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Numerical Data Displays and Spread: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Library books on a dot plot

Fourteen students record how many library books they borrowed this month: $3$ , $1$ , $4$ , $2$ , $0$ , $3$ , $3$ , $5$ , $2$ , $1$ , $3$ , $4$ , $2$ , $3$ .

Draw a dot plot of the data. Use it to find the mode, and the number of students who borrowed at least $4$ books.
1. Hint 1
  A dot plot needs a number line that reaches from the smallest value to the largest, with one dot for each student.
2. Hint 2
  Work through the list once, placing a dot above each value as you meet it, and stack repeated values in a column.
3. Hint 3
  Check that your dots add up to $14$ before reading anything off the plot.
Answer

The dot plot has $1, 2, 3, 5, 2, 1$ dots above $0, 1, 2, 3, 4, 5$ . The mode is $3$ books, and $3$ students borrowed at least $4$ books.

Full solution

Draw a number line from $0$ to $5$ , since the values run from $0$ books to $5$ books.

Count each value as you place its dots.
The value $0$ gets $1$ dot, $1$ gets $2$ , $2$ gets $3$ , $3$ gets $5$ , $4$ gets $2$ , and $5$ gets $1$ .

The dots must account for every student.
$1+2+3+5+2+1=14$

The tallest stack is the $5$ dots above $3$ , so the mode is $3$ books.

At least $4$ books means the stacks above $4$ and $5$ .
$2+1=3$

So $3$ students borrowed at least $4$ books.

The fourteen students as a dot plot, one dot per student.

Answer

The dot plot has $1, 2, 3, 5, 2, 1$ dots above $0, 1, 2, 3, 4, 5$ . The mode is $3$ books, and $3$ students borrowed at least $4$ books.

Key idea

A dot plot keeps every value, so counts such as the mode can be read straight from the stacks.
Problem 2 A season of points

A basketball player's points in $11$ games were $18$ , $9$ , $24$ , $15$ , $12$ , $21$ , $14$ , $16$ , $11$ , $19$ , $25$ .

Find the five-number summary and the interquartile range of the points.
1. Hint 1
  The median and the quartiles are found by place, so the order of the list matters first.
2. Hint 2
  With an odd count, the middle value belongs to neither half.
3. Hint 3
  Each half here has five values, so each quartile is the middle value of its half.
Answer

The five-number summary is $9, 12, 16, 21, 25$ , and the IQR is $9$ points.

Full solution

Sort the points from least to greatest: $9$ , $11$ , $12$ , $14$ , $15$ , $16$ , $18$ , $19$ , $21$ , $24$ , $25$ .

There are $11$ values, so the median is the $6$ th value, which is $16$ .

Leave the $16$ out of both halves.
The lower half is $9, 11, 12, 14, 15$ and the upper half is $18, 19, 21, 24, 25$ .

The middle of the lower half is $12$ , so $Q_1=12$ .
The middle of the upper half is $21$ , so $Q_3=21$ .

The smallest value is $9$ and the largest is $25$ , so the five-number summary is $9, 12, 16, 21, 25$ .

The IQR is the distance between the quartiles.
$\text{IQR}=21-12$ $\text{IQR}=9$

Answer

The five-number summary is $9, 12, 16, 21, 25$ , and the IQR is $9$ points.

Key idea

For an odd count, leave the median out and take the middle of each half to find the quartiles.
Problem 3 Eight bean plants

The heights of eight bean plants, in centimeters, are $14$ , $22$ , $9$ , $17$ , $25$ , $12$ , $19$ , $30$ .

Find the five-number summary and the interquartile range of the heights.
1. Hint 1
  Put the heights in order before you look for any middle.
2. Hint 2
  With an even count, no value is left out when you split the list into halves.
3. Hint 3
  Each half has four heights, so each quartile is the mean of two values.
Answer

The five-number summary is $9, 13, 18, 23.5, 30$ , in centimeters, and the IQR is $10.5$ centimeters.

