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QUESTION IMAGE

the following data represent the number of potholes on 35 randomly sele…

Question

the following data represent the number of potholes on 35 randomly selected 1-mile stretches of highway around a particular city. complete parts (a) and (b).
number of potholes
1 3 3 1 4
7 5 1 3 6
1 2 2 1 2
7 1 6 2 7
1 6 4 4 1
1 5 3 6 2
3 2 4 1 3
(b) construct a relative frequency histogram of the data. choose the correct answer.
a. histogram, b. histogram, c. histogram, d. histogram

Explanation:

Response

To solve part (b) of constructing a relative frequency histogram, we first need to determine the frequency of each number of potholes (1, 2, 3, 4, 5, 6, 7) in the given data set, then calculate the relative frequency (frequency divided by the total number of observations, which is 35).

Step 1: Count the frequency of each number of potholes
  • Number of potholes = 1: Let's count how many times 1 appears. By going through the data: 1, 3, 3, 1, 4, 7, 5, 1, 3, 6, 1, 2, 2, 1, 2, 7, 1, 6, 2, 7, 1, 6, 4, 4, 1, 1, 5, 3, 6, 2, 3, 2, 4, 1, 3. Counting the 1s: Let's list them:
  • Row 1: 1, 1
  • Row 2: 1
  • Row 3: 1, 1
  • Row 4: 1
  • Row 5: 1, 1
  • Row 6: 1
  • Row 7: 1

Wait, maybe a better way: Let's count each number:

  • 1: Let's count: 1 (1st), 1 (4th), 1 (8th), 1 (11th), 1 (14th), 1 (17th), 1 (21st), 1 (25th), 1 (26th), 1 (34th). Wait, maybe I made a mistake. Let's count again carefully:

The data is:

Row 1: 1, 3, 3, 1, 4 (two 1s)

Row 2: 7, 5, 1, 3, 6 (one 1)

Row 3: 1, 2, 2, 1, 2 (two 1s)

Row 4: 7, 1, 6, 2, 7 (one 1)

Row 5: 1, 6, 4, 4, 1 (two 1s)

Row 6: 1, 5, 3, 6, 2 (one 1)

Row 7: 3, 2, 4, 1, 3 (one 1)

Total 1s: 2 + 1 + 2 + 1 + 2 + 1 + 1 = 10

  • Number of potholes = 2: Let's count:

Row 1: none

Row 2: none

Row 3: 2, 2, 2 (three 2s)

Row 4: 2 (one 2)

Row 5: none

Row 6: 2 (one 2)

Row 7: 2 (one 2)

Total 2s: 3 + 1 + 1 + 1 = 6

  • Number of potholes = 3: Let's count:

Row 1: 3, 3 (two 3s)

Row 2: 3 (one 3)

Row 3: none

Row 4: none

Row 5: none

Row 6: 3 (one 3)

Row 7: 3, 3 (two 3s)

Total 3s: 2 + 1 + 1 + 2 = 6

  • Number of potholes = 4: Let's count:

Row 1: 4 (one 4)

Row 2: none

Row 3: none

Row 4: none

Row 5: 4, 4 (two 4s)

Row 6: none

Row 7: 4 (one 4)

Total 4s: 1 + 2 + 1 = 4

  • Number of potholes = 5: Let's count:

Row 1: none

Row 2: 5 (one 5)

Row 3: none

Row 4: none

Row 5: none

Row 6: 5 (one 5)

Row 7: none

Total 5s: 1 + 1 = 2

  • Number of potholes = 6: Let's count:

Row 1: none

Row 2: 6 (one 6)

Row 3: none

Row 4: 6 (one 6)

Row 5: 6 (one 6)

Row 6: 6 (one 6)

Row 7: none

Total 6s: 1 + 1 + 1 + 1 = 4

  • Number of potholes = 7: Let's count:

Row 1: none

Row 2: 7 (one 7)

Row 3: none

Row 4: 7, 7 (two 7s)

Row 5: none

Row 6: none

Row 7: none

Total 7s: 1 + 2 = 3

Let's verify the total frequency: 10 (for 1) + 6 (for 2) + 6 (for 3) + 4 (for 4) + 2 (for 5) + 4 (for 6) + 3 (for 7) = 10 + 6 + 6 + 4 + 2 + 4 + 3 = 35, which matches the number of observations. Good.

Step 2: Calculate the relative frequency for each number of potholes

Relative frequency = Frequency / Total number of observations (35)

  • For 1: \( \frac{10}{35} \approx 0.2857 \)
  • For 2: \( \frac{6}{35} \approx 0.1714 \)
  • For 3: \( \frac{6}{35} \approx 0.1714 \)
  • For 4: \( \frac{4}{35} \approx 0.1143 \)
  • For 5: \( \frac{2}{35} \approx 0.0571 \)
  • For 6: \( \frac{4}{35} \approx 0.1143 \)
  • For 7: \( \frac{3}{35} \approx 0.0857 \)

Now, let's analyze the relative frequency histogram options:

  • The relative frequency for 1 is the highest (≈0.2857), then 2 and 3 (≈0.1714), then 4 and 6 (≈0.1143), then 7 (≈0.0857), then 5 (≈0.0571).

Looking at the options:

  • Option A: The tallest bar is around 0.3 (which matches 1's relative frequency ≈0.2857), then the next bars for 2 and 3 are around 0.17 (matches), then 4 and 6 around 0.11, 7 around 0.08, 5 around 0.05. This seems to match.
  • Option B: The tallest bar is at the end (7), which doesn't match our relative frequencies (7 has the third lowest relative frequency).
  • Option C:…

Answer:

A.