Question

an automotive service company has franchises in two different parts of a city. the number of oil changes made daily, for 25 days, is given below. complete parts a) through f) below.

east store
37 58 27 32 43
20 37 29 25 32
44 29 54 38 54
39 69 37 44 46
15 27 51 42 46

west store
42 43 43 40 34
42 37 39 42 32
39 41 42 36 48
28 30 37 43 33
36 49 31 42 46

fill in the frequency distribution for the west store below.

oil changes | days
15-19 | 0
20-24 | 0
25-29 | 1
30-34 | 5
35-39 | 6
40-44 | 10
45-49 | 3
50-54 | 0
55-59 | 0
60-64 | 0
65-69 | 0

b) draw a histogram indicating the number of oil changes at each store.
choose the histogram for the east store below.

Explanation:

Construct frequency distribution for East Store

Using the Frequency Distribution knowledge point
We count the occurrences of daily oil changes for the East Store in each class interval:

• \(15\text{--}19\): \(15\) (1 value) \(

ightarrow\) Frequency = \(1\)

• \(20\text{--}24\): \(20\) (1 value) \(

ightarrow\) Frequency = \(1\)

• \(25\text{--}29\): \(27, 29, 29, 27\) (4 values) \(

ightarrow\) Frequency = \(4\)

• \(30\text{--}34\): \(32, 32, 38\) (Wait, \(38\) is in \(35\text{--}39\). Let's list values: \(32, 32\)) (2 values) \(

ightarrow\) Frequency = \(2\)

• \(35\text{--}39\): \(37, 37, 38, 39, 37\) (5 values) \(

ightarrow\) Frequency = \(5\)

• \(40\text{--}44\): \(43, 44, 44, 42\) (4 values) \(

ightarrow\) Frequency = \(4\)

• \(45\text{--}49\): \(46, 46\) (2 values) \(

ightarrow\) Frequency = \(2\)

• \(50\text{--}54\): \(58, 54, 54, 51\) (4 values) \(

ightarrow\) Frequency = \(4\)

• \(55\text{--}59\): None (0 values) \(

ightarrow\) Frequency = \(0\)

• \(60\text{--}64\): None (0 values) \(

ightarrow\) Frequency = \(0\)

• \(65\text{--}69\): \(69\) (1 value) \(

ightarrow\) Frequency = \(1\)

Match frequency distribution to histogram options

Using the Histogram Interpretation knowledge point
Let's check the heights of the bars in the given options for the East Store:

• Option A: Peak is around \(5\) at interval \(60\text{--}64\). This is incorrect.
• Option B:
• \(15\text{--}19\): \(1\)
• \(20\text{--}24\): \(1\)
• \(25\text{--}29\): \(4\)
• \(30\text{--}34\): \(2\)
• \(35\text{--}39\): \(5\)
• \(40\text{--}44\): \(4\)
• \(45\text{--}49\): \(2\)
• \(50\text{--}54\): \(4\)
• \(55\text{--}59\): \(0\)
• \(60\text{--}64\): \(0\)
• \(65\text{--}69\): \(1\)

This matches our calculated frequency distribution perfectly.

• Option C: Peak is at \(10\) around \(40\text{--}44\), which matches the West Store's distribution.
• Option D: Peak is at \(10\) around \(65\text{--}69\). This is incorrect.

Answer:

• A. The histogram with a peak of 5 in the 60-64 range.
• B. The histogram showing frequencies of 1, 1, 4, 2, 5, 4, 2, 4, 0, 0, 1 across the intervals from 15 to 69. (Correct answer)
• C. The histogram with a peak of 10 in the 40-44 range.
• D. The histogram with a peak of 10 in the 65-69 range.