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) through (d).

number of potholes
1 4 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 7 1 3

(a) construct a frequency distribution of the data.

potholes	frequency
2
3
4
5
6
7

Explanation:

Step1: Count frequency of 1

Count how many times 1 appears in the data. The data points with 1: 1,1,1,1,1,1,1,1 (let's list them: first row 1,1; second row 1; third row 1; fourth row 1; fifth row 1,1; sixth row 1; seventh row 1. Wait, let's count properly:

First row: 1,4,3,1,4 → two 1s

Second row:7,5,1,3,6 → one 1

Third row:1,2,2,1,2 → two 1s

Fourth row:7,1,6,2,7 → one 1

Fifth row:1,6,4,4,1 → two 1s

Sixth row:1,5,3,6,2 → one 1

Seventh row:3,2,7,1,3 → one 1

Total for 1: 2+1+2+1+2+1+1 = 10

Step2: Count frequency of 2

Data points with 2:

First row: none

Second row: none

Third row:2,2,2 → three 2s

Fourth row:2 → one 2

Fifth row: none

Sixth row:2 → one 2

Seventh row:2 → one 2

Wait, let's list all 2s:

Third row: 2,2,2 (three)

Fourth row: 2 (one)

Sixth row: 2 (one)

Seventh row: 2 (one)

Wait, third row: 1,2,2,1,2 → three 2s (positions 2,3,5)

Fourth row:7,1,6,2,7 → one 2 (position 4)

Sixth row:1,5,3,6,2 → one 2 (position 5)

Seventh row:3,2,7,1,3 → one 2 (position 2)

Total: 3 + 1 + 1 + 1 = 6

Step3: Count frequency of 3

Data points with 3:

First row:3 → one 3

Second row:3 → one 3

Fourth row: none

Fifth row: none

Sixth row:3 → one 3

Seventh row:3,3 → two 3s

Wait, let's list:

First row:1,4,3,1,4 → one 3 (position 3)

Second row:7,5,1,3,6 → one 3 (position 4)

Sixth row:1,5,3,6,2 → one 3 (position 3)

Seventh row:3,2,7,1,3 → two 3s (positions 1,5)

Also, wait, second row: 7,5,1,3,6 → 3 is there. Any other 3s?

Wait, first row: 3

Second row: 3

Sixth row: 3

Seventh row: 3,3

Wait, also, is there a 3 in other rows? Let's check third row: no. Fourth row: no. Fifth row: no.

Wait, let's count all 3s:

First row: 1

Second row: 1

Sixth row: 1

Seventh row: 2

Wait, that's 1+1+1+2=5? Wait, no, maybe I missed. Let's list all data points:

Row 1: 1,4,3,1,4 → 3 (1)

Row 2:7,5,1,3,6 → 3 (1)

Row 3:1,2,2,1,2 → no

Row 4:7,1,6,2,7 → no

Row 5:1,6,4,4,1 → no

Row 6:1,5,3,6,2 → 3 (1)

Row 7:3,2,7,1,3 → 3,3 (2)

Wait, also, is there a 3 in row 2? Yes. Row 1: yes. Row 6: yes. Row 7: two. Wait, that's 1+1+1+2=5? Wait, no, maybe I made a mistake. Let's count again:

All data points:

1,4,3,1,4,

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,

3,2,7,1,3

Now list all 3s:

From row 1: 3 (1)

Row 2: 3 (1)

Row 6: 3 (1)

Row 7: 3, 3 (2)

Wait, that's 1+1+1+2=5? Wait, no, row 7 has two 3s, row 1,2,6 have one each. So total 1+1+1+2=5? Wait, but let's check again. Wait, row 7: 3,2,7,1,3 → two 3s. Row 1: 3. Row 2: 3. Row 6: 3. So total 1+1+1+2=5? Wait, maybe I missed a 3. Let's check all numbers:

Looking at all data:

1,4,3,1,4,

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,

3,2,7,1,3

Now count 3s:

1. 3 (row1, col3)

2. 3 (row2, col4)

3. 3 (row6, col3)

4. 3 (row7, col1)

5. 3 (row7, col5)

Yes, that's 5. So frequency of 3 is 5.

Step4: Count frequency of 4

Data points with 4:

Row1:4,4 → two 4s

Row5:4,4 → two 4s

Any others?

Row1: 1,4,3,1,4 → two 4s (cols2,5)

Row5:1,6,4,4,1 → two 4s (cols3,4)

Any other 4s? Let's check other rows:

Row2: no

Row3: no

Row4: no

Row6: no

Row7: no

So total 2+2=4. So frequency of 4 is 4.

Step5: Count frequency of 5

Data points with 5:

Row2:5 → one 5

Row6:5 → one 5

Any others?

Row2:7,5,1,3,6 → 5 (col2)

Row6:1,5,3,6,2 → 5 (col2)

Any other 5s? No. So total 1+1=2. Frequency of 5 is 2.

Step6: Count frequency of 6

Data points with 6:

Row2:6 → one 6

Row4:6 → one 6

Row5:6 → one 6

Row6:6 → one 6

Wait, let's list:

Row2:7,5,1,3,6 → 6 (col5)

Row4:7,1,6,2,7 → 6 (col3)

Row5:1,6,4,4,1 → 6 (col2)

Row6:1,5,3,6,2 → 6 (col4)

Any other 6s? No. So total 1+1+1+1=4. Frequency of 6 is 4.

Step7: Count frequency of 7

Data points with 7:

Row2:7 → one 7

Row4:7,7 → two 7s

Row7:7 → one 7

Wait, let's list:

Row2:7,5,1,3,6 → 7 (col1)

Row4:7,1,6,2,7 → 7,7 (cols1,5)

Row7:3,2,7,1,3 → 7 (col3)

Any other 7s? No. So total 1+2+1=4. Frequency of 7 is 4.

Wait, let's verify total frequency: 10 (for 1) +6 (for 2)+5 (for3)+4 (for4)+2 (for5)+4 (for6)+4 (for7) = 10+6=16, 16+5=21, 21+4=25, 25+2=27, 27+4=31, 31+4=35. Which matches the number of data points (35). Good.

Answer:

Potholes	Frequency
2	6
3	5
4	4
5	2
6	4
7	4