S4, S5 - Frequency tables and comparing distributions

S4, S5 - Frequency tables and comparing distributions

Work out averages from frequency tables and estimate a mean from grouped measurements. Compare distributions using a centre and a spread, interpret their displays, and use sample statistics to describe a population without claiming more than the data support.

The mean from a frequency table

Two bus services both advertise a mean delay of 66 minutes. Could one still be much more dependable? We will compare their distributions after learning how to recover an average from a compact table.

You already know

Mean is total divided by the number of observations. Median uses ordered positions, mode uses frequency, and range uses the smallest and largest values.

Suppose a table says that 44 people each own 11 bike. That row contributes 44 people and 44 bikes. If 33 people each own 22 bikes, their row contributes 33 people and 66 bikes. Keep those two totals separate.

Mean from a frequency table

For value xx with frequency ff,

mean=∑fx∑f\text{mean}=\frac{\sum fx}{\sum f}

Here ∑\sum means “add all the entries in this column”. Multiply each value by its frequency, add the products, then divide by the total frequency.

Not on the formulae sheet: learn it.

A product column is a useful organiser, not a new kind of data.

Worked example 1: counting bikes and people

Question. Bike counts 0,1,2,30,1,2,3 have frequencies 2,4,3,12,4,3,1. Work out the mean number of bikes. Non-calculator.

Plan. Add the fxfx products for the bike total, then divide by the number of people.

xxfffxfx
002200
114444
223366
331133
∑f=2+4+3+1=10number of people∑fx=0+4+6+3=13number of bikesmean=13÷10=1.3bikes per person\begin{aligned} \sum f&=2+4+3+1=10 &&\text{number of people}\\ \sum fx&=0+4+6+3=13 &&\text{number of bikes}\\ \text{mean}&=13\div10=1.3 &&\text{bikes per person} \end{aligned}

Why this step? The table has four rows but represents ten people. Dividing by four would average the row labels, not the observations.

Answer. 1.31.3 bikes per person.

Check. Expanding the table gives 0,0,1,1,1,1,2,2,2,30,0,1,1,1,1,2,2,2,3, also totalling 1313.

Change only the zero row's frequency from 22 to 77. The bike total stays fixed, but more people share it.

Worked example 2: extra zero observations

Question. Bike counts 0,1,2,30,1,2,3 now have frequencies 7,4,3,17,4,3,1. Work out the mean to 22 decimal places. Calculator allowed.

Plan. Update the total frequency but keep the unchanged product total.

∑f=7+4+3+1=15five extra people∑fx=0×7+1×4+2×3+3×1=13no extra bikesmean=13÷15=0.8666…calculator expression=0.87round at the end\begin{aligned} \sum f&=7+4+3+1=15 &&\text{five extra people}\\ \sum fx&=0\times7+1\times4+2\times3+3\times1=13 &&\text{no extra bikes}\\ \text{mean}&=13\div15=0.8666\ldots &&\text{calculator expression}\\ &=0.87 &&\text{round at the end} \end{aligned}

Why this step? Zero observations add to the denominator even though their products are zero.

Answer. 0.870.87 bikes per person, to 22 decimal places.

Check. Adding people with no bikes must reduce the original mean of 1.31.3.

A correct table can still lead to the wrong denominator.

Watch out

Wrong: (0+1+2+3)÷4=1.5(0+1+2+3)\div4=1.5. This ignores frequency. Right: total the fxfx column and divide by the total frequency, not the number of rows.

Use both totals in the next table.

Combining groups with different sizes

A group mean is shorthand for a total. Before combining groups, unwrap that shorthand. A class of 3030 must have more influence than a class of 1010.

A combined mean

For two groups of sizes n1,n2n_1,n_2 with means m1,m2m_1,m_2,

combined mean=n1m1+n2m2n1+n2\text{combined mean}=\frac{n_1m_1+n_2m_2}{n_1+n_2}

The groups must measure the same quantity in the same units and must not overlap.

Not on the formulae sheet: learn it.

This is the ordinary mean formula after recovering each group's total. It is not permission to average averages without checking group sizes.

Worked example 3: equal-sized groups

Question. Ten pupils have mean score 66. Another ten have mean score 88. Work out the combined mean. Non-calculator.

Plan. Recover both score totals, add them and divide by all twenty pupils.

first total=10×6=60mean times group sizesecond total=10×8=80same methodcombined mean=(60+80)÷20=7all scores, all pupils\begin{aligned} \text{first total}&=10\times6=60 &&\text{mean times group size}\\ \text{second total}&=10\times8=80 &&\text{same method}\\ \text{combined mean}&=(60+80)\div20=7 &&\text{all scores, all pupils} \end{aligned}

Why this step? Each pupil counts equally. Equal group sizes happen to give equal weights to the two means.

Answer. 77.

