Applications: Measuring Data

The mean shares the total of all the prices equally among the houses, so a single very expensive house lifts it for every house in the street. The median is the middle price once the prices are in order, and it moves only when the middle of the list moves. The mode is the price that occurs most often. Working all three out first, and then asking which one a reader would recognize, is the whole of this question.

1. Put the prices in order and work in thousands of dollars, which keeps the arithmetic short: 250, 320, 380, 380, 450, 520, 1200. The answers are turned back into dollars at the end.
2. Add them for the mean: 250 + 320 + 380 + 380 + 450 + 520 + 1200 = 3500 thousand. There are 7 prices, so the mean is 3500 ÷ 7 = 500 thousand, that is $500 000.
3. (a) With 7 prices in order the median is the 4th of them, which is 380 thousand. The price 380 thousand occurs twice and every other price once, so the mode is 380 thousand as well. The mean is $500 000, the median is $380 000 and the mode is $380 000.
4. Count the prices below the mean. The prices 250, 320, 380, 380 and 450 are all less than 500, and only 520 and 1200 are above it, so 5 of the 7 houses sold for less than the advertised mean.
5. (b) The median, $380 000, describes an ordinary house in the street better. The one price of $1 200 000 is 700 thousand above the next highest, and on its own it lifts the mean by 700 ÷ 7 = 100 thousand. Check: leave that house out and the other six average 2300 ÷ 6 ≈ 383 thousand, which is close to the median.

answer(a) the mean is $500 000, the median is $380 000 and the mode is $380 000; (b) 5 of the 7 houses sold for less than the mean, and the median describes an ordinary house better

techniqueMeasures of Central Tendency · Median and Mode

examsO-Level · SAT · GCSE Higher

A frequency table lists each value once and says how often it happened, so the total is not the sum of the values but the sum of value times frequency. Set the table out in three columns, x for the goals, f for the matches and fx for the goals those matches produced. The mean is the total of the fx column divided by the total of the f column, and the median is found by counting along the frequencies until the middle match is reached.

1. Write the table in three columns: the goals x, the number of matches f, and the product fx, which is the number of goals scored in the matches of that row.
2. Fill in the fx column: 0 × 3 = 0, 1 × 5 = 5, 2 × 6 = 12, 3 × 3 = 9, 4 × 2 = 8 and 5 × 1 = 5.
3. (a) Total the two columns: 3 + 5 + 6 + 3 + 2 + 1 = 20 matches and 0 + 5 + 12 + 9 + 8 + 5 = 39 goals. The mean is 39 ÷ 20 = 1.95 goals in a match.
4. For the median, 20 values put it halfway between the 10th and the 11th. Counting along the frequencies, 3 matches had 0 goals and 3 + 5 = 8 had 1 or fewer, so the 9th to the 14th matches all had 2 goals. The 10th and the 11th are both 2, so the median is 2 goals. The largest frequency is 6, at 2 goals, so the mode is 2 goals as well.
5. (b) More than 1.95 goals means 2 goals or more, which is 6 + 3 + 2 + 1 = 12 matches. The median is 2 goals, the mode is 2 goals and 12 of the 20 matches were above the mean. Check: 8 matches were below the mean and 8 + 12 = 20.

answer(a) the team scored 39 goals in the 20 matches, so the mean is 1.95 goals in a match; (b) the median is 2 goals, the mode is 2 goals, and 12 of the 20 matches were above the mean

techniqueAverages from a Frequency Table · Median and Mode

examsO-Level · SAT · GCSE Higher

Once the times have been grouped, no individual time is left in the table, so the mean cannot be found exactly. The standard estimate gives everyone in a class the value at the middle of that class, and then the calculation is an ordinary frequency-table mean. The modal class needs no arithmetic at all: it is the class with the largest frequency, and it is an interval, not a single time.

