Question

question 1 (multiple choice worth 2 points)
(histograms and line plots mc)
the histogram shows data collected about the number of passengers using city bus transportation at a specific time of day.
histogram: city bus transportation, x-axis: number of passengers (bins 1–10, 11–20, 21–30, 31–40, 42–50), y-axis: frequency (2, 4, 5, 6, 3)
which of the following data sets best represents what is displayed in the histogram?
options:
○ (4, 5, 7, 8, 10, 12, 13, 15, 18, 21, 23, 28, 32, 34, 36, 40, 41, 41, 42, 42)
○ (4, 7, 11, 13, 14, 19, 22, 24, 26, 27, 29, 31, 33, 35, 36, 38, 40, 42, 42, 42)
○ (4, 5, 7, 8, 12, 13, 15, 18, 19, 21, 24, 25, 26, 28, 29, 30, 32, 33, 35, 42)
○ (4, 6, 11, 12, 16, 18, 21, 24, 25, 26, 28, 29, 30, 32, 35, 36, 38, 41, 41, 42)

Explanation:

Step1: Analyze each bin's frequency

• Bin \(1 - 10\): Frequency \(2\) → 2 data points in \(1 - 10\).
• Bin \(11 - 20\): Frequency \(4\) → 4 data points in \(11 - 20\).
• Bin \(21 - 30\): Frequency \(5\) → 5 data points in \(21 - 30\).
• Bin \(31 - 40\): Frequency \(6\) → 6 data points in \(31 - 40\).
• Bin \(41 - 50\) (note: \(42 - 50\) includes \(41\)? Wait, histogram bin \(42 - 50\) frequency \(3\) → 3 data points ≥41 (since \(42 - 50\) and \(41\) is close, but check data sets).

Step2: Check each option's count per bin

• Option 1:
• \(1 - 10\): \(4,5,7,8,10\) → Wait, no, count: \(4,5,7,8,10\) → 5? No, wait first option: \((4,5,7,8,10,12,13,15,18,21,23,28,32,34,36,40,41,41,42,42)\)
• \(1 - 10\): \(4,5,7,8,10\) → 5? No, wait histogram bin \(1 - 10\) frequency 2. Wait, no, I miscounted. Wait first option: let's list:
• \(1 - 10\): \(4,5,7,8,10\) → 5? No, wait the first option has 20 elements? Wait no, let's count:
• \(1 - 10\): \(4,5,7,8,10\) → 5? No, wait the histogram bin \(1 - 10\) has frequency 2. Wait, no, I made a mistake. Wait the first option: \((4,5,7,8,10,12,13,15,18,21,23,28,32,34,36,40,41,41,42,42)\) → count:
• \(1 - 10\): \(4,5,7,8,10\) → 5? No, that's 5. Wait no, the histogram bin \(1 - 10\) has frequency 2. Wait, no, maybe I misread the histogram. Wait the histogram: \(1 - 10\) frequency 2, \(11 - 20\) frequency 4, \(21 - 30\) frequency 5, \(31 - 40\) frequency 6, \(42 - 50\) frequency 3.

Wait let's re - check the first option:

• \(1 - 10\): \(4,5,7,8,10\) → 5? No, that's 5. Wait no, the first option's elements: \(4,5,7,8,10\) (5), \(12,13,15,18\) (4, which matches \(11 - 20\) frequency 4), \(21,23,28\) → wait no, \(21 - 30\) should have 5. Wait first option: \(21,23,28\) → only 3? No, wait I miscounted. Wait the first option: \(4,5,7,8,10\) (5), \(12,13,15,18\) (4), \(21,23,28\) → no, wait the first option has \(21,23,28\) and then \(32,34,36,40,41,41\) (wait \(31 - 40\) should have 6). Wait no, let's take the first option:

Count per bin:

• \(1 - 10\): numbers between 1 - 10: \(4,5,7,8,10\) → 5? No, the histogram has frequency 2. Wait, I think I made a mistake. Wait the first option: let's list all elements:

1. \(4,5,7,8,10\) (5 elements? No, the histogram bin \(1 - 10\) has frequency 2. So this is wrong.

Wait the second option: \((4,7,11,13,14,19,22,24,26,27,29,31,33,35,36,38,40,42,42,42)\)

• \(1 - 10\): \(4,7\) → 2 (matches frequency 2)
• \(11 - 20\): \(11,13,14,19\) → 4 (matches frequency 4)
• \(21 - 30\): \(22,24,26,27,29\) → 5 (matches frequency 5)
• \(31 - 40\): \(31,33,35,36,38,40\) → 6 (matches frequency 6)
• \(41 - 50\) (or \(42 - 50\)): \(42,42,42\) → 3 (matches frequency 3)

Yes, this matches the histogram's frequencies.

Answer:

The correct option is the second one: (4, 7, 11, 13, 14, 19, 22, 24, 26, 27, 29, 31, 33, 35, 36, 38, 40, 42, 42, 42)