QUESTION IMAGE
Question
- 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 | 3789
11 | 47
12 | 19
13 |
14 | 89
15 | 34445
16 | 239
17 | 235
18 | 326
19 | 1
20 | 3
21 | 0
22 | 2
key: 14|8 is a team with 148 home runs.
(a) 173 (b) 210 (c) 222 (d) 229 (e) 257
- which of the following boxplots best matches the distribution shown in the histogram?
(a) data
(b) data
(c) data
(d) data
(e) data
Question 125
Step 1: Recall the outlier formula
To find outliers, we use the interquartile range (IQR) method. The formula for outliers is: values above \( Q_3 + 1.5 \times \text{IQR} \) or below \( Q_1 - 1.5 \times \text{IQR} \) are outliers. First, we need to find \( Q_1 \), \( Q_3 \), and IQR from the stemplot.
Step 2: Organize the data from the stemplot
The stemplot data (home runs) for 30 teams:
- 09: 0,9 (wait, no, key is 14|8 is 148, so stems are tens place, leaves are ones. Wait, let's correct:
Stem 09: 09? Wait, no, stem 10: 3,7,8,9 (so 103, 107, 108, 109)
Stem 11: 4,7 (114, 117)
Stem 12: 1,9 (121, 129)
Stem 13: (no leaves? Wait, original stemplot:
09: 15? Wait, maybe I misread. Let's re-express the stemplot correctly with key 14|8 = 148. So stem is tens (and maybe hundreds? Wait, 14|8 is 148, so stem is 14 (tens and hundreds? Wait, 148: 1 (hundreds), 4 (tens), 8 (ones). Wait, maybe stem is the first two digits? Wait, 09: 15? No, maybe the stemplot is:
Stem 09: 1,5? No, the key is 14|8 is 148, so stem is 14 (i.e., 140s), leaf 8 is 8, so 148. So let's list all data points:
Stem 09: Wait, maybe the stem is the tens digit? No, 14|8 is 148, so stem is 14 (two digits), leaf is 8. So:
Stem 09: 1,5? Wait, the stemplot as given:
09 | 15
10 | 3789
11 | 47
12 | 19
13 | (empty)
14 | 89
15 | 34445
16 | 239
17 | 235
18 | 326
19 | 1
20 | 3
21 | 0
22 | 2
Wait, maybe the stem is the tens digit, and for stem 09, it's 9 (tens) with leaves 1,5? No, key is 14|8 is 148, so stem is 14 (i.e., 140-149), leaf 8 is 8, so 148. So:
Stem 09: 9 (tens) with leaves 1,5? No, 09 would be 9 (tens) but 09 is 90s? Wait, maybe the stem is the number of tens, so stem 9: 90s, stem 10: 100s, etc. But key is 14|8 is 148, so stem 14: 140s, leaf 8: 8, so 148. So let's list all data:
Stem 09: 91, 95 (wait, 09|15: 91, 95)
Stem 10: 103, 107, 108, 109
Stem 11: 114, 117
Stem 12: 121, 129
Stem 13: (no data)
Stem 14: 148, 149
Stem 15: 153, 154, 154, 154, 155
Stem 16: 162, 163, 169
Stem 17: 172, 173, 175
Stem 18: 183, 182, 186 (wait, 18|326: 183, 182, 186? Wait, leaf order? Maybe 18|3 2 6: 183, 182, 186? Or 18|326: 183, 182, 186? Wait, maybe leaves are in order, so 18|3 2 6: 183, 182, 186 (but usually leaves are in ascending order, so maybe 18|2 3 6: 182, 183, 186). Let's assume leaves are in ascending order.
Stem 19: 191
Stem 20: 203
Stem 21: 210
Stem 22: 222
Now, we have 30 data points. Let's list them in order:
91, 95,
103, 107, 108, 109,
114, 117,
121, 129,
148, 149,
153, 154, 154, 154, 155,
162, 163, 169,
172, 173, 175,
182, 183, 186,
191,
203,
210,
222
Wait, let's count:
Stem 09: 2
Stem 10: 4 (total 6)
Stem 11: 2 (8)
Stem 12: 2 (10)
Stem 13: 0 (10)
Stem 14: 2 (12)
Stem 15: 5 (17)
Stem 16: 3 (20)
Stem 17: 3 (23)
Stem 18: 3 (26)
Stem 19: 1 (27)
Stem 20: 1 (28)
Stem 21: 1 (29)
Stem 22: 1 (30). Yes, 30 data points.
Now, find \( Q_1 \) (25th percentile), \( Q_2 \) (median, 50th percentile), \( Q_3 \) (75th percentile).
For 30 data points, the median (Q2) is the average of the 15th and 16th terms.