1. Write down the middle of each class, halfway between its two ends: 5, 15, 25, 35 and 45 minutes.
2. Multiply each middle by its frequency: 5 × 4 = 20, 15 × 10 = 150, 25 × 14 = 350, 35 × 8 = 280 and 45 × 4 = 180.
3. (a) The largest frequency is 14, so the modal class is 20 < m ≤ 30 minutes. The totals are 40 staff and 20 + 150 + 350 + 280 + 180 = 980 minutes, so the estimated mean is 980 ÷ 40 = 24.5 minutes.
4. The estimate is only an estimate because every time was replaced by the middle of its class. To see how far out it could be, give everyone the lowest time their class allows: 0 × 4 + 10 × 10 + 20 × 14 + 30 × 8 + 40 × 4 = 780, and 780 ÷ 40 = 19.5 minutes.
5. (b) Now give everyone the highest time their class allows: 10 × 4 + 20 × 10 + 30 × 14 + 40 × 8 + 50 × 4 = 1180, and 1180 ÷ 40 = 29.5 minutes. The true mean therefore lies between 19.5 and 29.5 minutes, and 24.5 minutes is the estimate exactly halfway between those bounds.

answer(a) the modal class is 20 < m ≤ 30 minutes and the estimated mean is 24.5 minutes; (b) the mean is an estimate because each time was replaced by the middle of its class, and the true mean lies between 19.5 and 29.5 minutes

techniqueMean of Grouped Data · The Modal Class

examsO-Level · SAT · GCSE Higher

The mean is the total number of bicycles divided by the number of days, and both of those can be written in terms of the unknown frequency k. Setting that quotient equal to the stated mean gives one equation in k, and clearing the fraction leaves a linear equation to solve. Once k is known the table is complete and the median can be counted out in the ordinary way.

1. Count the days in terms of k: 4 + 6 + 9 + k + 5 = 24 + k days.
2. Count the bicycles in the same way: 0 × 4 + 1 × 6 + 2 × 9 + 3 × k + 4 × 5 = 6 + 18 + 3k + 20 = 44 + 3k bicycles.
3. The mean is the bicycles divided by the days, and that is 2, so 44 + 3k24 + k = 2. Multiply both sides by 24 + k: 44 + 3k = 2(24 + k) = 48 + 2k.
4. (a) Subtract 2k from both sides and then 44: k = 4. Check: the shop traded on 24 + 4 = 28 days and sold 44 + 12 = 56 bicycles, and 56 ÷ 28 = 2 as stated.
5. For the median there are 28 days, so it lies halfway between the 14th and the 15th. Counting along the frequencies, 4 days sold none, 4 + 6 = 10 days sold 1 or fewer and 10 + 9 = 19 days sold 2 or fewer. The 11th to the 19th days all sold 2, so the 14th and the 15th are both 2 and the median is 2 bicycles.
6. (b) More than the mean of 2 means 3 or 4 bicycles, which is 4 + 5 = 9 days. The median is 2 bicycles and the shop beat its mean on 9 of the 28 days.

answer(a) k = 4, so the shop traded on 28 days and sold 56 bicycles; (b) the median is 2 bicycles and the shop sold more than the mean on 9 days

techniqueFinding a Frequency from the Mean · Averages from a Frequency Table

examsO-Level · SAT · GCSE Higher

A cumulative frequency curve plots each upper class boundary against the number of students at or below it, so every reading answers the question "how many students are below this mark?". To find a mark, go in on the cumulative frequency axis at the student you want, across to the curve and down to the mark. The median is the n2th student, the quartiles are the n4th and the 3n4th, and a percentile is found the same way from the count that matches it.

1. Plot the points (10, 2), (20, 8), (30, 14), (40, 26), (50, 40), (60, 54), (70, 64), (80, 72), (90, 78) and (100, 80), starting from (0, 0), and join them with a smooth curve. Each point is plotted at the top of its class, because only then are all the students counted.
2. The median is the mark of the 802 = 40th student. Go in at 40 on the cumulative frequency axis, across to the curve and down: the mark is 50.
3. The lower quartile is the 804 = 20th student and the upper quartile is the 3 × 804 = 60th. Reading across at 20 gives 35 marks, and reading across at 60 gives 66 marks.
4. (a) The median is 50 marks, and the interquartile range is 66 − 35 = 31 marks. The middle half of the students scored between 35 and 66.
5. (b) The top 10% of 80 students is 8 students, so 80 − 8 = 72 students are below them. Read across at 72 and down: the mark is 80. A student needs 80 marks for a distinction. Check: the table says 72 students scored 80 or fewer, so exactly 8 scored more.

answer(a) the median is 50 marks and the interquartile range is 66 − 35 = 31 marks; (b) the lowest mark that earns a distinction is 80

techniqueQuartiles from the Curve · Cumulative Frequency · Percentiles

examsO-Level · GCSE Higher

An average on its own says where a set of values sits, and nothing at all about how tightly they are packed around it. Two sets can share a mean and a median and still behave quite differently, so the second measure is a measure of spread. The interquartile range is the width of the middle half of the values, from the lower quartile to the upper quartile, and it ignores the extremes at both ends.

