Data, Counting, and Probability: Chapter Test

Add the six counts and divide by six. Then sort them and average the middle pair, because with an even count no single value is central.

$\text{mean} = \frac{12 + 7 + 15 + 7 + 9 + 10}{6} = \frac{60}{6} = 10, \qquad \text{median} = \frac{9 + 10}{2} = 9.5$

Sorted, the six counts run $7, 7, 9, 10, 12, 15$, so the third and fourth are the pair to average.

The mode is whichever count occurs most often, and the range is the distance from the smallest count to the largest.

$\text{mode} = 7, \qquad \text{range} = 15 - 7 = 8$

Neither one is divided by anything. Only the mean owes a division by the number of mornings.

A mean is the total divided by the count, so the total is the mean times the count. Seven mornings at a mean of $11$ demand a total of $77$, and the six already recorded supply $60$.

$7 \times 11 - 60 = 77 - 60 = 17$

So the seventh morning brought $17$ goldfinches.

Compare the two totals the two means demand. Six mornings at a mean of $10$ need $60$; seven mornings at a mean of $11$ need $77$.

$77 - 60 = 17 = 11 + 6$

The split is the whole story. The new morning owes $11$ for itself, and it owes one more for each of the six earlier mornings, which used to be credited with $10$ each and are now credited with $11$ each. That is why a single added value can sit far above the mean it produces, and why the further the mean is dragged the further out that value has to be.

Every booking is counted in exactly one row, so the four frequencies add to the total. The difference compares the largest frequency with the smallest.

$14 + 9 + 6 + 11 = 40, \qquad 14 - 6 = 8$

The activity names are categories and the numbers beside them are frequencies, so only the second column is ever added.

Count the symbols first, treating a half symbol as half of one, then multiply the symbol count by the key.

$6.5 \times 20 = 130$

The half coin is worth $10$ dollars, half of the key's value, not one dollar and not nothing.

Read a single month off its own point, and read a change off the segment joining two neighbouring points.

$44 - 38 = 6 \text{ (largest rise)}, \qquad 41 - 33 = 8 \text{ (largest fall)}$

The five readings run $41$, $33$, $38$, $44$, $40$, so the four segments change by $-8$, $+5$, $+6$ and $-4$ in turn. June is the highest point, the steepest climb is May to June, and the steepest drop is March to April.

The guest ends up on one walk, and that walk sits in exactly one of the two Saturday groups, so the group sizes add.

$7 + 5 = 12$

No walk is printed in both groups, so nothing here is counted twice.

Each day is a stage. Sunday's own total comes the same way Saturday's did, and Sunday offers that same number of walks whichever Saturday walk was booked, so the two stage counts multiply.

$(7 + 5) \times (4 + 5) = 12 \times 9 = 108$

The two counts rest on different conditions.

$7 + 5 = 12 \quad \text{(one choice, separate groups)}, \qquad 12 \times 9 = 108 \quad \text{(two stages in sequence)}$

The first is the addition principle, and it demands groups with no member in common: a walk printed in both the morning and the afternoon list would be counted once from each and inflate the total. The second is the multiplication principle, and it demands that each stage offer the same number of choices whatever the earlier stage chose. That is exactly what the programme's promise that every walk runs regardless is there to guarantee. Cancel a Sunday walk whenever a particular Saturday walk is booked and the product stops being valid.

Every envelope holds exactly one thing, so the four counts add to $60$ and the thank-you cards are whatever is left over.

$60 - 9 - 6 - 27 = 18, \qquad P(\text{badge}) = \frac{27}{60} = \frac{9}{20} = 0.45 = 45\%$

The three forms are one number written three ways, not three different answers.

Count the favorable envelopes over the same total. No envelope holds a bus pass, so its favorable count is zero.

$P(\text{book token}) = \frac{9}{60} = \frac{3}{20} = 0.15, \qquad P(\text{bus pass}) = \frac{0}{60} = 0$

A probability of $0$ marks an impossible event, and nothing can go below it, just as nothing can go above the $1$ at the far end.

The formula divides favorable outcomes by total outcomes, and it is valid only when those outcomes are equally likely.

