Numerical Data Displays and Spread: Practice

There are $7$ values, so the median is the $4$th, $8$. The lower half is $2, 4, 6$ and the upper half is $10, 12, k$.

Since $k$ is at least $12$, it is the last value of the upper half, and the middle value of that half is $12$.

$\text{IQR} = 12 - 4 = 8$

The value of $k$ affects the range, $k - 2$, but not the quartiles.

With $9$ values the median is the $5$th, $5$. The lower half $1, 5, 5, 5$ gives $Q_1 = 5$, and the upper half $5, 5, 5, 12$ gives $Q_3 = 5$.

$\text{IQR} = 5 - 5 = 0, \qquad \text{range} = 12 - 1 = 11$

An IQR of $0$ means the middle of the data is piled on one value. It does not mean every value is equal: the $1$ and the $12$ still differ.

Compare the four sections. The left whisker runs from $10$ to $30$, and the right whisker from $38$ to $40$.

$30 - 10 = 20, \qquad 40 - 38 = 2$

The median, $36$, also sits near the right end of the box. The values crowd together at the high end and stretch far to the left, so the data is skewed left.

With $23$ values the median is the $12$th, and it is left out of both halves. So the lower half is the $1$st to $11$th values.

The lower half has $11$ values, an odd count, so its middle is a single value.

$\frac{11 + 1}{2} = 6$

So $Q_1$ is the $6$th value. Keeping the median in the lower half would give $12$ values and the mean of the $6$th and $7$th, which is the other method.

Adding a value changes the count, so the halves change too. The new sorted list $2, 4, 6, 8, 10, 20$ has an even count, so its halves are $2, 4, 6$ and $8, 10, 20$.

$Q_1 = 4, \qquad Q_3 = 10$

Moving an end value of a list of six or more values leaves its quartiles alone, as long as it stays at its end. Adding a whole new value is a different change, and here both quartiles moved.

The medians are both $63$ seconds, so their typical laps match. Compare the spreads of the middle halves.

$64 - 62 = 2, \qquad 68 - 58 = 10$

Ana's IQR is $2$ seconds and Ben's is $10$, so Ana's laps are much more consistent. One fast or slow lap does not decide who is faster.

Each bar's height is a count, so adding the heights gives the number of values exactly.

$\text{total} = \text{sum of the bar heights}$

The histogram does not show the values inside each interval, so the median, the largest value and the mean can only be placed in an interval or estimated.

Sort each age into its interval: $19$ in the first; $23, 28, 26, 21$ in the second; $35, 37, 30, 33, 39$ in the third; $41, 44, 48$ in the fourth; and $52, 57$ in the last.

$1 + 4 + 5 + 3 + 2 = 15$

The frequencies add to the $15$ people, so no age was missed or counted twice.

Each set has $9$ values, so the median is the $5$th value and the halves are the four values on each side.

For Set P, $Q_1 = \frac{2 + 5}{2} = 3.5$ and $Q_3 = \frac{5 + 8}{2} = 6.5$. For Set Q, $Q_1 = \frac{2 + 3}{2} = 2.5$ and $Q_3 = \frac{7 + 8}{2} = 7.5$.

$6.5 - 3.5 = 3, \qquad 7.5 - 2.5 = 5$

The same range can hide very different middles: Set P's middle values are packed at $5$.

Read the values in order: $3$, $4$, $4$, $5$, $5$, $5$, $6$, $7$, $7$, $9$, $12$. With $11$ values the median is the $6$th value, $5$, and it is left out of both halves.

The lower half $3, 4, 4, 5, 5$ has middle value $4$, and the upper half $6, 7, 7, 9, 12$ has middle value $7$.

$3,\ 4,\ 5,\ 7,\ 12$

The box always runs from $Q_1$ to $Q_3$, so its width is always their difference.

$\text{box width} = Q_3 - Q_1 = \text{IQR}$

The median can sit anywhere inside the box, and the whiskers can differ. The box need not hold exactly half either: for $1, 2, 3, \ldots, 11$, $Q_1 = 3$ and $Q_3 = 9$, so the box holds $7$ of the $11$ values.

The data is already sorted, with $7$ values, so the median is the $4$th value, $10$. Kai's lower half, $3, 7, 8, 10$, included that median.

Leaving the median out, the lower half is $3, 7, 8$.

$Q_1 = \text{median of } 3, 7, 8 = 7$