Let's list the ordered data with indices (1 to 30):
1: 91
2: 95
3: 103
4: 107
5: 108
6: 109
7: 114
8: 117
9: 121
10: 129
11: 148
12: 149
13: 153
14: 154
15: 154
16: 154
17: 155
18: 162
19: 163
20: 169
21: 172
22: 173
23: 175
24: 182
25: 183
26: 186
27: 191
28: 203
29: 210
30: 222
Median (Q2): (15th + 16th)/2 = (154 + 154)/2 = 154.
Q1: 25th percentile, which is the median of the first 15 data points (indices 1-15).
First 15 data points:
1:91, 2:95, 3:103, 4:107, 5:108, 6:109, 7:114, 8:117, 9:…
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Step 1: Recall the outlier formula
To find outliers, we use the interquartile range (IQR) method. The formula for outliers is: values above \( Q_3 + 1.5 \times \text{IQR} \) or below \( Q_1 - 1.5 \times \text{IQR} \) are outliers. First, we need to find \( Q_1 \), \( Q_3 \), and IQR from the stemplot.
Step 2: Organize the data from the stemplot
The stemplot data (home runs) for 30 teams:
- 09: 0,9 (wait, no, key is 14|8 is 148, so stems are tens place, leaves are ones. Wait, let's correct:
Stem 09: 09? Wait, no, stem 10: 3,7,8,9 (so 103, 107, 108, 109)
Stem 11: 4,7 (114, 117)
Stem 12: 1,9 (121, 129)
Stem 13: (no leaves? Wait, original stemplot:
09: 15? Wait, maybe I misread. Let's re-express the stemplot correctly with key 14|8 = 148. So stem is tens (and maybe hundreds? Wait, 14|8 is 148, so stem is 14 (tens and hundreds? Wait, 148: 1 (hundreds), 4 (tens), 8 (ones). Wait, maybe stem is the first two digits? Wait, 09: 15? No, maybe the stemplot is:
Stem 09: 1,5? No, the key is 14|8 is 148, so stem is 14 (i.e., 140s), leaf 8 is 8, so 148. So let's list all data points:
Stem 09: Wait, maybe the stem is the tens digit? No, 14|8 is 148, so stem is 14 (two digits), leaf is 8. So:
Stem 09: 1,5? Wait, the stemplot as given:
09 | 15
10 | 3789
11 | 47
12 | 19
13 | (empty)
14 | 89
15 | 34445
16 | 239
17 | 235
18 | 326
19 | 1
20 | 3
21 | 0
22 | 2
Wait, maybe the stem is the tens digit, and for stem 09, it's 9 (tens) with leaves 1,5? No, key is 14|8 is 148, so stem is 14 (i.e., 140-149), leaf 8 is 8, so 148. So:
Stem 09: 9 (tens) with leaves 1,5? No, 09 would be 9 (tens) but 09 is 90s? Wait, maybe the stem is the number of tens, so stem 9: 90s, stem 10: 100s, etc. But key is 14|8 is 148, so stem 14: 140s, leaf 8: 8, so 148. So let's list all data:
Stem 09: 91, 95 (wait, 09|15: 91, 95)
Stem 10: 103, 107, 108, 109
Stem 11: 114, 117
Stem 12: 121, 129
Stem 13: (no data)
Stem 14: 148, 149
Stem 15: 153, 154, 154, 154, 155
Stem 16: 162, 163, 169
Stem 17: 172, 173, 175
Stem 18: 183, 182, 186 (wait, 18|326: 183, 182, 186? Wait, leaf order? Maybe 18|3 2 6: 183, 182, 186? Or 18|326: 183, 182, 186? Wait, maybe leaves are in order, so 18|3 2 6: 183, 182, 186 (but usually leaves are in ascending order, so maybe 18|2 3 6: 182, 183, 186). Let's assume leaves are in ascending order.
Stem 19: 191
Stem 20: 203
Stem 21: 210
Stem 22: 222
Now, we have 30 data points. Let's list them in order:
91, 95,
103, 107, 108, 109,
114, 117,
121, 129,
148, 149,
153, 154, 154, 154, 155,
162, 163, 169,
172, 173, 175,
182, 183, 186,
191,
203,
210,
222
Wait, let's count:
Stem 09: 2
Stem 10: 4 (total 6)
Stem 11: 2 (8)
Stem 12: 2 (10)
Stem 13: 0 (10)
Stem 14: 2 (12)
Stem 15: 5 (17)
Stem 16: 3 (20)
Stem 17: 3 (23)
Stem 18: 3 (26)
Stem 19: 1 (27)
Stem 20: 1 (28)
Stem 21: 1 (29)
Stem 22: 1 (30). Yes, 30 data points.
Now, find \( Q_1 \) (25th percentile), \( Q_2 \) (median, 50th percentile), \( Q_3 \) (75th percentile).
For 30 data points, the median (Q2) is the average of the 15th and 16th terms.
