times m:\ 53.25},\ \allowbreak \allowbreak {0},\ \allowbreak \allowbreak {262},\ \allowbreak \allowbreak {286},\ \allowbreak \allowbreak {155},\ \allowbreak \allowbreak {250.5},\ \allowbreak \allowbreak {358},\ \allowbreak \allowbreak {95.5} \)
\[ \begin{aligned} &\textcolor{#1f5fbf}{53.25 + 0 + 262 + 286} \\ &{+}\ \textcolor{#1f5fbf}{155 + 250.5 + 358 + 95.5} = \textcolor{#1f5fbf}{1460.25} \end{aligned} \]
Add every class total
Why: x̄'s estimated numerator
\[ 1 + 0 + 4 + 4 + 2 + 3 + 4 + 1 = 19 \]
Add the frequencies
Why: n counts every student
\[ \textcolor{#1f5fbf}{1460.25} \div 19 \approx \textcolor{#6b7280}{76.86} \]
Divide by 19
Why: each student's equal share
Figure (svg): Example 2.30's classes with dots at midpoints and a grey line at the estimate 76.86
\[ 19 \times \textcolor{#6b7280}{76.86} \approx \textcolor{#1f5fbf}{1460.3} \]
Check: multiply back
Why: returns about 1460.25
Worked example
Figure (svg): Example 2.30's classes as boxes with counts, each class's f dots stacked at its midpoint
\[ \textcolor{#1f5fbf}{53.25} - \textcolor{#1f5fbf}{50},\ \textcolor{#1f5fbf}{56.5} - \textcolor{#1f5fbf}{53.25} = \textcolor{#1f5fbf}{3.25},\ \textcolor{#1f5fbf}{3.25} \]
Measure class 1's reach
Why: the widest class slides furthest
Figure (svg): Example 2.30's classes with dots at midpoints; class 50–56.5 highlighted
\[ 19 - 1 = 18 \]
Count the other grades
Why: all sit in 6-wide classes
Figure (svg): The same picture with the six occupied 6-wide classes highlighted
\[ 18 \times \textcolor{#1f5fbf}{3} = \textcolor{#1f5fbf}{54} \]
Give each a reach of 3
Why: half a 6-wide class
\[ \textcolor{#1f5fbf}{3.25} + \textcolor{#1f5fbf}{54} = \textcolor{#1f5fbf}{57.25} \]
Add class 1's reach
Why: the pool's largest possible shift
\[ \textcolor{#1f5fbf}{54} \div \textcolor{#1f5fbf}{3} = 18 \]
Check: share the reach sum
Why: returns the 18 grades
Worked example
Figure (svg): Example 2.30's classes as boxes with counts, each class's f dots stacked at its midpoint
\( {\textstyle{\textstyle\sum} fm = 1460.25},\quad\allowbreak \allowbreak {\text{shift} = 57.25} \)
\[ \textcolor{#1f5fbf}{1460.25} - \textcolor{#1f5fbf}{57.25} = \textcolor{#1f5fbf}{1403} \]
Slide every grade down
Why: smallest pool the table allows
Figure (svg): Example 2.30's classes with every dot slid to its class's lower limit and hollow rings left at the midpoints
\[ \textcolor{#1f5fbf}{1403} \div 19 \approx \textcolor{#6b7280}{73.84} \]
Share among 19
Why: turns the lowest pool into x̄
Figure (svg): The dots at lower limits with a grey line at 73.84
\[ 19 \times \textcolor{#6b7280}{73.84} = \textcolor{#1f5fbf}{1402.96} \approx \textcolor{#1f5fbf}{1403} \]
Check: multiply back
Why: near 1403 after rounding
Worked example
Figure (svg): The dots at lower limits with a grey line at 73.84
\( {\textstyle{\textstyle\sum} fm = 1460.25},\quad\allowbreak \allowbreak {\text{shift} = 57.25},\quad\allowbreak \allowbreak {\text{lowest} \approx 73.84} \)
\[ \textcolor{#1f5fbf}{1460.25} + \textcolor{#1f5fbf}{57.25} = \textcolor{#1f5fbf}{1517.5} \]
Slide every grade up
Why: reach up matches reach down
Figure (svg): Example 2.30's classes with every dot slid to its class's upper limit and hollow rings left at the midpoints
