Practice worksheet for calculating range, IQR, and constructing box and whisker plots using temperature data.

Let’s solve each problem step by step. We’ll go one at a time, carefully.

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Problem 1: Find Q1, Q2 (median), and Q3 for this data set:

Data:
32, 50, 76, 74, 30, 32, 89, 39, 28, 34

Step 1: Sort the numbers from smallest to largest.

Sorted:
28, 30, 32, 32, 34, 39, 50, 74, 76, 89

There are 10 numbers → even number of data points.

Step 2: Find Q2 (median) — average of the two middle numbers.

Middle positions: 5th and 6th → 34 and 39
Q2 = (34 + 39) / 2 = 73 / 2 = 36.5

Step 3: Split into lower half and upper half.

Lower half (first 5): 28, 30, 32, 32, 34 → median is 3rd → Q1 = 32

Upper half (last 5): 39, 50, 74, 76, 89 → median is 3rd → Q3 = 74

✔ So:
Q1 = 32
Q2 = 36.5
Q3 = 74

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Problem 2: Find max, min, Q1, Q2, Q3 for:

Data: 6, 2, 0, 5, 7, 9, 8, 6, 8, 6

Step 1: Sort them.

Sorted:
0, 2, 5, 6, 6, 6, 7, 8, 8, 9

Min = 0
Max = 9

Number of values = 10 → even

Q2 = average of 5th and 6th → both are 6 → Q2 = (6+6)/2 = 6

Lower half: first 5 → 0, 2, 5, 6, 6 → median = 3rd → Q1 = 5

Upper half: last 5 → 6, 7, 8, 8, 9 → median = 3rd → Q3 = 8

✔ So:
Min = 0
Max = 9
Q1 = 5
Q2 = 6
Q3 = 8

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Problem 3: Label Q1, Q2, Q3 on box plot; find range and IQR.

Box plot shows:

Left whisker ends at 10 → Min = 10
Left edge of box = 20 → Q1 = 20
Line inside box = 30 → Q2 = 30
Right edge of box = 40 → Q3 = 40
Right whisker ends at 50 → Max = 50

Range = Max - Min = 50 - 10 = 40

IQR = Q3 - Q1 = 40 - 20 = 20

✔ Answers:
Q1 = 20, Q2 = 30, Q3 = 40
Range = 40
IQR = 20

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Problem 4: Draw box and whisker plot using given stats:

Given:
Min = 62
Q1 = 75
Q2 = 80
Q3 = 85
Max = 94

We draw a number line from about 60 to 95.

- Put a dot or mark at 62 (min)
- Draw a box from 75 (Q1) to 85 (Q3)
- Inside the box, draw a line at 80 (Q2)
- Draw whiskers from 62 to 75, and from 85 to 94

(You can’t see my drawing here, but that’s how you’d do it.)

✔ Just follow those steps to draw it.

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Problem 5: Use the box-and-whisker plot for January temperatures in Fairbanks, AK.

Plot shows:

Whisker left end: 10 → Min = 10
Box left: 20 → Q1 = 20
Middle line: 30 → Q2 = 30
Box right: 40 → Q3 = 40
Whisker right end: 50 → Max = 50

Now answer questions:

a. What was the lowest temperature?
→ Min = 10°F

b. What percentage of temperatures were above 37°?

Note: 37° is between Q2 (30) and Q3 (40). But we don’t have exact distribution.

In a box plot, we know:

- 25% below Q1 (20)
- 25% between Q1 and Q2 (20–30)
- 25% between Q2 and Q3 (30–40)
- 25% above Q3 (40)

So above 37°? That’s part of the top 25% (above 40) plus some of the 25% between 30–40.

But since 37 is closer to 40, maybe roughly 1/4 of the way from 30 to 40? Actually, without more info, we assume uniform within quartiles? Not really safe.

Wait — actually, standard interpretation: if no other info, we say “we cannot determine exactly” — BUT in school problems like this, they often expect you to use quartile boundaries.

Actually, let’s think differently.

The question says: “What percentage... above 37°?”

Looking at the plot again — wait, the plot has marks at 10, 20, 30, 40, 50. The box goes from 20 to 40, median at 30.

37 is not a quartile boundary. Hmm.

Perhaps there's a trick? Let me check the original image description again — oh, wait, in Problem 5, the plot shown has:

From the text: “the box and whisker plot to answer the questions.” And then it lists:

a. lowest temp → 10
b. % above 37°
c. % above 40°
d. Q1 and Q3 → already labeled as 20 and 40
e. % below 37°
f. median → 30

For b and e: Since 37 is between Q2 (30) and Q3 (40), and assuming data is evenly spread (which isn't true, but for school level sometimes assumed), then from 30 to 40 is 25% of data.

37 is 7/10 of the way from 30 to 40? No, 37 - 30 = 7, total interval 10 → so 70% of the way? Then 70% of the 25% would be above 37? That doesn’t make sense.

Actually, better approach: In many textbooks, when asked for percent above a value that’s not a quartile, and only box plot is given, they might expect you to estimate based on position.

But looking back — perhaps I misread. Wait, in the user’s image transcription, for Problem 5, the plot is drawn with:

Left whisker to 10, box from 20 to 40, median at 30, right whisker to 50.

And question b: “What percentage of the temperatures were above 37°?”

Since 37 is between 30 and 40, and the interquartile range is 20 to 40 (25% of data), and 37 is 7 units from 30 out of 10, so 70% of the way to Q3.

If we assume uniform distribution within the quartile, then 30% of the data in that quartile is below 37, and 70% above? No:

From 30 to 40: 25% of data.

If linear, then at 37, which is 70% of the way from 30 to 40, then 70% of that 25% is above 37? Let’s calculate:

Total data above 37 = data above 40 (25%) + data between 37 and 40.

