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Sagot :
Answer:
a) For Pawnee:
Range = 11, Mean = 3.13, Mean deviation = 2.44
For Chickpee:
Range = 7, Mean = 4.25, Mean deviation = 1.59
b) Pawnee Location has fewer lost days.
c) variation of the Chickpee is less than Pawnee location
Step-by-step explanation:
Data given:
For Pawnee:
3 1 0 0 2 11 5 3
Max = 11
Min = 0
Range = Max - Min
Range = 11-0 = 11
Range = 11
For Chickpee:
3 6 8 4 4 5 3 1
Max = 8
Min = 1
Range = Max - Min
Range = 8 - 1
Range = 7
For mean of Pawnee:
Mean = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3)/8
Mean = 25/8
Mean = 3.125
For mean deviation, we find the difference from the mean of each point.
Note: all the negative values will be taken as 0.
Differences from the mean of each point =
(3-3.13)=0.13, (1-3.31) = 2.13 ,
(0-3.13)=0, (0-3.13)=3.13,
(2-3.13)=1.13, (11-3.13)=7.87,
(5-3.13)=1.87, (3-3.13)=0.13
So, our differences are:
(0.13, 2.13, 0, 3.13, 1.13, 7.87, 1.87, 0.13 )
Mean deviation = Sum of all differences/8
Mean deviation = (0.13+2.13+3.13+3.13+ 1.13+ 7.87+ 1.87+ 0.13)/8
Mean deviation = 2.44
For mean of Chickpee:
Mean = (3+6+8+4+4+5+3+1)/8
Mean = 34/8
Mean = 4.25
Similarly, we need to find mean deviation for Chickpee, for that we need to find the differences first as done above for Pawnee.
Note: all the negative values will be taken as 0
Differences from the mean:
(3-4.25) = 1.25, (6-4.25) =1.75,
(8-4.25) = 3.75, (4-4.25)=0.25,
(4-4.25)=0.25, (5-4.25)= 0.75,
(3-4.25)=1.25, (1-4.25) =3.25
So, the differences are:
Differences = (1.25,1.75,3.75,0.25,0.25,0.75,1.25,3.25)
Mean deviation = (1.25+ 1.75+ 3.75+ 0.25+ 0.25+ 0.75+ 1.25+ 3.25)/8
Mean deviation = 1.59
b) which location has fewer lost days:
This can be found out by the total number of days of Pawnee and chickpee.
Pawnee = Sum of values = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3) = 25 days
Chickpee = Sum of values = (3+6+8+4+4+5+3+1) = 34 days
Hence, Pawnee Location has fewer lost days.
C) which location has less variation?
Formula for variation = ∑([tex]\frac{x^{2} }{8}[/tex]) - ([tex]mean^{2}[/tex])
where x = values of the set.
For Pawnee:
[tex]mean^{2}[/tex] = [tex]3.13^{2}[/tex] = 9.8
Sum of square of all the data points of Pawnee = 169
Variation = 169/8 - 98 = 11.325
Similarly,
For Chickpee:
[tex]mean^{2}[/tex] = [tex]4.25^{2}[/tex] = 18.06
Sum of square of all the data points of Chickpee = 176
Variation = 176/8 - 18.06
Variation = 22-18.06
Variation = 4
Hence, variation of the Chickpee is less than Pawnee location
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