Data, Counting, and Probability: Chapter Test

Read each bar's end against the scale, which runs from $0$ in steps of $5$. Tuesday reaches $20$ and Wednesday reaches $10$, and the word more asks for the difference.

$20 - 10 = 10$

Because the scale starts at $0$, the bar lengths really are proportional to the values, so this difference is a fair comparison.

A notebook is a sequence of two choices, a color and then a ruling, and every color comes in every ruling, so the stage counts multiply.

$6 \times 3 = 18$

The six colors form six groups of three rulings each, which is exactly what multiplication counts.

Sort first, because the median is the middle in order. The sorted distances are $4, 5, 7, 7, 9, 10, 12, 14$. With an even count no single value is central, so average the fourth and fifth.

$\frac{7 + 9}{2} = \frac{16}{2} = 8$

The median need not be one of the recorded distances, and here it is not.

Taking a sock without looking from identical socks makes all $9$ equally likely, so the sample space has $9$ outcomes and the favorable ones are the $3$ navy socks.

$P(\text{navy}) = \frac{3}{9} = \frac{1}{3}$

The denominator is the whole drawer, not the part left over after the navy socks are set aside.

A slice is a share of the whole, so its count is its percent of the total.

$\frac{22}{100} \times 450 = 0.22 \times 450 = 99$

So $99$ visitors arrived by train. The $22$ becomes a count only after it is multiplied by the total.

Fill the three positions as three stages. Any of the $9$ photographs can take the first position; once it is hung it is unavailable, so $8$ remain for the second and $7$ for the third.

$9 \times 8 \times 7 = 504$

The counts shrink because a photograph already on the wall cannot be hung again.

The two orchards share a center and differ in spread. The range is the distance from the smallest weight to the largest, so a larger range means the weights stretch further apart.

$\text{range}_A = 8 \text{ g}, \qquad \text{range}_B = 60 \text{ g}$

Equal means say the two sets balance at the same point; they say nothing about how far the values sit from it, which is exactly what the range reports.

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

$5.5 \times 8 = 44$

The half symbol contributes $4$ loaves, half of the key's $8$, not one loaf.

A mean is the total divided by the count, so the total is the mean times the count. Four days at a mean of $26$ need a total of $104$, and the first three supply $23 + 31 + 18 = 72$.

$4 \times 26 - 72 = 104 - 72 = 32$

So the fourth ride must be $32$ kilometers.

The horizontal axis here is the eight years, which are ordered, so a segment joining two readings stands for a real change between them.

$\text{rising segment} \Rightarrow \text{more trees}, \qquad \text{falling segment} \Rightarrow \text{fewer trees}$

That is exactly what a line graph is built for. The other displays hold the same numbers, but none of them shows the change from one year to the next at a glance.

Three stages in sequence: the song, then the presenter, then the co-presenter. The two roles go to different people, so the second role has one fewer choice than the first.

$5 \times 8 \times 7 = 280$

The song stage offers the same $5$ whatever the team does, and the co-presenter stage offers the same $7$ whoever presents, so the stage counts multiply.

Read each bar against the scale first. A bar halfway between two marks that are $10$ apart is $5$ above the lower mark, so the four values are $30$, $45$, $20$ and $15$.

$\frac{30 + 45 + 20 + 15}{4} = \frac{110}{4} = 27.5$

The mean is $27.5$ parcels per bar, and it need not be a value any bar actually reaches.

Each day is a single choice from two separate groups, so within a day the group sizes add. The two days are then a sequence of stages, so the day totals multiply.

$(6 + 4) \times (5 + 3) = 10 \times 8 = 80$

A workshop is either pottery or printmaking and never both, which is what licenses the addition inside each day.

A slice takes the same fraction of one full turn as its category takes of the data, and the total is the sum of the four frequencies, $27 + 18 + 21 + 6 = 72$.

$\frac{18}{72} \times 360^\circ = \frac{1}{4} \times 360^\circ = 90^\circ$

A quarter of the members gives a quarter turn, the right angle the eye expects.

Add and divide for the mean; sort and count in for the median. Sorted, the pages read are $11, 12, 12, 13, 14, 15, 70$.

$\text{mean} = \frac{147}{7} = 21, \qquad \text{median} = 13$

The mean uses the size of every value, so the single reader on $70$ pages drags it above six of the seven students. The median uses only position, so it stays among the ordinary readers and describes a typical student better.

Each person is a stage, and every person still has all three answers available whatever anyone before them said, so the same factor repeats once per person.

$3 \times 3 \times 3 \times 3 = 3^4 = 81$

The base is the number of choices at one stage and the exponent is the number of stages, so it is $3^4$ and not $4^3$.

The lunch is a sequence of two stages, so the multiplication principle sizes both the sample space and the favorable set.

$P = \frac{2 \times 1}{6 \times 5} = \frac{2}{30} = \frac{1}{15}$

There are $30$ equally likely lunches, and the favorable ones are the $2$ that pair one of the two vegetarian soups with the single rye bread.

Both centers are worth having here. The six takings total $1800$ dollars, and sorted they run $180, 190, 195, 205, 210, 820$.

$\text{mean} = \frac{1800}{6} = 300, \qquad \text{median} = \frac{195 + 205}{2} = 200$

So the advertised figure really is the mean, and the arithmetic is honest. What it is not is typical: the festival day has pulled it above five of the six days, and on skewed data like this the median is the more trustworthy center. The range of $820 - 180 = 640$ dollars measures spread, so it is no substitute for either.

The four slices account for the whole circle, so the three labeled slices are exactly the households that do not heat with gas.

$P(\text{not gas}) = 0.31 + 0.09 + 0.14 = 0.54, \qquad 0.54 \times 600 = 324$

The complement rule reaches the same value the other way: the gas slice is $1 - 0.54 = 0.46$, and $1 - 0.46 = 0.54$. A percent becomes a count only after it is multiplied by the total.

Two spins give $4 \times 4 = 16$ equally likely ordered results. Totals of at least $6$ come from $(2,4), (3,3), (4,2)$ for a total of $6$, from $(3,4), (4,3)$ for $7$, and from $(4,4)$ for $8$, which is $6$ results.

$P_{\text{theory}} = \frac{6}{16} = 0.375, \qquad P_{\text{exp}} = \frac{86}{200} = 0.43$

The two differ, which is what a run of $200$ pairs should be expected to do. By the law of large numbers the experimental value settles toward the theoretical one as the number of pairs grows, so a gap this size is what a limited run is expected to produce and does not by itself show the spinner is unfair.