1. Both lists are already in order. Add each one: route A gives 28 + 29 + 29 + 30 + 30 + 30 + 31 + 31 + 32 = 270 and route B gives 22 + 25 + 27 + 29 + 30 + 31 + 33 + 35 + 38 = 270. With 9 days each, both means are 270 ÷ 9 = 30 minutes.
2. With 9 times in order the median is the 5th. For route A that is 30 minutes and for route B it is 30 minutes, so the medians agree as well. Neither average separates the two routes.
3. The lower quartile is the median of the four times below the middle one. Route A: 29 + 292 = 29 minutes. Route B: 25 + 272 = 26 minutes.
4. (a) The upper quartile is the median of the four times above the middle one. Route A: 31 + 312 = 31 minutes, so its interquartile range is 31 − 29 = 2 minutes. Route B: 33 + 352 = 34 minutes, so its interquartile range is 34 − 26 = 8 minutes.
5. (b) Take route A. The two routes average the same 30 minutes, but the middle half of route A's journeys covers only 2 minutes while route B's covers 8. The ranges tell the same story: 32 − 28 = 4 minutes against 38 − 22 = 16 minutes, and a 38 minute journey on route B would make the commuter late.

answer(a) both routes have a mean of 30 minutes and a median of 30 minutes; route A has an interquartile range of 2 minutes and route B one of 8 minutes; (b) route A, because its times are packed far more tightly around the same average

techniqueRange and IQR · Measures of Spread · Comparing Data Sets

examsO-Level · SAT · GCSE Higher

A box plot draws the five-number summary: the box runs from the lower quartile to the upper quartile with the median marked inside it, and the whiskers reach out to the smallest and the largest value. Drawn on one scale, two box plots can be compared at a glance, and the shape of a single box answers a different question: if the median sits in the middle of its box and the whiskers are about equal, the values are roughly symmetric about the middle.

1. Draw one scale in minutes long enough for both clubs, from about 35 to 70. For each club draw the box from the lower quartile to the upper quartile, mark the median inside it, and draw the whiskers out to the smallest and the largest time.
2. The interquartile ranges come straight off the boxes: club A has 52 − 44 = 8 minutes and club B has 55 − 40 = 15 minutes.
3. (a) Club B has the lower median, 43 minutes against club A's 48 minutes, so a typical club B runner is faster. Club A's interquartile range is 8 minutes against club B's 15, so club A's runners are much more alike; the slowest club B runner took 70 minutes, 12 minutes behind the slowest in club A.
4. For club A, compare the two halves of the box: 48 − 44 = 4 minutes below the median and 52 − 48 = 4 minutes above it. The whiskers match too, 44 − 38 = 6 minutes and 58 − 52 = 6 minutes.
5. (b) Club A is roughly symmetric, because each half of its box is 4 minutes wide and each whisker is 6 minutes long. Club B is not: its box is 43 − 40 = 3 minutes wide below the median and 55 − 43 = 12 minutes wide above it, and the upper whisker runs 70 − 55 = 15 minutes against 40 − 36 = 4 minutes at the other end, so club B's times are stretched toward the slow end.

answer(a) club A has a median of 48 minutes and an interquartile range of 8 minutes, club B a median of 43 minutes and an interquartile range of 15 minutes, so club B is faster in the middle and club A is far more consistent; (b) club A is roughly symmetric, with 4 minutes on each side of its median and whiskers of 6 minutes, while club B is stretched toward the slow end, with 3 minutes below its median against 12 minutes above it

techniqueBox and Whisker · Judging Normality from a Box Plot · Comparing Data Sets

examsO-Level · SAT · GCSE Higher

The standard deviation measures how far the values sit from their own mean. Each distance from the mean is squared, so that a weight below the mean counts as much as one above it, the squares are averaged over the number of bottles, and the square root turns the answer back into grams. Two machines can hit the target on average and still behave very differently, and the standard deviation is what separates them.