Let's list the ordered data with indices (1 to 30):
1: 91
2: 95
3: 103
4: 107
5: 108
6: 109
7: 114
8: 117
9: 121
10: 129
11: 148
12: 149
13: 153
14: 154
15: 154
16: 154
17: 155
18: 162
19: 163
20: 169
21: 172
22: 173
23: 175
24: 182
25: 183
26: 186
27: 191
28: 203
29: 210
30: 222
Median (Q2): (15th + 16th)/2 = (154 + 154)/2 = 154.
Q1: 25th percentile, which is the median of the first 15 data points (indices 1-15).
First 15 data points:
1:91, 2:95, 3:103, 4:107, 5:108, 6:109, 7:114, 8:117, 9:121, 10:129, 11:148, 12:149, 13:153, 14:154, 15:154
Median of these 15 (8th term, since (15+1)/2=8): 8th term is 117. So Q1 = 117.
Q3: 75th percentile, median of the last 15 data points (indices 16-30).
Last 15 data points:
16:154, 17:155, 18:162, 19:163, 20:169, 21:172, 22:173, 23:175, 24:182, 25:183, 26:186, 27:191, 28:203, 29:210, 30:222
Median of these 15 (8th term, (15+1)/2=8): 8th term is 175. So Q3 = 175.
IQR = Q3 - Q1 = 175 - 117 = 58.
Now, upper outlier boundary: Q3 + 1.5IQR = 175 + 1.558 = 175 + 87 = 262? Wait, that can't be, since the maximum is 222. Wait, maybe I made a mistake in data ordering.
Wait, maybe the stem 09 is 9 (tens) with leaves 1 and 5, so 91 and 95. Then stem 10: 103,107,108,109 (correct). Stem 11:114,117 (correct). Stem 12:121,129 (correct). Stem 13: empty. Stem 14:148,149 (correct). Stem 15:153,154,154,154,155 (correct). Stem 16:162,163,169 (correct). Stem 17:172,173,175 (correct). Stem 18:183,182,186? Wait, maybe the leaves are 3,2,6, but in ascending order, it's 2,3,6, so 182,183,186 (correct, indices 24,25,26). Stem 19:191 (27), stem 20:203 (28), stem 21:210 (29), stem 22:222 (30). So data is correct.
Wait, maybe the stem 09 is 09 (i.e., 9) with leaves 1 and 5, so 91 and 95. Then the first 15 data points:
1:91, 2:95, 3:103, 4:107, 5:108, 6:109, 7:114, 8:117, 9:121, 10:129, 11:148, 12:149, 13:153, 14:154, 15:154. So Q1 is the 8th term: 117. Correct.
Q3: last 15 data points (indices 16-30):
16:154, 17:155, 18:162, 19:163, 20:169, 21:172, 22:173, 23:175, 24:182, 25:183, 26:186, 27:191, 28:203, 29:210, 30:222. So 8th term is 175. Correct.
IQR = 175 - 117 = 58. Then upper outlier: 175 + 1.5*58 = 175 + 87 = 262. But the options are 173,210,222,229,257. None is 262. So I must have misinterpreted the stemplot.
Wait, maybe the stem is the tens digit, and for stem 09, it's 9 (tens) with leaves 1 and 5, but maybe the stem is 0 (tens) with leaves 9,1,5? No, key is 14|8=148, so stem is 14 (two digits), leaf 8. So stem 09: 09 (i.e., 9) with leaves 1 and 5: 91,95. Stem 10:10 with leaves 3,7,8,9:103,107,108,109. Stem 11:11 with leaves 4,7:114,117. Stem 12:12 with leaves 1,9:121,129. Stem 13:13 with no leaves. Stem 14:14 with leaves 8,9:148,149. Stem 15:15 with leaves 3,4,4,4,5:153,154,154,154,155. Stem 16:16 with leaves 2,3,9:162,163,169. Stem 17:17 with leaves 2,3,5:172,173,175. Stem 18:18 with leaves 3,2,6:183,182,186 (but sorted:182,183,186). Stem 19:19 with leaf 1:191. Stem 20:20 with leaf 3:203. Stem 21:21 with leaf 0:210. Stem 22:22 with leaf 2:222.
Wait, maybe the stem is the number of home runs in tens, so stem 9: 90-99, stem 10:100-109, etc. Then the data is:
91,95,
103,107,108,109,
114,117,
121,129,
148,149,
153,154,154,154,155,
162,163,169,
172,173,175,
182,183,186,
191,
203,
210,
222.
Now, let's recalculate Q1, Q2, Q3.
n=30, so:
Q1 is at position (30+1)0.25 = 7.75, so 7th + 0.75(8th -7th)
7th term:114, 8th term:117. So Q1 = 114 + 0.75*(117-114) = 114 + 2.25 = 116.25 ≈116 (but maybe using integer positions, since 30 is even, Q1 is median of first 15, which is (7th +8th)/2? Wait, no, for n=30, the first quartile is the median of the first 15 data points (positions 1-15). The first 15 data points:
1:91,2:95,3:103,4:107,5:108,6:109,7:114,8:117,9:121,10:129,11:148,12:149,13:153,14:154,15:1