\[ \textcolor{#1f5fbf}{1517.5} \div 19 \approx \textcolor{#6b7280}{79.87} \]
Share among 19
Why: turns the highest pool into x̄
Figure (svg): The dots at upper limits with a grey line at 79.87
\[ (\textcolor{#6b7280}{73.84} + \textcolor{#6b7280}{79.87}) \div 2 \approx \textcolor{#6b7280}{76.86} \]
Check: halfway between the bounds
Why: the estimate sits there
Trap
\[ \textcolor{#1f5fbf}{{\textstyle\sum} fm} = \textcolor{#1f5fbf}{1460.25} \]
Start from Σfm
Why: this pool is correct
\[ \textcolor{#1f5fbf}{1460.25} \div 8 \approx \textcolor{#6b7280}{182.53} \]
Divide by 8
Why: treats classes as grades
\[ \textcolor{#6b7280}{182.53} > \textcolor{#1f5fbf}{98.5} \]
Compare with 98.5
Why: fails: above all
Figure (svg): A grade axis from 50 to 200 with the classes' range 50 to 98.5 shaded blue and a grey line at 182.53, far to its right
\[ n = 19 \]
Add frequencies
Why: n counts students
\[ \textcolor{#1f5fbf}{1460.25} \div 19 \approx \textcolor{#6b7280}{76.86} \]
Divide by 19
Why: one share per grade
Figure (svg): The same axis with a grey line at 76.86 inside the shaded range of grades
\[ \textcolor{#6b7280}{73.84} < \textcolor{#6b7280}{76.86} < \textcolor{#6b7280}{79.87} \]
Check the bounds
Why: inside the range
Faded example
Figure (svg): Try It 2.30's table: hours 0–3.5, 3.5–7.5, 7.5–11.5, 11.5–15.5, 15.5–19.5 with f = 3, 7, 12, 7, 9
Try It 2.30: teenagers' weekly video-game hours, grouped.
Fill in the blanks
m for 7.5–11.5 = 9.5; Σfm = 409.75; n = 38; x̄ ≈ 10.78 hours
Why: Midpoints 1.75, 5.5, 9.5, 13.5, 17.5; f × m gives 5.25, 38.5, 114, 94.5, 157.5, which add to 409.75. The frequencies add to 38, and 409.75 ÷ 38 ≈ 10.78 hours: an estimate.
Worked example
Figure (svg): Try It 2.30's table: hours 0–3.5, 3.5–7.5, 7.5–11.5, 11.5–15.5, 15.5–19.5 with f = 3, 7, 12, 7, 9
\[ \begin{array}{lcl} \textcolor{#1f5fbf}{0 + 3.5} = \textcolor{#1f5fbf}{3.5} & & \textcolor{#1f5fbf}{11.5 + 15.5} = \textcolor{#1f5fbf}{27} \\ \textcolor{#1f5fbf}{3.5 + 7.5} = \textcolor{#1f5fbf}{11} & & \textcolor{#1f5fbf}{15.5 + 19.5} = \textcolor{#1f5fbf}{35} \\ \textcolor{#1f5fbf}{7.5 + 11.5} = \textcolor{#1f5fbf}{19} & & \end{array} \]
Add each class's limits
Why: each midpoint needs its limit sum
\[ \begin{array}{lcl} \textcolor{#1f5fbf}{3.5} \div 2 = \textcolor{#1f5fbf}{1.75} & & \textcolor{#1f5fbf}{27} \div 2 = \textcolor{#1f5fbf}{13.5} \\ \textcolor{#1f5fbf}{11} \div 2 = \textcolor{#1f5fbf}{5.5} & & \textcolor{#1f5fbf}{35} \div 2 = \textcolor{#1f5fbf}{17.5} \\ \textcolor{#1f5fbf}{19} \div 2 = \textcolor{#1f5fbf}{9.5} & & \end{array} \]
Halve each sum
Why: completes each class's halfway rule
Figure (svg): Try It 2.30's table: hours 0–3.5, 3.5–7.5, 7.5–11.5, 11.5–15.5, 15.5–19.5 with f = 3, 7, 12, 7, 9; midpoints filled
\[ \textcolor{#1f5fbf}{17.5} + \textcolor{#1f5fbf}{17.5} = \textcolor{#1f5fbf}{35} \]
Check: double the top m
Why: back to its limit sum
Worked example
Figure (svg): Try It 2.30's table: hours 0–3.5, 3.5–7.5, 7.5–11.5, 11.5–15.5, 15.5–19.5 with f = 3, 7, 12, 7, 9; midpoints filled
\( {m = 1.75,\ 5.5,\ 9.5,\ 13.5,\ 17.5} \)
