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The data below gives the amount of gasoline (in liters) consumed by a tourist
bus in 40 days.

23 42 39 21 29 21 17 51
21 54 28 17 19 40 21 49
48 22 21 19 23 28 20 48
48 20 46 25 20 31 32 48
20 21 58 25 28 45 60 29
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1. Construct a frequency distribution table for this data using 7 classes with 14 as
the starting point for the lowest class.

2. Find the mean of the distribution.

3. Add a column for the "less than cumulative frequency" distribution table and then find the median.

4. Find the mode of the distribution.
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NOTE: Show you solution below ​

Sagot :

Kenjinxidn

[tex]\sf{﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌}[/tex]

[tex]\rm{\huge{ANSWER:}}[/tex]

⓵ Frequency Distribution Table

Class | Frequency

14-20 | 8

21-27 | 10

28-34 | 6

35-41 | 4

42-48 | 6

49-55 | 4

56-62 | 2

⓶ Mean

Sum: 23 + 42 + 39 + 21 + 29 + 21 + 17 + 51 + 21 + 54 + 28 + 17 + 19 + 40 + 21 + 49 + 48 + 22 + 21 + 19 + 23 + 28 + 20 + 48 + 48 + 20 + 46 + 25 + 20 + 31 + 32 + 48 + 20 + 21 + 58 + 25 + 28 + 45 + 60 + 29 = 1044

Total observations: [tex]\underline{\sf\red{40}}[/tex]

Mean: [tex]\large{\frac{1044}{40}}[/tex] = [tex]\underline{\sf\red{26.1}}[/tex]

⓷ Median

Class | Frequency | Less Cumulative Frequency

14-20 | 8 | 8

21-27 | 10 | 18

28-34 | 6 | 24

35-41 | 4 | 28

42-48 | 6 | 34

49-55 | 4 | 38

56-62 | 2 | 40

Median: [tex]\underline{\sf\red{28}}[/tex] (since 50% of the data points are below this value)

⓸ Mode

❏ The mode is the value that appears most frequently.

❏ In this case, the value 21 appears 3 times, which is the highest frequency.

❏ Therefore, the mode is [tex]\underline{\sf\red{21}}[/tex].

[tex]\sf\small{╰─▸ ❝ @[kenjinx]}[/tex]

[tex]\sf{﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌}[/tex]

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