125. the stemplot shows the number of home runs hit by each of the 30 major league baseball teams in a single season. home run totals above what value should be considered outliers?
09 | 15
10 | 37
11 | 4789
12 | 19
13 |
14 |
15 | 89
16 | 84445
17 | 239
18 | 223
19 | 356
20 | 1
21 | 3
22 | 0
2 |
key: 1|48 is a team with 148 home runs.
(a) 173 (b) 210 (c) 222 (d) 229 (e) 257

Question

Accepted Answer

To determine the outlier using the stemplot, we first need to find the quartiles (Q1, Q2, Q3) and then use the interquartile range (IQR) method. Outliers are typically values below $ Q1 - 1.5 \times IQR $ or above $ Q3 + 1.5 \times IQR $.

### Step 1: Order the Data (from the stemplot)
The stemplot gives the number of home runs for 30 teams. Let's list the data:
- Stem 09: 15
- Stem 10: 37, 89
- Stem 11: 19
- Stem 12: (no data)
- Stem 13: (no data)
- Stem 14: 89
- Stem 15: 89, 4445 (wait, correction: stemplot key is "14|8 is 148", so each stem is tens place, leaf is ones. Let's re-express correctly:
  - 09|15: 91, 95? Wait, no—stemplot stem is the leading digits. Wait, the key says "14|8 is a team with 148 home runs", so stem is the first two digits? Wait, no, "14|8" means 14 (stem) and 8 (leaf), so 148. So:
  - 09|15: 91, 95
  - 10|37 89: 103, 107, 108, 109
  - 11|19: 111, 119
  - 14|89: 148, 149
  - 15|89 4445: Wait, no—leaf is single digit? Wait, maybe the stemplot is:
    - 09: 1, 5 (so 91, 95)
    - 10: 3, 7, 8, 9 (103, 107, 108, 109)
    - 11: 1, 9 (111, 119)
    - 14: 8, 9 (148, 149)
    - 15: 8, 9, 4, 4, 4, 5? No, that can't be. Wait, the problem says 30 teams. Let's count the number of leaves:
    - 09: 2 leaves (1,5)
    - 10: 4 leaves (3,7,8,9)
    - 11: 2 leaves (1,9)
    - 12: 0
    - 13: 0
    - 14: 2 leaves (8,9)
    - 15: 6 leaves? Wait, no—maybe the stemplot is:
      - 09: 15 (1 leaf? No, key is 14|8 is 148, so stem is two digits, leaf is one. So:
        - 09|1 5: 91, 95 (2)
        - 10|3 7 8 9: 103, 107, 108, 109 (4)
        - 11|1 9: 111, 119 (2)
        - 12| (0)
        - 13| (0)
        - 14|8 9: 148, 149 (2)
        - 15|8 9 4 4 4 5: Wait, no—maybe 15|8, 15|9, 15|4, 15|4, 15|4, 15|5? No, that's 6. Then 16|2 3 9: 162, 163, 169 (3)
        - 17|2 2 3: 172, 173, 173 (3)
        - 18|3 5 6: 183, 185, 186 (3)
        - 19|1: 191 (1)
        - 20|3: 203 (1)
        - 21|0: 210 (1)
        - 22|2: 222 (1)
        - Wait, let's count total: 2+4+2+2+6+3+3+3+1+1+1+1= 27? No, problem says 30. Maybe I misread the stemplot. Let's check the original:

Original stemplot:
```
09 | 15
10 | 3789
11 | 19
12 |
13 |
14 | 89
15 | 894445
16 | 239
17 | 223
18 | 356
19 | 1
20 | 3
21 | 0
22 | 2
```
Ah, now I see: each stem is two digits, leaf is one digit (for 09, 10, 11) or multiple? Wait, no—"14|89" is two leaves (8,9), "15|894445" is six leaves (8,9,4,4,4,5), "16|239" is three (2,3,9), "17|223" is three (2,2,3), "18|356" is three (3,5,6), "19|1" is one, "20|3" is one, "21|0" is one, "22|2" is one.

Now count:
- 09: 2 (1,5) → 91, 95
- 10: 4 (3,7,8,9) → 103, 107, 108, 109
- 11: 2 (1,9) → 111, 119
- 14: 2 (8,9) → 148, 149
- 15: 6 (8,9,4,4,4,5) → 158, 159, 154, 154, 154, 155
- 16: 3 (2,3,9) → 162, 163, 169
- 17: 3 (2,2,3) → 172, 173, 173
- 18: 3 (3,5,6) → 183, 185, 186
- 19: 1 (1) → 191
- 20: 1 (3) → 203
- 21: 1 (0) → 210
- 22: 1 (2) → 222

Wait, that's 2+4+2+2+6+3+3+3+1+1+1+1 = 27. Missing 3. Maybe 12| and 13| have leaves? No, the stemplot shows 12| and 13| empty. Maybe the stem is single digit? Let's re-interpret: stem is tens place, leaf is ones. So:

- Stem 9: 1,5 → 91, 95 (2)
- Stem 10: 3,7,8,9 → 103, 107, 108, 109 (4)
- Stem 11: 1,9 → 111, 119 (2)
- Stem 12: (0)
- Stem 13: (0)
- Stem 14: 8,9 → 148, 149 (2)
- Stem 15: 8,9,4,4,4,5 → 158, 159, 154, 154, 154, 155 (6)
- Stem 16: 2,3,9 → 162, 163, 169 (3)
- Stem 17: 2,2,3 → 172, 173, 173 (3)
- Stem 18: 3,5,6 → 183, 185, 186 (3)
- Stem 19: 1 → 191 (1)
- Stem 20: 3 → 203 (1)
- Stem 21: 0 → 210 (1)
- Stem 22: 2 → 222 (1)

Still 27. Maybe the original stemplot has more leaves. Alternatively, use the IQR method with the given options. Let's list the data in order:

91, 95, 103, 107, 108, 109, 111, 119, 148, 149, 154, 154, 154, 155, 158, 159, 162, 163, 169, 172, 173, 173, 183, 185, 186, 191, 203, 210, 222, [missing 3 values? Maybe 229, 257, and another?]

Wait, the options are 173, 210, 222, 229, 257. Let's assume the data is ordered, and n=30. So median (Q2) is the average of 15th and 16th terms.

Let's list the data with 30 terms (adding 229, 257, and one more, say 223? No, options include 229, 257). Let's proceed with n=30:

- Position of Q1: $ \frac{n+1}{4} = 7.75 $ (so average of 7th and 8th terms)
- Position of Q3: $ \frac{3(n+1)}{4} = 23.25 $ (average of 23rd and 24th terms)

Let's list the ordered data (correctly, assuming 30 terms):

1. 91
2. 95
3. 103
4. 107
5. 108
6. 109
7. 111
8. 119
9. 148
10. 149
11. 154
12. 154
13. 154
14. 155
15. 158
16. 159
17. 162
18. 163
19. 169
20. 172
21. 173
22. 173
23. 183
24. 185
25. 186
26. 191
27. 203
28. 210
29. 222
30. 229 (or 257? Wait, options have 229 and 257)

Wait, let's recalculate positions:

For n=30 (even), median (Q2) is average of 15th and 16th terms:

15th term: let's count:

1-6: 91,95,103,107,108,109 (6)
7-8: 111,119 (2) → total 8
9-10: 148,149 (2) → total 10
11-14: 154,154,154,155 (4) → total 14
15: 158
16: 159

So Q2 = (158 + 159)/2 = 158.5

Q1: median of first 15 terms (positions 1-15). n=15 (odd), so Q1 is 8th term (since (15+1)/2=8). 8th term is 119.

Q3: median of last 15 terms (positions 16-30). n=15 (odd), so Q3 is 16 + 7 = 23rd term (since (15+1)/2=8, so 16+7=23). 23rd term: let's count from 16th term (159):

16:159, 17:162, 18:163, 19:169, 20:172, 21:173, 22:173, 23:183

So Q3 = 183

IQR = Q3 - Q1 = 183 - 119 = 64

Outlier cutoff: $ Q3 + 1.5 \times IQR = 183 + 1.5 \times 64 = 183 + 96 = 279 $. Wait, that's too high. Maybe my data ordering is wrong.

Alternative approach: Maybe the stem is two digits (tens and hundreds), leaf is ones. So:

- 09|15: 91, 95
- 10|3789: 103, 107, 108, 109
- 11|19: 111, 119
- 14|89: 148, 149
- 15|894445: 158, 159, 154, 154, 154, 155
- 16|239: 162, 163, 169
- 17|223: 172, 173, 173
- 18|356: 183, 185, 186
- 19|1: 191
- 20|3: 203
- 21|0: 210
- 22|2: 222
- 22|9: 229 (missing leaf)
- 25|7: 257 (missing leaf)

Now n=30. Let's order all 30 terms:

1. 91
2. 95
3. 103
4. 107
5. 108
6. 109
7. 111
8. 119
9. 148
10. 149
11. 154
12. 154
13. 154
14. 155
15. 158
16. 159
17. 162
18. 163
19. 169
20. 172
21. 173
22. 173
23. 183
24. 185
25. 186
26. 191
27. 203
28. 210
29. 222
30. 229
31. Wait, no—n=30, so maybe 257 is the 30th term. Let's adjust:

29. 229
30. 257

Now recalculate Q1, Q2, Q3:

- n=30, so Q2 (median) is average of 15th and 16th terms:
  15th term: 158
  16th term: 159
  Q2 = (158 + 159)/2 = 158.5

- Q1: median of first 15 terms (positions 1-15). n=15 (odd), so Q1 is 8th term: 119

- Q3: median of last 15 terms (positions 16-30). n=15 (odd), so Q3 is 16 + 7 = 23rd term:
  Positions

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QUESTION IMAGE

Question

Explanation:

Step 1: Order the Data (from the stemplot)

Answer:

Step 1: Order the Data (from the stemplot)