1. Find each mean. Machine A: 497 + 499 + 500 + 501 + 503 = 2500, and 2500 ÷ 5 = 500 g. Machine B: 494 + 498 + 500 + 502 + 506 = 2500, and 2500 ÷ 5 = 500 g. Both machines average the target weight exactly.
2. Machine A, distance from the mean and its square: −3 gives 9, −1 gives 1, 0 gives 0, 1 gives 1 and 3 gives 9. The squares total 9 + 1 + 0 + 1 + 9 = 20, so the variance is 20 ÷ 5 = 4 and the standard deviation is √4 = 2 g.
3. Machine B, the same two columns: −6 gives 36, −2 gives 4, 0 gives 0, 2 gives 4 and 6 gives 36. The squares total 36 + 4 + 0 + 4 + 36 = 80, so the variance is 80 ÷ 5 = 16 and the standard deviation is √16 = 4 g.
4. (a) Both means are 500 g. Machine A has a standard deviation of 2 g and machine B one of 4 g, so machine B's weights sit twice as far from the target on average.
5. (b) A bottle is kept when it weighs between 495 g and 505 g. Machine A's lightest bottle is 497 g and its heaviest is 503 g, so all five are kept. Machine B's 494 g and 506 g bottles are outside the limits, so 2 bottles are rejected, both from machine B, and machine B is the machine to service. Check: every rejected bottle is more than one standard deviation of machine B from the target, and machine A's whole range, 497 g to 503 g, sits inside the limits.

answer(a) both machines have a mean of 500 g, and the standard deviations are 2 g for machine A and 4 g for machine B; (b) 2 bottles are rejected, both from machine B, so machine B should be serviced

techniqueStandard Deviation: Ungrouped · Standard Deviation · Comparing Data Sets

examsO-Level · SAT

Grouped data is handled by giving every battery in a class the value at the middle of that class, so a class of 7 batteries becomes the same middle value counted 7 times. The mean is then the usual frequency-table mean, and the variance is the average of the squared distances from that mean, again weighted by the frequencies. Everything that follows is an estimate, because the individual lives are gone.

1. Write down the middle of each class: 10, 14, 18, 22 and 26 hours.
2. Multiply each middle by its frequency: 10 × 2 = 20, 14 × 7 = 98, 18 × 12 = 216, 22 × 7 = 154 and 26 × 2 = 52. The totals are 30 batteries and 20 + 98 + 216 + 154 + 52 = 540 hours, so the estimated mean is 540 ÷ 30 = 18 hours.
3. Now the distance of each middle from the mean, and its square times the frequency: 10 − 18 = −8, and 2 × 64 = 128; 14 − 18 = −4, and 7 × 16 = 112; 18 − 18 = 0, and 12 × 0 = 0; 22 − 18 = 4, and 7 × 16 = 112; 26 − 18 = 8, and 2 × 64 = 128.
4. (a) Those products total 128 + 112 + 0 + 112 + 128 = 480, so the variance is 480 ÷ 30 = 16 and the standard deviation is √16 = 4 hours. The estimated mean is 18 hours.
5. One standard deviation each side of the mean runs from 18 − 4 = 14 hours to 18 + 4 = 22 hours. The class 12 < h ≤ 16 is 4 hours wide and half of it lies in that range, which is half of 7, or 3.5 batteries. The class 16 < h ≤ 20 lies inside it completely, which is 12 batteries. Half of the class 20 < h ≤ 24 lies in it, another 3.5 batteries.
6. (b) That gives 3.5 + 12 + 3.5 = 19 batteries, so about 19 of the 30 last within one standard deviation of the mean. Check: that is a little under two thirds of the batch, which is what a single peaked batch usually gives.

answer(a) the estimated mean is 18 hours and the standard deviation is 4 hours; (b) about 19 of the 30 batteries

techniqueStandard Deviation: Grouped · Mean of Grouped Data

examsO-Level · SAT

The mean absolute deviation is the plainest measure of spread there is: the average distance of the values from their own mean. Each distance is taken without its sign, because a day $60 below the mean is as far from it as a day $60 above. Nothing is squared and no square root is needed, so the answer stays in dollars and can be read as "a typical day is this far from the mean".

