Stem and Leaf Plot Worksheets - Free Printable

Let’s solve this step by step.

We are given these average ages of players in Major League Baseball teams:

30.6, 29.5, 28.8, 28.2, 29.2, 28.6, 28.4, 28.7, 29.6, 29.7, 28.9, 28.6

First, let’s list them in order from smallest to largest to make it easier:

28.2
28.4
28.6
28.6
28.7
28.8
28.9
29.2
29.5
29.6
29.7
30.6

There are 12 numbers total → so 12 teams participated.

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Now, let’s answer each question one by one.

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Question 1: What is the highest average age among the teams?

Look at the sorted list — the last number is 30.6

✔ Answer: 30.6

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Question 2: How many teams participated in the Major League Baseball?

Count the numbers in the original list or sorted list — there are 12

✔ Answer: 12

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Question 3: What is the median of the given data?

Median = middle value(s) when data is ordered.

Since we have 12 numbers (even), the median is the average of the 6th and 7th numbers.

Sorted list again:

1st: 28.2
2nd: 28.4
3rd: 28.6
4th: 28.6
5th: 28.7
6th: 28.8 ←
7th: 28.9 ←
8th: 29.2
9th: 29.5
10th: 29.6
11th: 29.7
12th: 30.6

So, 6th = 28.8, 7th = 28.9

Median = (28.8 + 28.9) ÷ 2 = 57.7 ÷ 2 = 28.85

✔ Answer: 28.85

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Question 4: What is the average age of the players participating in the Major League Baseball?

This means: find the mean (average) of all 12 numbers.

Add them up:

Let’s add carefully:

Start with:

28.2 + 28.4 = 56.6
+28.6 = 85.2
+28.6 = 113.8
+28.7 = 142.5
+28.8 = 171.3
+28.9 = 200.2
+29.2 = 229.4
+29.5 = 258.9
+29.6 = 288.5
+29.7 = 318.2
+30.6 = 348.8

Total sum = 348.8

Number of teams = 12

Mean = 348.8 ÷ 12

Let’s divide:

12 × 29 = 348 → so 348.8 ÷ 12 = 29 + (0.8 ÷ 12) = 29 + 0.0666... ≈ 29.0667

But let’s do exact division:

348.8 ÷ 12 = ?

Multiply numerator and denominator by 10 to eliminate decimal:

3488 ÷ 120

Divide:

120 × 29 = 3480 → remainder 8

So 29 + 8/120 = 29 + 1/15 ≈ 29.0667

But since the original data has one decimal place, we can round to two decimals if needed.

Actually, let’s check our addition again to be sure.

List again:

28.2
28.4 → 56.6
28.6 → 85.2
28.6 → 113.8
28.7 → 142.5
28.8 → 171.3
28.9 → 200.2
29.2 → 229.4
29.5 → 258.9
29.6 → 288.5
29.7 → 318.2
30.6 → 348.8 ✔ Correct.

348.8 ÷ 12:

Do long division:

12 into 34 → 2 (24), remainder 10
Bring down 8 → 108
12 into 108 → 9 → 108, remainder 0
Bring down .8 → 8 tenths
12 into 8.0 → 0.666...

Wait — better way:

348.8 ÷ 12 = (348 ÷ 12) + (0.8 ÷ 12) = 29 + 0.0666... = 29.0666...

Rounded to two decimal places: 29.07

But maybe they want exact fraction? Or keep as is?

Looking at the data — all values are to one decimal place. So perhaps report mean to two decimals.

Alternatively, let's compute exactly:

348.8 / 12 = 3488 / 120 = simplify:

Divide numerator and denominator by 8: 436 / 15 = 29.0666...

So, 29.07 if rounded to two decimals.

But let me double-check using another method.

Group them:

(28.2 + 30.6) = 58.8
(28.4 + 29.7) = 58.1
(28.6 + 29.6) = 58.2
(28.6 + 29.5) = 58.1
(28.7 + 29.2) = 57.9
(28.8 + 28.9) = 57.7

Now add those:

58.8 + 58.1 = 116.9
+58.2 = 175.1
+58.1 = 233.2
+57.9 = 291.1
+57.7 = 348.8 ✔ Same total.

So yes, 348.8 ÷ 12 = 29.0666...

In school problems like this, often they expect you to round to nearest hundredth.

So 29.07

But wait — let’s see what the key says? The image shows “Key: 28|1 = ___” which suggests stem-and-leaf plot uses tens and units, but for mean, we’ll go with calculation.

Actually, looking back — the problem says “What is the average age...” — so mean.

I think 29.07 is fine, but let’s see if it divides evenly.

348.8 ÷ 12 = ?

12 × 29.0666... = 348.8 exactly? Let’s test:

12 × 29 = 348
12 × 0.0666... = 0.8 → yes, because 0.0666... = 1/15, and 12 × 1/15 = 12/15 = 0.8

So 29 + 1/15 = 29.0666...

As a decimal, it’s repeating. But for practical purposes, we can write 29.07

However, sometimes in such contexts, they might accept 29.1 if rounding to one decimal, but since inputs are to one decimal, output to two is standard.

Wait — let me check online or standard practice.

Actually, no need — I’ll go with 29.07

But let me calculate numerically:

348.8 ÷ 12:

12 × 29.0 = 348.0 → subtract → 0.8 left

0.8 ÷ 12 = 0.0666...

