Question

which histogram represents the data?
1, 2, 12, 14, 15, 16, 18, 24, 30, 34, 34, 36, 38, 40, 46, 46, 50

Explanation:

Step1: Determine data ranges

First, we identify the range of the data. The minimum value is 1 and the maximum is 50. Let's create bins (intervals) to group the data. A common approach is to use equal - width bins. Let's choose bins like 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60 (but since our max is 50, the last bin can be 40 - 50 and 50 - 60 will have 1 data point if needed).

Step2: Count data in each bin

• Bin 0 - 10: Data points 1, 2. So count = 2.
• Bin 10 - 20: Data points 12, 14, 15, 16, 18. Count = 5.
• Bin 20 - 30: Data points 24, 30. Count = 2.
• Bin 30 - 40: Data points 34, 34, 36, 38, 40. Wait, no, 30 is in 20 - 30, so 34, 34, 36, 38, 40? Wait, 30 is in 20 - 30, then 34, 34, 36, 38, 40: Wait, 30 is 30, so 30 is in 20 - 30. Then 34, 34, 36, 38, 40: Wait, 30 - 40: 30 is the start, so 30, 34, 34, 36, 38, 40? Wait, I made a mistake earlier. Let's re - do the binning properly. Let's use bins with width 10:
• Bin 1 - 10: 1, 2 (count = 2)
• Bin 11 - 20: 12, 14, 15, 16, 18 (count = 5)
• Bin 21 - 30: 24, 30 (count = 2)
• Bin 31 - 40: 34, 34, 36, 38, 40 (wait, 30 is in 21 - 30? No, 30 is 30, so 21 - 30: 24, 30 (count = 2). 31 - 40: 34, 34, 36, 38, 40 (count = 5? Wait, 34, 34, 36, 38, 40: that's 5? Wait 34 is 34, 34, 36, 38, 40: 5 data points? Wait no, 30 is in 21 - 30, then 34, 34, 36, 38, 40: 5 data points. Then 41 - 50: 46, 46, 50. Wait, 40 is in 31 - 40, so 46, 46, 50: count = 3? Wait no, 40 is in 31 - 40, so 41 - 50: 46, 46, 50. Wait, but 40 is in 31 - 40, so 41 - 50: 46, 46, 50 (count = 3). Wait, but let's check the original data again: 1, 2, 12, 14, 15, 16, 18, 24, 30, 34, 34, 36, 38, 40, 46, 46, 50.

Let's list all data points:
1 (0 - 10), 2 (0 - 10), 12 (10 - 20), 14 (10 - 20), 15 (10 - 20), 16 (10 - 20), 18 (10 - 20), 24 (20 - 30), 30 (20 - 30), 34 (30 - 40), 34 (30 - 40), 36 (30 - 40), 38 (30 - 40), 40 (30 - 40), 46 (40 - 50), 46 (40 - 50), 50 (40 - 50)

Now count:

• 0 - 10: 1, 2 → 2
• 10 - 20: 12,14,15,16,18 → 5
• 20 - 30: 24,30 → 2
• 30 - 40: 34,34,36,38,40 → 5 (wait, 30 is in 20 - 30, so 34,34,36,38,40: 5 data points)
• 40 - 50: 46,46,50 → 3
• 50 - 60: 0 (since 50 is in 40 - 50? Wait, 50 is the upper limit, so 40 - 50 includes 40 ≤ x < 50? No, usually, for histograms, the bins are like [a, b), so 0 - 10: [0,10), 10 - 20: [10,20), 20 - 30: [20,30), 30 - 40: [30,40), 40 - 50: [40,50), 50 - 60: [50,60)

So:

• [0,10): 1,2 → 2
• [10,20):12,14,15,16,18 → 5
• [20,30):24,30 → 2 (since 30 is in [20,30))
• [30,40):34,34,36,38,40 → 5 (34 ≥ 30 and < 40)
• [40,50):46,46,50? Wait, 40 is in [30,40), 46 is ≥40 and <50, 50 is in [50,60). So [40,50):46,46 → 2, [50,60):50 → 1

Wait, maybe the bins in the histogram are 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50. Let's recalculate with [0,10), [10,20), [20,30), [[SSE Completed, Client Connection Error][SSE onError error]

Answer:

Step1: Determine data ranges

First, we identify the range of the data. The minimum value is 1 and the maximum is 50. Let's create bins (intervals) to group the data. A common approach is to use equal - width bins. Let's choose bins like 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60 (but since our max is 50, the last bin can be 40 - 50 and 50 - 60 will have 1 data point if needed).