Full solution

Sort the heights.
$9,12,14,17,19,22,25,30$

With $8$ values, the median is the mean of the $4$ th and $5$ th values.
$\text{median}=\frac{17+19}{2}$ $\text{median}=18$

The lower half is $9, 12, 14, 17$ .
Its middle values are $12$ and $14$ , so $Q_1$ is their mean.
$Q_1=\frac{12+14}{2}$ $Q_1=13$

The upper half is $19, 22, 25, 30$ .
Its middle values are $22$ and $25$ .
$Q_3=\frac{22+25}{2}$ $Q_3=23.5$

With the minimum $9$ and the maximum $30$ , the five-number summary is $9, 13, 18, 23.5, 30$ .

Subtract the quartiles to find the IQR.
$\text{IQR}=23.5-13$ $\text{IQR}=10.5$

Answer

The five-number summary is $9, 13, 18, 23.5, 30$ , in centimeters, and the IQR is $10.5$ centimeters.

Key idea

When a half has an even count, its quartile is the mean of its two middle values, just like a median.
Problem 4 Sixteen days of weather

The daily high temperatures for $16$ days, in degrees Fahrenheit, were $58$ , $63$ , $71$ , $66$ , $59$ , $74$ , $68$ , $62$ , $77$ , $65$ , $70$ , $61$ , $69$ , $73$ , $64$ , $67$ .

Make a frequency table using the intervals $55$ to $59$ , $60$ to $64$ , $65$ to $69$ , $70$ to $74$ , and $75$ to $79$ . Describe the histogram you would draw from it, and say which interval holds the median temperature.
1. Hint 1
  Each whole-number temperature belongs to exactly one interval, so tally each one once.
2. Hint 2
  The frequencies should add to the number of days.
3. Hint 3
  With $16$ values the median sits between the $8$ th and $9$ th values. Count up through the intervals from the lowest.
Answer

The frequencies are $2$ , $4$ , $5$ , $4$ , $1$ . The histogram has five touching bars of those heights, in order from $55$ to $59$ up to $75$ to $79$ , and the median is in the interval $65$ to $69$ .

Full solution

Tally each temperature in its interval.
The interval $55$ to $59$ gets $58$ and $59$ .
The interval $60$ to $64$ gets $63$ , $62$ , $61$ , $64$ .

The interval $65$ to $69$ gets $66$ , $68$ , $65$ , $69$ , $67$ .
The interval $70$ to $74$ gets $71$ , $74$ , $70$ , $73$ , and $75$ to $79$ gets only $77$ .

So the frequencies are $2$ , $4$ , $5$ , $4$ , $1$ , and they account for every day.
$2+4+5+4+1=16$

The histogram has one bar per interval with those heights.
The bars touch and stay in number order, because the intervals sit side by side on the temperature scale.

The median is the mean of the $8$ th and $9$ th temperatures in order.
The first two intervals hold $2+4=6$ values, and the first three hold $6+5=11$ .

So the $8$ th and $9$ th values are both in $65$ to $69$ , and so is the median.

Answer

The frequencies are $2$ , $4$ , $5$ , $4$ , $1$ . The histogram has five touching bars of those heights, in order from $55$ to $59$ up to $75$ to $79$ , and the median is in the interval $65$ to $69$ .

Key idea

Counting up through a histogram's bars finds the interval of each middle value; when an even count's two middle values land in different intervals, the histogram cannot tell which interval holds the median.
Problem 5 Nine swimmers

Nine swimmers' times for one length of the pool, in seconds, are $34$ , $41$ , $29$ , $37$ , $45$ , $32$ , $38$ , $52$ , $36$ .

Draw a box plot of the times. Then compare the lengths of its two whiskers and say what the comparison tells you about the fastest and the slowest swimmers.
1. Hint 1
  A box plot is built from the five-number summary, so find those five numbers first.
2. Hint 2
  With nine values, the middle one is left out of both halves, and each half then has four values.
3. Hint 3
  Each whisker runs from a quartile out to the fastest or the slowest time, so compare how wide a stretch each one covers.
Answer

Five-number summary $29, 33, 37, 43, 52$ seconds: a box from $33$ to $43$ with a line at $37$ , and whiskers to $29$ and $52$ . The right whisker spans $9$ seconds and the left only $4$ : the slowest times are more spread out than the fastest.

