Numerical Data Displays and Spread: Practice

Sort first: $6$, $9$, $9$, $11$, $12$, $14$, $15$, $17$, $18$, $21$, $25$. With $11$ values the median is the $6$th value, $14$, and it is left out of both halves.

The lower half $6, 9, 9, 11, 12$ has middle value $9$, and the upper half $15, 17, 18, 21, 25$ has middle value $18$.

$6,\ 9,\ 14,\ 18,\ 25$

Sort first: $3$, $5$, $6$, $7$, $8$, $10$, $12$, $15$. With an even count, nothing is left out: the lower half is $3, 5, 6, 7$ and the upper half is $8, 10, 12, 15$.

Each half has an even count, so each quartile is the mean of that half's two middle values.

$Q_1 = \frac{5 + 6}{2} = 5.5, \qquad Q_3 = \frac{10 + 12}{2} = 11$

Sort first: $18$, $19$, $22$, $24$, $25$, $27$, $30$, $35$, $41$. The median is the $5$th value, $25$, and it is left out of both halves.

The lower half $18, 19, 22, 24$ gives $Q_1 = \frac{19 + 22}{2} = 20.5$, and the upper half $27, 30, 35, 41$ gives $Q_3 = \frac{30 + 35}{2} = 32.5$.

$\text{IQR} = 32.5 - 20.5 = 12$

The range would be $41 - 18 = 23$, and keeping the $25$ in both halves gives $8$, the other method.

With $20$ values the median is the mean of the $10$th and $11$th values in order. Count up through the bars from the left.

$3 + 6 = 9, \qquad 3 + 6 + 8 = 17$

The first two bars hold the $1$st to $9$th values, and the third bar holds the $10$th to $17$th, so both middle values, and the median, are in $20$ to $29$.

A score of $70$ or more lies in one of the last three intervals, so add those three bars.

$9 + 6 + 4 = 19$

The other $4 + 7 = 11$ students scored below $70$, and $19 + 11 = 30$ checks the total.

With $40$ sorted heights, the median is the mean of the $20$th and $21$st, and the upper half is the $21$st to $40$th heights. Its median, $Q_3$, is the mean of the $30$th and $31$st.

No two heights are equal, so the heights strictly between the median and $Q_3$ are the $21$st to $30$th.

$30 - 21 + 1 = 10$

That is a quarter of the class, whatever the length of that section on the plot.

The $50$ is still the largest value, so every value keeps its place in the sorted list. With $7$ values, $Q_1$ is the $2$nd value and $Q_3$ is the $6$th, which are still $6$ and $10$.

$\text{IQR} = 10 - 6 = 4, \qquad \text{range} = 50 - 4 = 46$

The range uses the maximum itself, so only the range changes.

The medians are the middle numbers, $18$ for Team X and $16$ for Team Y. The IQRs come from the quartiles.

$22 - 14 = 8, \qquad 17 - 15 = 2$

So Team X has both the higher median and the larger IQR. Team Y has the larger range, $18$ against $16$, only because of its maximum of $30$.

With the labels $10$ to $20$ and $20$ to $30$, a value of exactly $20$ would fit both intervals and could be counted twice.

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

The labels $10$ to $19$ and $20$ to $29$ give every whole number one interval, and the bars still touch.

Read the values off the plot in order: $1$, $1$, $1$, $2$, $2$, $2$, $2$, $3$, $3$, $4$, $4$, $6$. With $12$ values the median is the mean of the $6$th and $7$th, which are both $2$.

The upper half is the last six values, $2, 3, 3, 4, 4, 6$, and its two middle values are $3$ and $4$.

$Q_3 = \frac{3 + 4}{2} = 3.5$

With $500$ different values, a dot plot would be a long row of single dots. A histogram groups the times into equal intervals and shows how many fall in each.

$500 \text{ values} \to \text{a few interval counts}$

A circle graph shows parts of a whole, and a line graph shows change over time, so neither suits a list of travel times.

Each whisker runs from an end of the data to a quartile, so each covers about a quarter of the places in the sorted list. A longer whisker spreads its quarter over a wider stretch.

$\text{longer section} \ne \text{more values}$

So the low values are spread out and the high values are packed together. The lengths say nothing about how many values each whisker holds.