Check. The answer is midway between 66 and 88, as the group sizes match.

Change only the second group's size from 1010 to 3030. Its mean should now pull the answer more strongly.

Worked example 4: unequal-sized groups

Question. Ten pupils have mean score 66. Another thirty have mean score 88. Work out the combined mean. Non-calculator.

Plan. Recalculate the second total and the total number of pupils.

first total=10×6=60unchanged groupsecond total=30×8=240larger groupcombined mean=(60+240)÷40=7.5divide by all forty\begin{aligned} \text{first total}&=10\times6=60 &&\text{unchanged group}\\ \text{second total}&=30\times8=240 &&\text{larger group}\\ \text{combined mean}&=(60+240)\div40=7.5 &&\text{divide by all forty} \end{aligned}

Why this step? Thirty pupils contribute three times as many observations as ten. Averaging the two means would give the groups equal influence instead.

Answer. 7.57.5.

Check. The answer lies between 66 and 88, nearer 88 because more pupils belong to that group.

The size check often catches an incorrect combined mean before you finish writing it.

Watch out

Wrong: (6+8)÷2=7(6+8)\div2=7 for unequal-sized groups. Right: recover the totals first. Averaging group means directly works when the group sizes are equal, not in general.

The next groups share the same measurement and units, so their totals can be combined.

Median and mode from frequencies

A frequency table already puts repeated values into bundles. You can locate the middle without writing every observation out, but you must count through those bundles.

For values 0,1,2,30,1,2,3 with frequencies 3,4,5,23,4,5,2, the zeros occupy positions 11 to 33, the ones occupy 44 to 77, the twos occupy 88 to 1212, and the threes occupy 1313 to 1414. These running totals locate observations; they are not new data values.

Key idea

Find the median's observation positions using the total frequency, then locate their values. For the mode, choose the value with the greatest frequency.

The median-position rule is not on the formulae sheet: learn it. A vertical line chart supplies the same table: its horizontal positions are values and its heights are frequencies.

Worked example 5: middle values in different rows

Question. Values 0,1,2,30,1,2,3 have frequencies 3,4,5,23,4,5,2. Work out the median and mode. Non-calculator.

Plan. Locate positions 77 and 88 among the fourteen observations.

total frequency=3+4+5+2=14count observationsposition 7 has value 1ones end at position 7position 8 has value 2twos start at position 8median=(1+2)÷2=1.5average middle valuesmode=2greatest frequency is 5\begin{aligned} \text{total frequency}&=3+4+5+2=14 &&\text{count observations}\\ \text{position 7}&\text{ has value }1 &&\text{ones end at position 7}\\ \text{position 8}&\text{ has value }2 &&\text{twos start at position 8}\\ \text{median}&=(1+2)\div2=1.5 &&\text{average middle values}\\ \text{mode}&=2 &&\text{greatest frequency is 5} \end{aligned}

Why this step? Two middle positions can fall in different rows. Do not choose whichever row has the larger frequency.

Answer. Median 1.51.5, mode 22.

Check. There are seven observations at most 11, and seven at least 22.

Change only the last frequency from 22 to 33. There is now one middle observation.

Worked example 6: an odd total frequency

Question. Values 0,1,2,30,1,2,3 have frequencies 3,4,5,33,4,5,3. Work out the median and mode. Non-calculator.

Plan. Find position 88 in the new total of fifteen observations.

total frequency=3+4+5+3=15one extra observationmedian position=(15+1)÷2=8single middle positionmedian=2position 8 is in the twosmode=2frequency 5 is still largest\begin{aligned} \text{total frequency}&=3+4+5+3=15 &&\text{one extra observation}\\ \text{median position}&=(15+1)\div2=8 &&\text{single middle position}\\ \text{median}&=2 &&\text{position 8 is in the twos}\\ \text{mode}&=2 &&\text{frequency 5 is still largest} \end{aligned}

Why this step? An extra observation can change which position defines the median, even when the most frequent value stays the same.

Answer. Median 22, mode 22.

Check. Seven observations occur before position 88, and seven after it.

The frequency itself is not the mode.

Watch out

Wrong: mode =5=5 because 55 is the largest frequency. Right: mode =2=2, the value occurring five times. Likewise, median position 88 does not mean median value 88.

Here both middle positions fall in one bundle.

Grouped data and estimates of the mean

A table saying eight people waited between 1010 and 2020 minutes has hidden their individual waits. You cannot recover the exact total. Use each class midpoint as a representative value and be honest that the answer is an estimate.

Class interval, midpoint and modal class

A class interval groups values between boundaries. For 10<t⩽2010<t\leqslant20, include 2020, but exclude 1010, so adjoining classes do not overlap. Its midpoint is (10+20)÷2=15(10+20)\div2=15. For equal-width classes, the modal class is the class with the greatest frequency; it is an interval, not its midpoint.

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