Full solution

Sort the times.
$29,32,34,36,37,38,41,45,52$

The median is the $5$ th value, $37$ , and it is left out of both halves.

The lower half $29, 32, 34, 36$ gives $Q_1$ as the mean of $32$ and $34$ , which is $33$ .
The upper half $38, 41, 45, 52$ gives $Q_3$ as the mean of $41$ and $45$ , which is $43$ .

So the five-number summary is $29, 33, 37, 43, 52$ .

Draw a number line from about $25$ to $55$ .
Draw a box from $33$ to $43$ with a line across it at $37$ , then whiskers out to $29$ and to $52$ .

The left whisker covers $33-29=4$ seconds and the right whisker covers $52-43=9$ seconds.

The two fastest times, $29$ and $32$ , lie below $Q_1$ , and the two slowest, $45$ and $52$ , lie above $Q_3$ .
The right whisker spans $9$ seconds against the left whisker's $4$ , so the slowest times are more spread out than the fastest.

The nine times as a box plot.

Answer

Five-number summary $29, 33, 37, 43, 52$ seconds: a box from $33$ to $43$ with a line at $37$ , and whiskers to $29$ and $52$ . The right whisker spans $9$ seconds and the left only $4$ : the slowest times are more spread out than the fastest.

Key idea

A longer whisker means the values beyond that quartile are more spread out, not that there are more of them.
Problem 6 Reading a commute box plot

The box plot shows the commute times, in minutes, of $80$ workers. The times were measured to the second, so no two workers have the same time.

How many of the workers commute for at least $15$ minutes? How many commute for between $20$ and $28$ minutes? What is the interquartile range? A worker says that more people commute between $28$ and $45$ minutes than between $15$ and $20$ minutes, because that part of the plot is longer. Is the worker right?

Commute times of 80 workers, in minutes.

Text description of this figure

A box plot on a number line from 5 to 50 minutes, marked every 5. The left whisker runs from 8 to 15. The box runs from 15 to 28, with a line across it at 20. The right whisker runs from 28 to 45. The five values 8, 15, 20, 28 and 45 are labeled above the plot.
1. Hint 1
  Read the five numbers from the plot: the two whisker ends, the two box edges, and the line inside the box.
2. Hint 2
  With $80$ different times, each quartile and the median fall between two neighboring places in the sorted list. Work out which places.
3. Hint 3
  The length of a section measures how spread out its times are, not how many workers it holds.
Answer

$60$ workers commute for at least $15$ minutes, and $20$ commute for between $20$ and $28$ minutes. The IQR is $13$ minutes. The worker is wrong: each of those two stretches holds $20$ workers.

Full solution

The plot shows a minimum of $8$ , $Q_1=15$ , a median of $20$ , $Q_3=28$ , and a maximum of $45$ .

With $80$ sorted times, the lower half is the $1$ st to $40$ th.
So $Q_1$ is the mean of the $20$ th and $21$ st times, the median is the mean of the $40$ th and $41$ st, and $Q_3$ is the mean of the $60$ th and $61$ st.

No two times are equal, so the $20$ th time is below $15$ and the $21$ st is above it.
The workers at $15$ minutes or more are the $21$ st to $80$ th.
$80-20=60$

In the same way, the times between $20$ and $28$ are the $41$ st to $60$ th, which is $20$ workers.

The IQR is the width of the box.
$\text{IQR}=28-15$ $\text{IQR}=13$

The times from $15$ to $20$ are the $21$ st to $40$ th, and the times from $28$ to $45$ are the $61$ st to $80$ th, so each stretch holds $20$ workers.

The right whisker is longer only because those commute times are more spread out, so the worker is wrong.

Answer

$60$ workers commute for at least $15$ minutes, and $20$ commute for between $20$ and $28$ minutes. The IQR is $13$ minutes. The worker is wrong: each of those two stretches holds $20$ workers.

Key idea

When no values repeat, each section of a box plot holds about a quarter of the sorted list (exactly a quarter when the count divides by $4$ ), whatever its length.
Problem 7 The sunflower typo

The heights of seven sunflowers, in centimeters, are $150$ , $162$ , $158$ , $171$ , $165$ , $168$ , $160$ . When the data was typed up, the $171$ was entered as $1710$ by mistake.

