Numerical Data Displays and Spread: Additional Practice (Set 2)

With six values the median is the mean of the $3$rd and $4$th, so $\frac{9 + x}{2} = 11$. That gives $9 + x = 22$, so $x = 13$.

The halves are $4, 6, 9$ and $13, 15, 18$, so $Q_1 = 6$ and $Q_3 = 15$.

$\text{IQR} = 15 - 6 = 9$

Read each median from the line inside its box: $5$ for North and $6$ for South. Read the quartiles from the box edges.

$6 - 4 = 2, \qquad 10 - 3 = 7$

South's median is higher and its IQR is larger, so South's typical reader reads a little more, but South's hours are far more spread out.

Most of the dots crowd together between $40$ and $50$, which is a cluster. The few low values stretch the data out on the left.

$\text{tail toward } 12 \Rightarrow \text{skewed left}$

The tail points toward the low values, so the data is skewed left.

The $12$ values are sorted. The median is the mean of the $6$th and $7$th values, $\frac{13 + 15}{2} = 14$.

The lower half $5, 7, 8, 10, 12, 13$ has middle pair $8$ and $10$, and the upper half $15, 16, 18, 21, 22, 30$ has middle pair $18$ and $21$.

$Q_1 = \frac{8 + 10}{2} = 9, \qquad Q_3 = \frac{18 + 21}{2} = 19.5$

Adding $5$ to every score keeps the scores in the same order. Each of the median, $Q_1$ and $Q_3$ is a score or the mean of two scores, so each goes up by $5$.

$(Q_3 + 5) - (Q_1 + 5) = Q_3 - Q_1$

The whole picture slides $5$ to the right without stretching, so the center moves and the spread does not.

With $20$ values the lower half is the first $10$. Its median, $Q_1$, is the mean of the $5$th and $6$th values.

$1 + 3 = 4, \qquad 1 + 3 + 8 = 12$

The first two bars hold only the $1$st to $4$th values, so the $5$th and $6$th values, and $Q_1$, are in $20$ to $29$.

A box plot draws only the five-number summary, and the IQR is the width of its box.

$\text{IQR} = Q_3 - Q_1$

It does not show how many values there are, what they add to, or how often any one value repeats.

The mean uses the size of every value, so a few very large values pull it toward the long right tail. The median depends only on position, so it barely moves.

$\text{large values} \Rightarrow \text{mean pulled up}$

The median can never be greater than $Q_3$, whatever the shape.

Seven weights sit between $12$ and $18$, and the $41$ sits far above all of them.

$\text{range} = 41 - 12 = 29, \qquad \text{IQR} = 17.5 - 14.5 = 3$

The range is stretched by the single $41$, while the IQR describes how tightly the other parcels sit together.

With five sorted values the median is the $3$rd and is left out, so the halves are the first two values and the last two. Each quartile is the mean of a pair.

$\frac{1 + 5}{2} = 3, \qquad \frac{8 + 10}{2} = 9$

The list $1, 3, 6, 9, 11$ gives $Q_1 = 2$ and $Q_3 = 10$ by this method. It gives $3$ and $9$ only if the median is kept in both halves.

A time of exactly $20$ fits both $10$ to $20$ and $20$ to $30$, so it could be counted twice. With the labels $10$ to $19$ and $20$ to $29$, every whole number has one interval.

$20 \to 20 \text{ to } 29 \text{ only}$

The intervals sit side by side on a number line, in a fixed order, so the bars should touch.

Shifting a whole box plot $10$ units to the right adds $10$ to each of its five numbers, so the median is $10$ higher.

$(Q_3 + 10) - (Q_1 + 10) = Q_3 - Q_1$

The quartiles move together, so the IQR is unchanged. A box plot shows nothing about how many values there are, so the counts cannot be compared.