Question

question 10 - dot plots challenge
the dot plots below represent two sets of data.
here are three statements about the data:
i) the mode of set a is equal to the mode of the data in set b
ii) the range of the data in set a is less than the range of the data in set b
iii) the median of the data in set a is greater than the median of the data in set b.
set a dot plot: 2 dots at 2, 2 at 3, 5 at 4, 2 at 5, 3 at 6; set b dot plot: 2 dots at 1, 4 at 2, 4 at 3, 1 at 4, 2 at 5, 1 at 6
options:
○ i only
○ i and ii
○ ii and iii
○ iii only

Explanation:

Step1: Analyze Mode (Statement I)

- Mode is the most frequent value.
- Set A: Count dots at each value. At 4, there are 5 dots (most).
- Set B: Count dots at each value. At 2 and 3, there are 4 dots each (most, bimodal). Wait, no—wait, Set A: values are 2 (2), 3 (2), 4 (5), 5 (2), 6 (3). Wait, no, let's count correctly. Set A: x=2: 2 dots, x=3: 2 dots, x=4: 5 dots, x=5: 2 dots, x=6: 3 dots. So mode is 4 (highest frequency 5). Set B: x=1: 2, x=2: 4, x=3: 4, x=4:1, x=5:2, x=6:1. So modes are 2 and 3 (frequency 4). Wait, that can't be. Wait, maybe I miscounted. Wait, Set A: the dot plot: 2 has 2, 3 has 2, 4 has 5, 5 has 2, 6 has 3. Set B: 1 has 2, 2 has 4, 3 has 4, 4 has 1, 5 has 2, 6 has 1. Wait, but maybe the original problem's Set A and B have different counts. Wait, maybe I made a mistake. Wait, let's re-express. Wait, maybe the mode of Set A is 4, and Set B: wait, maybe the dots at 2 and 3 are 4 each, but maybe the question's Set B has mode 4? No, no. Wait, maybe the initial analysis is wrong. Wait, let's check again. Wait, maybe the user's dot plot: Set A: 2 (2), 3 (2), 4 (5), 5 (2), 6 (3). Set B: 1 (2), 2 (4), 3 (4), 4 (1), 5 (2), 6 (1). Wait, but maybe the mode of Set A is 4, and Set B: wait, no—maybe the mode of Set B is also 4? No, that's not. Wait, maybe I misread the dot plot. Alternatively, maybe the mode of Set A is 4, and Set B: let's count the dots again. Set B: x=1: 2 dots, x=2: 4 dots, x=3: 4 dots, x=4:1, x=5:2, x=6:1. So modes are 2 and 3 (frequency 4). Set A: x=4 has 5 dots (highest). So mode of A is 4, mode of B is 2 and 3? That would make statement I false. But that contradicts. Wait, maybe I made a mistake. Wait, maybe the Set B's dots at 4? No, the plot shows Set B: 1 (2), 2 (4), 3 (4), 4 (1), 5 (2), 6 (1). Wait, maybe the original problem's Set A and B have different counts. Alternatively, maybe the mode of Set A is 4, and Set B's mode is 4? No, that's not. Wait, perhaps the user's dot plot is different. Wait, maybe I should proceed to range and median.

Step2: Analyze Range (Statement II)

- Range = Max - Min.
- Set A: Min=2, Max=6. Range = 6 - 2 = 4.
- Set B: Min=1, Max=6. Range = 6 - 1 = 5.
- So Set A's range (4) < Set B's range (5). So statement II is true.

Step3: Analyze Median (Statement III)

- Median is the middle value (or average of two middle) when data is ordered.
- First, find number of data points in each set.
- Set A: Count dots: 2 (x=2) + 2 (x=3) + 5 (x=4) + 2 (x=5) + 3 (x=6) = 2+2+5+2+3=14. So 14 data points. Median is average of 7th and 8th term (since 14 is even: (n/2)th and (n/2 +1)th). Order the data: 2,2,3,3,4,4,4,4,4,5,5,6,6,6. Wait, no: x=2: 2, x=3:2, x=4:5, x=5:2, x=6:3. So ordered data: [2,2,3,3,4,4,4,4,4,5,5,6,6,6]. 7th term: 4, 8th term:4. Median = (4+4)/2 =4.
- Set B: Count dots: 2 (x=1) +4 (x=2) +4 (x=3) +1 (x=4) +2 (x=5) +1 (x=6) = 2+4+4+1+2+1=14. Also 14 data points. Ordered data: [1,1,2,2,2,2,3,3,3,3,4,5,5,6]. 7th term:3, 8th term:3. Median = (3+3)/2=3.
- So Median of A (4) > Median of B (3). So statement III is true.

Wait, but earlier mode analysis: maybe I made a mistake. Let's recheck mode. Set A: x=4 has 5 dots (highest). Set B: x=2 and x=3 have 4 dots (highest). So mode of A is 4, mode of B is 2 and 3. So statement I: "mode of A equals mode of B" is false? But that contradicts. Wait, maybe the dot plot is different. Wait, maybe Set B's mode is 4? No, the dots at 2 and 3 are 4 each. Wait, maybe the original problem's Set A and B have different counts. Alternatively, maybe I misread the dot plot. Wait, the problem says "the mode of Set A is equal to the mode of Set B". Let's check again. Set A: x=4 has 5 dots. Set B: x=2 and x=3 have 4 dots. So mode of A is 4, mode of B is 2 and 3. So statement I is false. But then II and III are true. So the answer would be II and III. But let's confirm.

Wait, maybe the dot plot for Set B: x=4 has more dots? No, the plot shows Set B: 1 (2), 2 (4), 3 (4), 4 (1), 5 (2), 6 (1). So x=4 has 1 dot. So mode of B is 2 and 3 (frequency 4). Set A: x=4 has 5 (frequency 5). So mode of A is 4, mode of B is 2 and 3. So statement I is false. Then II: range of A is 6-2=4, range of B is 6-1=5. So 4 <5: true. III: median of A is 4, median of B is 3: 4>3: true. So II and III are true. So the correct option is "II and III".

Answer:

II and III (the option with "II and III")