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

a) construct a frequency distribution for each store.

fill in the frequency distribution for the east store below.

oil changes | days
15-19 |
20-24 |
25-29 |
30-34 |
35-39 |
40-44 |

oil changes | days
45-49 |
50-54 |
55-59 |
60-64 |
65-69 |

Explanation:

Sort the East Store data

Using the Frequency Distribution knowledge point
The raw daily oil changes for the East Store (25 days) are:
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.

Sorted in ascending order:
15, 20, 25, 27, 27, 29, 29, 32, 32, 37, 37, 37, 38, 39, 42, 43, 44, 44, 46, 46, 51, 54, 54, 58, 69.

Count frequencies for each class interval

Using the Frequency Distribution knowledge point

• 15–19: 15 (1 value) \(

ightarrow 1\)

• 20–24: 20 (1 value) \(

ightarrow 1\)

• 25–29: 25, 27, 27, 29, 29 (5 values) \(

ightarrow 5\)

• 30–34: 32, 32 (2 values) \(

ightarrow 2\)

• 35–39: 37, 37, 37, 38, 39 (5 values) \(

ightarrow 5\)

• 40–44: 42, 43, 44, 44 (4 values) \(

ightarrow 4\)

• 45–49: 46, 46 (2 values) \(

ightarrow 2\)

• 50–54: 51, 54, 54 (3 values) \(

ightarrow 3\)

• 55–59: 58 (1 value) \(

ightarrow 1\)

• 60–64: None (0 values) \(

ightarrow 0\)

• 65–69: 69 (1 value) \(

ightarrow 1\)

Total count verification: \(1 + 1 + 5 + 2 + 5 + 4 + 2 + 3 + 1 + 0 + 1 = 25\).

Answer:

Oil changes	Days
20–24	1
25–29	5
30–34	2
35–39	5
40–44	4
45–49	2
50–54	3
55–59	1
60–64	0
65–69	1