1. Find the mean first, because every distance is measured from it: 190 + 210 + 230 + 270 + 290 + 310 = 1500, and 1500 ÷ 6 = $250.
2. Take the distance of each day's takings from $250, ignoring the sign: 60, 40, 20, 20, 40 and 60 dollars.
3. (a) Those distances total 60 + 40 + 20 + 20 + 40 + 60 = 240, so the mean absolute deviation is 240 ÷ 6 = $40. The mean is $250 and on a typical day the takings are about $40 from it. Check: the three days above the mean are 20 + 40 + 60 = $120 above it in total and the three below are $120 below it, so the distances must total 2 × 120 = 240.
4. A day is unusual when its distance from $250 is more than $40. Comparing the six distances with 40: 60 is more, 40 is not more, 20 is not, 20 is not, 40 is not and 60 is more.
5. (b) So 2 of the 6 days were unusual, the $190 day and the $310 day, each $60 from the mean. The other four days were $40 or less from it.

answer(a) the mean is $250 and the mean absolute deviation is $40; (b) 2 of the 6 days were unusual

techniqueMean Absolute Deviation · Measures of Spread

examsO-Level · SAT

A rate is always a rate out of something, and the something matters. Within a group the comparison is straightforward: divide the recoveries by the cases of that kind. The overall rate of a clinic is not the average of its two group rates; it is a weighted average, weighted by how many cases of each kind the clinic treated. When the two clinics treat very different mixes of cases, the overall comparison can point the opposite way from both of the group comparisons.

1. Mild cases: clinic A recovered 160200 = 80% and clinic B recovered 4550 = 90%. Clinic B is ahead.
2. Severe cases: clinic A recovered 1050 = 20% and clinic B recovered 60200 = 30%. Clinic B is ahead again.
3. (a) All cases: clinic A recovered 160 + 10 = 170 of its 250 patients, which is 170250 = 68%, and clinic B recovered 45 + 60 = 105 of its 250, which is 105250 = 42%. So clinic B leads on mild cases and on severe cases, and trails badly overall.
4. The reason is the mix. Severe cases recover far less often at either clinic, and clinic B treats 200 of them against clinic A's 50, while clinic A's work is 200 mild cases against clinic B's 50. Each overall rate is pulled toward the rate of the kind of case that clinic mostly sees.
5. (b) Send the patient to clinic B, because whichever kind of case the patient turns out to be, clinic B recovers a greater share of them. Check by giving clinic B clinic A's mix of cases: 200 mild at 90% is 180 recoveries and 50 severe at 30% is 15, a total of 195 out of 250, which is 78%, well above clinic A's 68%.

answer(a) clinic A recovered 80% of its mild cases, 20% of its severe cases and 68% of all its cases; clinic B recovered 90%, 30% and 42%; (b) clinic B, because it recovers a greater share of both kinds of case, and its overall rate is lower only because 200 of its 250 cases were severe while 200 of clinic A's were mild, and on clinic A's mix of cases clinic B would recover 78%

techniqueSimpson’s Paradox · Comparing Data Sets

examsO-Level · SAT · GCSE Higher

All three averages can be worked out from a frequency table, but only one of them names something a shop can buy. The mean of a set of sizes need not be a size at all; the median and the mode always are, because both are values that actually occurred. Once the modal size is known, the order is a proportion: the share of the sales that size took, applied to the number of pairs being ordered.

1. Total the pairs: 4 + 8 + 14 + 20 + 10 + 4 = 60 pairs. Then multiply each size by its frequency: 5 × 4 = 20, 6 × 8 = 48, 7 × 14 = 98, 8 × 20 = 160, 9 × 10 = 90 and 10 × 4 = 40, which total 456.
2. The mean size is 456 ÷ 60 = 7.6.
3. (a) For the median, 60 pairs put it halfway between the 30th and the 31st. Counting along the frequencies, 4, then 12, then 26, then 46 pairs are reached, so the 27th to the 46th pairs were all size 8 and the median is size 8. The largest frequency is 20, at size 8, so the modal size is 8 as well. The mean is 7.6, the median is 8 and the mode is 8.
4. Size 8 took 20 of the 60 pairs sold, which is 2060 = 13 of the week's sales. Keeping the same proportions, the order should hold 13 of 120 pairs in size 8.
5. (b) 13 × 120 = 40 pairs of size 8. The mode is the average of use here, because 7.6 is not a size any shoe is made in and no pair of that size could be ordered. Check: ordering every size in the same proportion doubles each frequency, and 2 × 20 = 40.

answer(a) over the 60 pairs sold the mean is 7.6, the median is size 8 and the modal size is 8; (b) 40 pairs of size 8, and the mode is the average of use, because no shoe is made in size 7.6

techniqueMedian and Mode · Measures of Central Tendency · Averages from a Frequency Table

examsO-Level · SAT · GCSE Higher