Find the range and the interquartile range of the correct heights and of the typed heights. Which measure of spread does the mistake affect, and why?
1. Hint 1
  Sort each list first. Is the mistyped value still in the same place in the sorted order?
2. Hint 2
  With seven values, each half has three values, so each quartile is the middle value of its half.
3. Hint 3
  Think about which values each measure actually uses.
Answer

Correct heights: range $21$ and IQR $10$ centimeters. Typed heights: range $1560$ and IQR still $10$ centimeters. Only the range is affected, since it uses the maximum and the quartiles do not.

Full solution

Sort the correct heights.
$150,158,160,162,165,168,171$

The median is the $4$ th value, $162$ .
The lower half $150, 158, 160$ gives $Q_1=158$ , and the upper half $165, 168, 171$ gives $Q_3=168$ .

So the IQR is $168-158=10$ , and the range is
$171-150=21.$

In the typed list, $1710$ is still the largest value, so every height keeps its place.
The quartiles are still the $2$ nd and $6$ th values, $158$ and $168$ , so the IQR is still $10$ .

The range now uses $1710$ as the maximum.
$\text{range}=1710-150$ $\text{range}=1560$

So the mistake changes the range enormously and leaves the IQR alone.
This works because the list has at least six values, so the quartiles come from the middle of each half and never use the largest value, and because $1710$ stays at the end of the list.

Answer

Correct heights: range $21$ and IQR $10$ centimeters. Typed heights: range $1560$ and IQR still $10$ centimeters. Only the range is affected, since it uses the maximum and the quartiles do not.

Key idea

In a list of six or more values, an end value pushed further out, staying at its end, stretches the range but leaves the quartiles and the IQR unchanged.
Problem 8 Two study groups

Two study groups record their homework times, in minutes, for one evening.

Group P: $20$ , $25$ , $25$ , $30$ , $30$ , $30$ , $35$ , $35$ , $40$ . Group Q: $5$ , $15$ , $20$ , $30$ , $35$ , $45$ , $50$ , $55$ , $60$ .

Compare the two groups by center and by spread, using medians, interquartile ranges and ranges, and write one sentence describing how their evenings differ.
1. Hint 1
  Both lists are already sorted and have nine values each, so each median is a single value.
2. Hint 2
  Leave each median out, and each half then has four values.
3. Hint 3
  Compare one measure at a time: first the centers, then the spreads.
Answer

Group P has median $30$ , IQR $10$ and range $20$ . Group Q has median $35$ , IQR $35$ and range $55$ . Group Q's typical time is a little longer, but its times are far more spread out, while Group P's times are all close to half an hour.

Full solution

Group P's median is its $5$ th value, $30$ .
Its lower half $20, 25, 25, 30$ gives $Q_1=25$ , and its upper half $30, 35, 35, 40$ gives $Q_3=35$ .

So Group P's IQR is $35-25=10$ and its range is $40-20=20$ .

Group Q's median is its $5$ th value, $35$ .
Its lower half $5, 15, 20, 30$ gives $Q_1$ as the mean of $15$ and $20$ , which is $17.5$ .

Its upper half $45, 50, 55, 60$ gives $Q_3$ as the mean of $50$ and $55$ , which is $52.5$ .
$\text{IQR}=52.5-17.5$ $\text{IQR}=35$

Group Q's range is $60-5=55$ .

The centers are close, $30$ against $35$ minutes.
The spreads are not: Group Q's IQR is more than three times Group P's.
Group P's students all worked for about half an hour, while Group Q's times ran from a few minutes to an hour.

Answer

Group P has median $30$ , IQR $10$ and range $20$ . Group Q has median $35$ , IQR $35$ and range $55$ . Group Q's typical time is a little longer, but its times are far more spread out, while Group P's times are all close to half an hour.

Key idea

Two data sets with similar centers can still have very different spreads, so compare both.
Problem 9 The longest whisker

A box plot of the finishing times, in minutes, of $40$ runners in a school fun run has the five-number summary $4, 12, 15, 17, 20$ . The times were measured to the second, so no two runners have the same time.