Step2: Count data in each bin

• Bin 0 - 10: Data points 1, 2. So count = 2.
• Bin 10 - 20: Data points 12, 14, 15, 16, 18. Count = 5.
• Bin 20 - 30: Data points 24, 30. Count = 2.
• Bin 30 - 40: Data points 34, 34, 36, 38, 40. Wait, no, 30 is in 20 - 30, so 34, 34, 36, 38, 40? Wait, 30 is in 20 - 30, then 34, 34, 36, 38, 40: Wait, 30 is 30, so 30 is in 20 - 30. Then 34, 34, 36, 38, 40: Wait, 30 - 40: 30 is the start, so 30, 34, 34, 36, 38, 40? Wait, I made a mistake earlier. Let's re - do the binning properly. Let's use bins with width 10:
• Bin 1 - 10: 1, 2 (count = 2)
• Bin 11 - 20: 12, 14, 15, 16, 18 (count = 5)
• Bin 21 - 30: 24, 30 (count = 2)
• Bin 31 - 40: 34, 34, 36, 38, 40 (wait, 30 is in 21 - 30? No, 30 is 30, so 21 - 30: 24, 30 (count = 2). 31 - 40: 34, 34, 36, 38, 40 (count = 5? Wait, 34, 34, 36, 38, 40: that's 5? Wait 34 is 34, 34, 36, 38, 40: 5 data points? Wait no, 30 is in 21 - 30, then 34, 34, 36, 38, 40: 5 data points. Then 41 - 50: 46, 46, 50. Wait, 40 is in 31 - 40, so 46, 46, 50: count = 3? Wait no, 40 is in 31 - 40, so 41 - 50: 46, 46, 50. Wait, but 40 is in 31 - 40, so 41 - 50: 46, 46, 50 (count = 3). Wait, but let's check the original data again: 1, 2, 12, 14, 15, 16, 18, 24, 30, 34, 34, 36, 38, 40, 46, 46, 50.

Let's list all data points:
1 (0 - 10), 2 (0 - 10), 12 (10 - 20), 14 (10 - 20), 15 (10 - 20), 16 (10 - 20), 18 (10 - 20), 24 (20 - 30), 30 (20 - 30), 34 (30 - 40), 34 (30 - 40), 36 (30 - 40), 38 (30 - 40), 40 (30 - 40), 46 (40 - 50), 46 (40 - 50), 50 (40 - 50)

Now count:

• 0 - 10: 1, 2 → 2
• 10 - 20: 12,14,15,16,18 → 5
• 20 - 30: 24,30 → 2
• 30 - 40: 34,34,36,38,40 → 5 (wait, 30 is in 20 - 30, so 34,34,36,38,40: 5 data points)
• 40 - 50: 46,46,50 → 3
• 50 - 60: 0 (since 50 is in 40 - 50? Wait, 50 is the upper limit, so 40 - 50 includes 40 ≤ x < 50? No, usually, for histograms, the bins are like [a, b), so 0 - 10: [0,10), 10 - 20: [10,20), 20 - 30: [20,30), 30 - 40: [30,40), 40 - 50: [40,50), 50 - 60: [50,60)

So:

• [0,10): 1,2 → 2
• [10,20):12,14,15,16,18 → 5
• [20,30):24,30 → 2 (since 30 is in [20,30))
• [30,40):34,34,36,38,40 → 5 (34 ≥ 30 and < 40)
• [40,50):46,46,50? Wait, 40 is in [30,40), 46 is ≥40 and <50, 50 is in [50,60). So [40,50):46,46 → 2, [50,60):50 → 1

Wait, maybe the bins in the histogram are 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50. Let's recalculate with [0,10), [10,20), [20,30), [[SSE Completed, Client Connection Error][SSE onError error]