Numerical Data Displays and Spread: Additional Practice (Set 1)

The left edge of the box is $Q_1 = 64$, which sits about a quarter of the way along the sorted list. With no repeated scores, roughly a quarter of the class scored below it.

$1 - \tfrac14 = \tfrac34$

So roughly three quarters of the class scored $64$ or more: the box and the right whisker together.

Sort first: $27$, $29$, $31$, $33$, $35$, $38$, $40$, $42$, $45$, $47$, $50$, $55$. With $12$ values the median is the mean of the $6$th and $7$th, $\frac{38 + 40}{2} = 39$.

The lower half is the first six values, with middle pair $31$ and $33$. The upper half is the last six, with middle pair $45$ and $47$.

$Q_1 = \frac{31 + 33}{2} = 32, \qquad Q_3 = \frac{45 + 47}{2} = 46$

So the summary is $27, 32, 39, 46, 55$.

Sort first: $4, 8, 9, 11, 15, 20$. Six values split into a lower half $4, 8, 9$ and an upper half $11, 15, 20$, with nothing left out.

$Q_1 = 8, \qquad Q_3 = 15$

So the IQR is $15 - 8 = 7$. The range would be $20 - 4 = 16$.

The $80$ is still the largest price, so every price keeps its place. The median is still the $4$th value, $5$, and with $7$ values $Q_1$ and $Q_3$ are still the $2$nd and $6$th values, $4$ and $7$.

$\text{IQR} = 7 - 4 = 3 \text{ before and after}$

The range uses the maximum, so it grows from $5$ to $77$. The mean uses every value's size, so it grows too, from $37 \div 7$ to $109 \div 7$.

Values of $10$ or more fill the last three bars.

$9 + 5 + 3 = 17$

Out of all $25$ values, that is $\frac{17}{25}$. The first two bars hold the other $2 + 6 = 8$.

The dots crowd together at the low end and trail off toward the high end, so the long tail points to the right.

$\text{long tail on the right} \Rightarrow \text{skewed right}$

Skew is named for the side the tail stretches toward, not for where the tall stacks are.

With $60$ sorted values, the lower half is the $1$st to $30$th, so $Q_1$ is the mean of the $15$th and $16$th, and the median is the mean of the $30$th and $31$st.

No two values are equal, so the values strictly between $20$ and $35$ are the $16$th to $30$th.

$30 - 16 + 1 = 15$

The numbers $20$ and $35$ are values on the scale, not counts.

Each list is already sorted and has an even count, so split it into two halves. For $1, 3, 5, 7, 9, 11$ the halves are $1, 3, 5$ and $7, 9, 11$.

$Q_3 - Q_1 = 9 - 3 = 6$

The others give $5 - 2 = 3$, then $\frac{12 + 14}{2} - \frac{4 + 6}{2} = 13 - 5 = 8$, and $6 - 6 = 0$.

With seven sorted values, the median is the $4$th value, $Q_1$ is the $2$nd and $Q_3$ is the $6$th, because each half has three values.

For $2$, $4$, $5$, $7$, $8$, $9$, $12$: the $2$nd value is $4$, the $4$th is $7$, and the $6$th is $9$.

$2,\ 4,\ 7,\ 9,\ 12$

The other lists give the summaries $2, 3, 7, 10, 12$, then $2, 4, 9, 11, 12$, then $2, 5, 7, 9, 12$.

With $7$ values, $Q_1$ is the $2$nd value, $3$, and the median, $Q_3$ and maximum are the $4$th, $6$th and $7$th values, all $8$.

$\text{min} = 1,\ Q_1 = 3,\ \text{median} = Q_3 = \text{max} = 8$

So the box runs from $3$ to $8$, the median line sits on its right edge, and the right whisker has zero length. Every value in those collapsed sections is $8$.

A histogram is for numerical data, grouped into intervals on a number line. Tree heights are numbers on a scale.

$\text{heights} \to 100 \text{ to } 109,\ 110 \text{ to } 119,\ \ldots$

Fruits, clubs and colors are categories, which a bar graph shows with separate bars.

With $14$ values, an even count, the lower half is the first $7$ values. Its middle is a single value.

$\frac{7 + 1}{2} = 4$

So $Q_1$ is the $4$th value. The mean of the $7$th and $8$th values is the median of the whole list.