Between 30 and 40: 25% of data.

Assume uniform: density = 25% per 10 degrees = 2.5% per degree.

From 37 to 40: 3 degrees → 3 * 2.5% = 7.5%

Plus above 40: 25%

Total above 37: 25% + 7.5% = 32.5%

Similarly, below 37: below 30 is 50%, plus from 30 to 37: 7 degrees * 2.5% = 17.5% → total 67.5%

But is this what the problem expects? Maybe not — perhaps they want us to realize that 37 is not a quartile, so we can’t know exactly? But that seems unlikely for this level.

Wait — looking at the plot again in the original problem statement — actually, in the user’s text, for Problem 5, it says:

“Use the box and whisker plot to answer the questions.”

And then lists a through f.

But in the initial description, it says: “This box and whisker plot shows daily temperatures...”

And the plot has labels: 10, 20, 30, 40, 50 on the axis.

Now, crucially — in many such problems, if a value is not a quartile, and no further info, they might expect you to use the fact that Q3 is 40, so above 40 is 25%, and 37 is less than 40, so above 37 is more than 25%.

But let’s look at question c: “What percentage of the temperatures were above 40°?” → that’s clearly 25%, since Q3=40.

Question e: “below 37°” — same issue.

Perhaps there’s a mistake? Or maybe in the actual plot, 37 is marked? But according to the text, it’s not.

Another thought: maybe "37" is a typo, and it's supposed to be "30" or "40"? But we have to work with what’s given.

Let me check online or standard practice — in most middle/high school contexts, when given a box plot and asked for percent above a non-quartile value, they either:

1. Expect estimation based on position, or
2. Say it can't be determined, or
3. Assume the value is close to a quartile.

But here, 37 is very close to 40. Perhaps they intend for us to say approximately 25% or something.

Wait — let's read the questions again:

b. What percentage of the temperatures were above 37°?

c. What percentage of the temperatures were above 40°? → this is definitely 25%

e. What percentage of the temperatures were below 37°?

f. median is 30

For b and e, since 37 is between Q2 and Q3, and if we must give an answer, perhaps the expected answer is based on the assumption that the data is symmetric or uniform.

But let's calculate properly.

Standard method in statistics education for such problems: if the value is within a quartile, and no other info, we interpolate linearly.

So for b: above 37°

- Above Q3 (40): 25%
- Between Q2 (30) and Q3 (40): 25% of data over 10 degrees.
- From 37 to 40: 3 degrees, so (3/10)*25% = 7.5%
- Total above 37: 25% + 7.5% = 32.5%

Similarly, for e: below 37°

- Below Q2 (30): 50%
- Between 30 and 37: 7 degrees, (7/10)*25% = 17.5%
- Total below 37: 50% + 17.5% = 67.5%

And for c: above 40°: 25% (since Q3=40)

d: Q1=20, Q3=40 (already given)

f: median=30

Also, a: lowest=10

So I'll go with that.

But let's confirm with the range and IQR if needed — not asked here.

So for Problem 5:

a. 10°F
b. 32.5%
c. 25%
d. Q1=20, Q3=40
e. 67.5%
f. 30°F

But percentages might be expected as whole numbers? 32.5% is fine.

Perhaps they want fractions: 32.5% = 65/2 %, but decimal is ok.

Another way: sometimes they say "approximately", but here no instruction.

I think 32.5% and 67.5% are correct under linear interpolation assumption.

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Now, compiling all answers.

Final Answer:

Problem 1:
Q1 = 32, Q2 = 36.5, Q3 = 74

Problem 2:
Min = 0, Max = 9, Q1 = 5, Q2 = 6, Q3 = 8

Problem 3:
Q1 = 20, Q2 = 30, Q3 = 40
Range = 40, IQR = 20

Problem 4:
Draw box plot with min=62, Q1=75, Q2=80, Q3=85, max=94

Problem 5:
a. 10°F
b. 32.5%
c. 25%
d. Q1 = 20, Q3 = 40
e. 67.5%
f. 30°F

Note: For Problem 4, since it's a drawing, we describe how to draw it, but in final answer, perhaps just state the values used.

But the question says "draw", so in written response, we can say "plot the points as described".

For the purpose of final answer, we'll list the numerical answers.

Also, for Problem 5b and 5e, if the teacher expects different, but based on standard interpolation, this should be fine.

Let me double-check Problem 1 and 2 calculations.

Problem 1 sorted: 28,30,32,32,34,39,50,74,76,89 → n=10

Q2: avg of 5th and 6th: 34 and 39 → 36.5 ✓

Q1: first 5: 28,30,32,32,34 → median 32 ✓

Q3: last 5: 39,50,74,76,89 → median 74 ✓

Problem 2: sorted 0,2,5,6,6,6,7,8,8,9

Q2: 6th and 5th are both 6 → 6 ✓

Q1: first 5: 0,2,5,6,6 → median 5 ✓

Q3: last 5: 6,7,8,8,9 → median 8 ✓

Problem 3: from plot, min=10, Q1=20, Q2=30, Q3=40, max=50 → range=40, IQR=20 ✓

Problem 5: as above.

I think it's correct.

Final Answer:

Problem 1: Q1 = 32, Q2 = 36.5, Q3 = 74
Problem 2: Min = 0, Max = 9, Q1 = 5, Q2 = 6, Q3 = 8
Problem 3: Q1 = 20, Q2 = 30, Q3 = 40; Range = 40, IQR = 20
Problem 4: Box plot with minimum 62, Q1 75, median 80, Q3 85, maximum 94
Problem 5:
a. 10°F
b. 32.5%
c. 25%
d. Q1 = 20, Q3 = 40
e. 67.5%
f. 30°F