Jordan says, "The left whisker is by far the longest part of the plot, so most of the runners finished in $4$ to $12$ minutes." Explain what is wrong with Jordan's reasoning, say how many runners finished in $4$ to $12$ minutes, and say what the long whisker does tell you.
1. Hint 1
  Think about what the five numbers of a box plot divide the data into.
2. Hint 2
  With $40$ different times, $Q_1$ falls between two neighboring places in the sorted list. Which places?
3. Hint 3
  If a section holds a quarter of the runners but covers a wide stretch of times, how close together are those times?
Answer

Jordan reads a section's length as its number of values. Only $10$ of the $40$ runners, a quarter, finished in $4$ to $12$ minutes, and the long whisker shows that the fastest quarter is spread out over $8$ minutes.

Full solution

With $40$ sorted times, the lower half is the $1$ st to $20$ th, so $Q_1$ is the mean of the $10$ th and $11$ th times.

No two times are equal, so the $10$ th time is below $12$ and the $11$ th is above it.
The runners who finished in $4$ to $12$ minutes are the $1$ st to $10$ th, which is $10$ runners, a quarter of the field and not most of it.

The left whisker is $12-4=8$ minutes long, while the other three sections cover $15-12=3$ , $17-15=2$ and $20-17=3$ minutes.

So the fastest quarter of the times is spread thinly over $8$ minutes, and the other three quarters are packed between $12$ and $20$ minutes.

Answer

Jordan reads a section's length as its number of values. Only $10$ of the $40$ runners, a quarter, finished in $4$ to $12$ minutes, and the long whisker shows that the fastest quarter is spread out over $8$ minutes.

Key idea

The length of a box plot section shows how spread out its values are, never how many values it holds.
Problem 10 Building a data set

Write a list of seven whole numbers whose five-number summary is $1, 3, 6, 8, 12$ .

Then describe every possible value of the $3$ rd and $5$ th numbers in the sorted list, and explain why those are the only possibilities.
1. Hint 1
  In a sorted list of seven values, work out which place holds the median and which places hold the two quartiles.
2. Hint 2
  With seven values, each half has three values, so each quartile is a single value in the list.
3. Hint 3
  The places that are not fixed by the summary still have to fit between their neighbors in the sorted order.
Answer

One list that works is $1$ , $3$ , $4$ , $6$ , $7$ , $8$ , $12$ . Any list of the form $1, 3, a, 6, b, 8, 12$ works, where the $3$ rd number $a$ is a whole number from $3$ to $6$ and the $5$ th number $b$ is a whole number from $6$ to $8$ .

Full solution

In a sorted list of seven values, the median is the $4$ th value.
The lower half is the $1$ st to $3$ rd values, with middle value the $2$ nd, and the upper half is the $5$ th to $7$ th, with middle value the $6$ th.

So the summary fixes five places: the $1$ st value is $1$ , the $2$ nd is $3$ , the $4$ th is $6$ , the $6$ th is $8$ , and the $7$ th is $12$ .

The list must stay sorted, so the $3$ rd value sits between the $2$ nd and the $4$ th.
It can be any whole number from $3$ to $6$ .

In the same way, the $5$ th value sits between the $4$ th and the $6$ th, so it can be any whole number from $6$ to $8$ .

Choosing $4$ and $7$ gives $1$ , $3$ , $4$ , $6$ , $7$ , $8$ , $12$ .
Checking it, the median is $6$ , the lower half $1, 3, 4$ has middle value $3$ , and the upper half $7, 8, 12$ has middle value $8$ , as required.

Answer

One list that works is $1$ , $3$ , $4$ , $6$ , $7$ , $8$ , $12$ . Any list of the form $1, 3, a, 6, b, 8, 12$ works, where the $3$ rd number $a$ is a whole number from $3$ to $6$ and the $5$ th number $b$ is a whole number from $6$ to $8$ .

Key idea

For a sorted list of seven values, a five-number summary fixes only five of the seven places, so many different lists share the same box plot.

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