\[ \begin{array}{lcl} 3 \times \textcolor{#1f5fbf}{1.75} = \textcolor{#1f5fbf}{5.25} & & 7 \times \textcolor{#1f5fbf}{13.5} = \textcolor{#1f5fbf}{94.5} \\ 7 \times \textcolor{#1f5fbf}{5.5} = \textcolor{#1f5fbf}{38.5} & & 9 \times \textcolor{#1f5fbf}{17.5} = \textcolor{#1f5fbf}{157.5} \\ 12 \times \textcolor{#1f5fbf}{9.5} = \textcolor{#1f5fbf}{114} & & \end{array} \]
Multiply each m by its f
Why: stand-ins replace unknown hours
Figure (svg): Try It 2.30's table: hours 0–3.5, 3.5–7.5, 7.5–11.5, 11.5–15.5, 15.5–19.5 with f = 3, 7, 12, 7, 9; midpoints filled; f × m filled
\[ \textcolor{#1f5fbf}{157.5} \div 9 = \textcolor{#1f5fbf}{17.5} \]
Check: undo the top × 9
Why: returns the top midpoint
Worked example
Figure (svg): Try It 2.30's table: hours 0–3.5, 3.5–7.5, 7.5–11.5, 11.5–15.5, 15.5–19.5 with f = 3, 7, 12, 7, 9; midpoints filled; f × m filled
\( {f \times m = 5.25,\ 38.5,\ 114,\ 94.5,\ 157.5} \)
\[ \textcolor{#1f5fbf}{5.25 + 38.5 + 114 + 94.5 + 157.5} = \textcolor{#1f5fbf}{409.75} \]
Add the class totals
Why: the estimated mean needs Σfm
\[ 3 + 7 + 12 + 7 + 9 = 38 \]
Add the frequencies
Why: n is the formula's divisor
\[ \textcolor{#1f5fbf}{409.75} \div 38 \approx \textcolor{#6b7280}{10.78} \]
Divide by 38
Why: each teenager's typical week
Figure (svg): Try It 2.30's five classes as boxes on an hours axis from 0 to 20 with counts 3, 7, 12, 7, 9, each class's dots stacked at its midpoint, and a grey line at the estimate 10.78
\[ 38 \times \textcolor{#6b7280}{10.78} = \textcolor{#1f5fbf}{409.64} \approx \textcolor{#1f5fbf}{409.75} \]
Check: multiply back
Why: close, after rounding
Pattern
\[ \textcolor{#6b7280}{\bar{x}} = \frac{\textcolor{#1f5fbf}{\sum x}}{n} = \frac{\textcolor{#1f5fbf}{\sum f x}}{\sum f},\qquad \text{median place} = (1 + n) \div 2 \]
Check
Figure (svg): Try It 2.28's 25 book counts as dot stacks on 0 to 12: 0, 0, 0, 1, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12, so stacks of 3, 1, 1, 2, 2, 2 at 0 to 5, 4 at 7, 3 at 8, 1 at 9, 2 each at 10, 11, 12
Check your understanding
Try It 2.28: a 26th student joins these 25 book counts. Which count makes the data bimodal?
Answer: A
Why: 7 appears 4 times, more than any other count. 8 appears 3 times, so one more 8 ties it: two modes, 7 and 8.
Worked example
Figure (svg): Try It 2.28's 25 book counts as dot stacks on 0 to 12: 0, 0, 0, 1, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12, so stacks of 3, 1, 1, 2, 2, 2 at 0 to 5, 4 at 7, 3 at 8, 1 at 9, 2 each at 10, 11, 12
\[ \textcolor{#1f5fbf}{7}: f = 4 \]
Count the 7s
Why: the current mode's frequency
\[ \textcolor{#1f5fbf}{0}: 3,\ \ \textcolor{#1f5fbf}{8}: 3,\ \ \text{others} \le 2 \]
Count the runners-up
Why: only a 3 can reach 4
\[ \textcolor{#1f5fbf}{8}: 3 + 1 = 4 \]
Add the newcomer at 8
Why: ties the 7s' count
Figure (svg): Try It 2.28's 25 book counts as dot stacks on 0 to 12: 0, 0, 0, 1, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12, so stacks of 3, 1, 1, 2, 2, 2 at 0 to 5, 4 at 7, 3 at 8, 1 at 9, 2 each at 10, 11, 12; a hollow dot added at 8
\[ \text{modes} = \textcolor{#1f5fbf}{7},\ \textcolor{#1f5fbf}{8} \]
List the tied values
Why: two modes: bimodal
\[ \textcolor{#1f5fbf}{7, 7, 7, 7};\ \ \textcolor{#1f5fbf}{8, 8, 8},\ \textcolor{#1f5fbf}{8} \]
Check: recount in the list
Why: four of each after joining
Check
Try It 2.29, adapted: a factory's 301 workers.