$P(\text{badge}) = \frac{27}{60} \quad \text{requires all } 60 \text{ envelopes to be equally likely}$

What the counting cannot do without is that equal likelihood itself. Here the stall supplies it with two facts: the envelopes are identical from the outside, and the customer takes one without looking. Those two are one route to it rather than the only one, which is why the test of any change is whether equal likelihood survives it. Standing the badge envelopes at the back of an unshuffled box, or printing them on heavier paper a customer can feel through the pile, leaves every count exactly as it was while making some envelopes easier to reach for than others. The counts would still be right and the probability would no longer follow from them.

Add all eight and divide by eight. For the median, sort them and average the fourth and fifth, since eight is an even count.

$\text{mean} = \frac{280}{8} = 35, \qquad \text{median} = \frac{16 + 18}{2} = 17$

Sorted, the eight lengths run $9, 12, 14, 16, 18, 21, 22, 168$.

Drop the $168$ and work again on the remaining seven.

$\text{mean} = \frac{112}{7} = 16, \qquad \text{median} = 16$

Sorted, the seven run $9, 12, 14, 16, 18, 21, 22$, so the fourth value is now the single middle one. The mean has moved $35 - 16 = 19$ minutes and the median $17 - 16 = 1$ minute.

Line the four figures up.

$\text{mean: } 35 \to 16, \qquad \text{median: } 17 \to 16$

The mean is the total shared out evenly, and the total records the actual size of every hire, so a single value of $168$ minutes adds $168$ to it and lifts the mean above seven of the eight hires. The median is fixed by position, and the forgotten bike occupies exactly one place at the top of the sorted list however long it happened to be; removing it shifts the middle by one place. That is why the median describes an ordinary hire on this record and the mean does not, and it is the general reason skewed data is usually reported with a median.

Every item is counted in exactly one category, so the four counts add to $160$ and the reference books are what is left. A share is the count divided by that same total.

$160 - 64 - 48 - 32 = 16, \qquad \frac{64}{160} = 40\%, \quad \frac{48}{160} = 30\%, \quad \frac{32}{160} = 20\%, \quad \frac{16}{160} = 10\%$

The four shares add to $100\%$, which is the check that nothing was missed and nothing counted twice.

A full circle is one complete turn, $360^\circ$, and a slice takes the same fraction of that turn as its category takes of the data.

$0.40 \times 360^\circ = 144^\circ, \qquad 0.10 \times 360^\circ = 36^\circ$

Subtracting two angles gives an angle. It answers a question about the drawing, not about the library.

$144^\circ - 36^\circ = 108^\circ, \qquad 108^\circ \times \frac{160}{360^\circ} = 48$

The whole circle stands for $160$ items across $360$ degrees, so each degree stands for $\frac{160}{360}$ of an item. Converting the $108$ degree gap gives $48$ items, which is exactly $64 - 16$ read straight off the log. The poster has reported the drawing's units as though they were the library's.

Fill the stands one at a time, treating each as a stage. A mask placed on a stand has left the store room, so every later stand offers one fewer than the last.

$12 \times 11 \times 10 = 1320$

The count stops after three factors because there are only three stands to fill.

With copies on the shelf, every stand still has all $12$ masks to choose from whatever the earlier stands took, so the same factor repeats once per stand.

$12 \times 12 \times 12 = 12^3 = 1728$

The exponent is the number of stages and the base is the number of choices at one stage, which is why this is $12^3$ and not $3^{12}$.

The two products agree on the first factor and part company after it.

$12 \times 11 \times 10 = 1320, \qquad 12 \times 12 \times 12 = 1728, \qquad 1728 - 1320 = 408$

Every display counted by the first is also counted by the second, because a display using three different masks is legal under both rules. What the second adds is exactly the displays that repeat a mask, and the arithmetic shows where they come in: the first stand offers $12$ in both counts, and each later stand offers the full $12$ rather than a shrinking $11$ and $10$. Shrinking counts and repeated counts are the same principle applied to stages with different rules about reuse.

The two slots are two stages, and the right slot offers all $5$ kinds whatever went into the left, so the stage counts multiply. The favorable boxes are counted the same way.

$5 \times 5 = 25, \qquad P(\text{both almond}) = \frac{2 \times 2}{25} = \frac{4}{25}$

Keeping the slots apart matters: a box is an ordered pair, so an almond on the left with a plain pastry on the right is a different box from the reverse.