So 29.0666... → to two decimals: look at third decimal is 6 ≥ 5 → round up second decimal: 29.07

✔ So 29.07

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Question 5: How many teams have the average age between 28 to 29?

Note: “between 28 to 29” — does this include 28 and 29? Usually in math problems, “between A and B” may or may not include endpoints. But here, since ages are like 28.2, etc., and 29.0 isn’t present, let’s interpret as greater than or equal to 28 and less than 29? Or including 29?

Look at the data:

All values are from 28.2 to 30.6

“Between 28 to 29” — likely means 28 ≤ age < 29, because if it included 29, then 29.2, 29.5, etc. would be over.

But let’s read carefully: “between 28 to 29”

In common usage, especially in stats, “between 28 and 29” often includes 28 but excludes 29, unless specified.

But let’s list all values that are ≥28 and <29:

From sorted list:

28.2, 28.4, 28.6, 28.6, 28.7, 28.8, 28.9 → that’s 7 values

Next is 29.2 which is >29

What about 29.0? Not in list.

Is 29.0 included? No team has exactly 29.0.

So if “between 28 to 29” means [28, 29), then 28.2 to 28.9 inclusive → 7 teams.

If it means [28, 29], then still same, since no 29.0.

But 29.2 is above 29, so not included.

Wait — 29.0 is not in the list, so whether inclusive or exclusive of 29 doesn't matter.

But what about 28.0? All are above 28.

So values from 28.2 to 28.9 are all between 28 and 29.

Count: positions 1 to 7 in sorted list:

1. 28.2
2. 28.4
3. 28.6
4. 28.6
5. 28.7
6. 28.8
7. 28.9 → that’s 7

Then 8. 29.2 → which is greater than 29, so not included.

So 7 teams

But let me confirm: is 29.0 considered "between 28 and 29"? Technically, 29.0 is not less than 29, so if strict, no.

But since no team has 29.0, it doesn't affect.

Another interpretation: sometimes "between 28 and 29" could mean 28 < x < 29, excluding endpoints.

But 28.2 > 28, and 28.9 < 29, so still 7.

If it included 29, then 29.2 is still out.

So regardless, only up to 28.9.

Thus, 7 teams

But wait — let's list all original data and count how many are >=28 and <29:

Original: 30.6 (no), 29.5 (no), 28.8 (yes), 28.2 (yes), 29.2 (no), 28.6 (yes), 28.4 (yes), 28.7 (yes), 29.6 (no), 29.7 (no), 28.9 (yes), 28.6 (yes)

So:

28.8 - yes
28.2 - yes
28.6 - yes
28.4 - yes
28.7 - yes
28.9 - yes
28.6 - yes

That’s seven: 28.2, 28.4, 28.6, 28.6, 28.7, 28.8, 28.9

Yes, 7 teams.

✔ Answer: 7

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Now, just to be thorough, the stem-and-leaf plot part — but the questions don’t ask us to draw it, only to use the data to answer questions. So we’re good.

Also, the key says “28|1 = ___” which probably means 28.1, but since we don’t have 28.1, it’s just an example.

So final answers:

1. Highest average age: 30.6
2. Number of teams: 12
3. Median: 28.85
4. Mean (average age): 29.07
5. Teams with avg age between 28 and 29: 7

For question 4, some might argue to leave as fraction or more precise, but 29.07 is appropriate.

Alternatively, since 348.8 / 12 = 29.0666..., and if we keep one decimal, it would be 29.1, but I think two decimals is better.

Let me see: the data has one decimal, so mean can have two.

Standard is to report mean with one more decimal than data, so yes, 29.07.

Some textbooks might say 29.1, but I'll stick with 29.07.

Actually, let's calculate exactly:

348.8 ÷ 12 = 3488 ÷ 120 = 436 ÷ 15 = 29 1/15 ≈ 29.0667 → rounds to 29.07

Yes.

Final Answer:

1. 30.6
2. 12
3. 28.85
4. 29.07
5. 7

But for the boxed answer, since it's multiple parts, I should present all.

The instruction says: "Provide only the final correct answer" — but since there are five questions, I need to provide all five.

Probably list them as per the questions.

So:

Final Answer:
1. 30.6
2. 12
3. 28.85
4. 29.07
5. 7

To match format, perhaps write clearly.

Since it's a single response, I'll box the entire set.

But the rule says "only the final correct answer", implying after reasoning.

I think for multi-part, we list them.

Perhaps:

\boxed{30.6} for Q1, but better to specify.

Looking at the user's request: "Final Answer: [Provide only the final correct answer]"

And since there are five answers, I should provide all five in order.

So:

Final Answer:
30.6, 12, 28.85, 29.07, 7

To be precise, let's write:

The answers are:
1. 30.6
2. 12
3. 28.85
4. 29.07
5. 7

But for boxing, perhaps:

\boxed{30.6}, \boxed{12}, \boxed{28.85}, \boxed{29.07}, \boxed{7}

Since it's text, I'll write:

Final Answer:
1. 30.6
2. 12
3. 28.85
4. 29.07
5. 7

I think that's clear.

One last check on Q4: is there a possibility they want the mean without rounding?

But 29.0666... is not nice, so 29.07 is fine.

Some might calculate as 348.8 / 12 = 29.0667, and round to 29.07.

Yes.

Q3: median is exactly 28.85, since (28.8 + 28.9)/2 = 57.7/2 = 28.85

Perfect.

Q5: 7 teams — confirmed.

So all good.