Check your understanding
Mode $25,000 (150 workers), median $50,000, mean $47,500. Which figure is no worker's pay?
Answer: A
Why: With 301 pays the median is the 151st pay, $50,000. The 150 pays of $25,000 sit below it, so they fill places 1 to 150, and every pay from place 151 up is at least $50,000. No pay lies between $25,000 and $50,000, where the mean sits.
Worked example
Figure (svg): A bar of 301 ordered places, not yet split
\[ 1 + 301 = 302 \]
Add places 1 and 301
Why: first move to the middle worker
\[ 302 \div 2 = 151 \]
Halve 302
Why: odd n gives one middle place
Figure (svg): A bar of 301 ordered places with a grey line at place 151
\[ \textcolor{#1f5fbf}{25} < \textcolor{#6b7280}{50} \]
Compare 25 with the 151st pay
Why: lower pays sort earlier
\[ \text{places } 1\text{–}150 = \textcolor{#1f5fbf}{25} \]
Fill the places before 151
Why: 150 pays, 150 places
Figure (svg): A bar of the 301 ordered pays in $ thousands: places 1 to 150 at 25, places 151 to 301 at 50 or more, with a grey line at place 151
\[ \textcolor{#1f5fbf}{25} < \textcolor{#6b7280}{47.5} < \textcolor{#6b7280}{50} \]
Place the mean
Why: tests whether any worker earns it
\[ \text{place } 150 = \textcolor{#1f5fbf}{25},\ \ 151 = \textcolor{#6b7280}{50} \]
Check: read places 150 and 151
Why: no pay sits between them
Worked example
Figure (svg): The town's incomes in $ thousands: full scale 0 to 5000 with 49 dots at 30 and one at 5000; a zoom from 0 to 200 with the 49 dots at 30
Incomes in $ thousands.
\[ 49 \times \textcolor{#1f5fbf}{30} = \textcolor{#1f5fbf}{1470} \]
Pool the 49 ordinary incomes
Why: counted form of repeated adding
\[ \textcolor{#1f5fbf}{1470} + \textcolor{#1f5fbf}{5000} = \textcolor{#1f5fbf}{6470} \]
Add the one large income
Why: Σx for all 50 residents
\[ \textcolor{#1f5fbf}{6470} \div 50 = \textcolor{#6b7280}{129.4} \]
Divide by 50
Why: the mean the leaflet reports
Figure (svg): The town's incomes in $ thousands: full scale 0 to 5000 with 49 dots at 30 and one at 5000; a zoom from 0 to 200 with the 49 dots at 30, a grey line at the mean 129.4
\[ 50 \times \textcolor{#6b7280}{129.4} = \textcolor{#1f5fbf}{6470} \]
Check: rebuild the pool
Why: undoes the equal share
Worked example
Figure (svg): The town's incomes in $ thousands: full scale 0 to 5000 with 49 dots at 30 and one at 5000; a zoom from 0 to 200 with the 49 dots at 30, a grey line at the mean 129.4
\[ 1 + 50 = 51 \]
Add places 1 and 50
Why: starts the median-place rule
\[ 51 \div 2 = 25.5 \]
Halve 51
Why: shows which incomes to read
\[ \text{25th} = \textcolor{#1f5fbf}{30},\ \ \text{26th} = \textcolor{#1f5fbf}{30} \]
Read places 25 and 26
Why: no income sits at place 25.5
\[ M = \textcolor{#6b7280}{30} \]
Take halfway: 30 and 30
Why: no gap between the two middles
Figure (svg): The town's incomes in $ thousands: full scale 0 to 5000 with 49 dots at 30 and one at 5000; a zoom from 0 to 200 with the 49 dots at 30, a grey line at the mean 129.4, a grey line at the median 30
\[ 49 \ge 26 \]
Check: the 30s reach past place 26
Why: both middle places are 30
Recap
OpenStax Introductory Statistics 2e, §2.5 Measures of the Center of the Data §2.5, pp. 98-104 — Examples 2.26–2.30 and the Try Its trace back here
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