Count the favorable boxes stage by stage over the same total of $25$. Two of the five kinds are almond, so three are not.

$P = \frac{2 \times 3}{25} = \frac{6}{25}$

The sample space has not changed, because the box is still filled the same way.

The two figures are counted over different sample spaces.

$\frac{2}{5} = P(\text{the left slot is almond}), \qquad \frac{4}{25} = P(\text{both slots are almond})$

Filling one slot has $5$ equally likely results; filling both has $5 \times 5 = 25$. The customer has kept the sample space of a single slot while asking a question about the whole box. The favorable boxes are the four that pair an almond on the left with an almond on the right, so the probability is $\frac{4}{25}$. It has to be the smaller of the two, because a box that satisfies both slots is harder to come by than one that satisfies one, and indeed $\frac{4}{25}$ is less than $\frac{10}{25} = \frac{2}{5}$.

Four of the twelve sectors give a prize of some kind. Winning nothing is everything else in the sample space, so its probability is what is left of the one whole.

$P(\text{something}) = \frac{1 + 3}{12} = \frac{1}{3}, \qquad P(\text{nothing}) = 1 - \frac{1}{3} = \frac{2}{3}$

Counting the eight blank sectors directly gives $\frac{8}{12} = \frac{2}{3}$ as well, because every sector belongs to one of the two events and none belongs to both.

An experimental probability divides the number of times the event happened by the number of trials, which here is the $150$ visits.

$\frac{9 + 45}{150} = \frac{54}{150} = \frac{9}{25} = 0.36, \qquad \frac{9}{150} = \frac{3}{50} = 0.06$

Pair the figures up before judging them.

$0.36 \text{ against } \tfrac{1}{3} \approx 0.333, \qquad 0.06 \text{ against } \tfrac{1}{12} \approx 0.083$

What settles this is the length of the run. A theoretical probability says what share of visits win in the long run, and a month of $150$ visits is a short one, so the recorded share scatters around the theoretical value rather than landing on it. Gaps of this size are the ordinary size of that scatter, which makes the record compatible with an app spinning exactly as described: it is not by itself evidence that the app is rigged, and neither is it proof that the app is fair. By the law of large numbers the scatter shrinks as the number of visits grows, so what would narrow the gaps is more visits, not a redesign of the spinner.

A bar's length measures its value only when the value axis begins at zero. Here it begins at $35$, so each bar draws only the amount above $35$.

$\frac{55 - 35}{40 - 35} = \frac{20}{5} = 4, \qquad \frac{55}{40} = 1.375$

The caption has compared the drawn lengths and reported them as memberships. Read off the axis, athletics has $55$ members and rowing has $40$, so the honest caption is that athletics has $15$ more members, a little under one and a half times as many.

A line graph earns its segments from an ordered horizontal axis, almost always time, where the points between two readings really exist.

$\text{athletics} \to \text{hockey} \quad \text{has no midpoint for the line to pass through}$

Be exact about what survives. The two ends of a segment are real readings, so an uphill segment does establish that hockey has more members than athletics, and a downhill one that rowing has fewer than netball. What may not be read is the segment between them: it is not a change, because nothing turned into anything; no height along it is a membership, because there is no club in between; and the run of segments is not a trend. The clearest sign that the shape carries no information of its own is that relisting the five clubs alphabetically, or by size, would change the shape completely while changing no membership. Drawn as five separate bars the same five numbers support the comparison the page actually wants.

Work out what the two graphics jointly determine before judging the conclusion.

$0.18 \times 9000 = 1620, \qquad \frac{1620}{40} = 40.50$

So rowing receives $1620$ dollars for its $40$ members, $40.50$ dollars each, and that is a fact about rowing alone. The words most generously rank rowing against the other four clubs, and neither graphic supplies what a ranking needs: the other four slices carry no labels here and only two of the five memberships are stated. A club holding a larger slice could beat rowing per member without contradicting anything on the page: a $25\%$ slice is $0.25 \times 9000 = 2250$ dollars, and across $45$ members that is $50$ dollars each, above rowing's $40.50$. Both figures are available, since the four unlabelled slices share the remaining $82\%$ and Graphic A leaves every other membership between $40$ and $55$. So the conclusion runs ahead of what is on the page.