The table compares the number of home runs hit by two baseball teams over a period of 8 years ,Then the mean absolute deviation of the data for team = 1.06.
Mean is the ratio of sum of the dataset to the sample size. Mathematically:
First take Team 1 :
Mean m = sum of the terms / number of the terms
sum of the terms = 106+119+136+125+120+103+112+131
sum of the terms = 952
number of the terms N = 8
So,
Mean m = 952/8
m = 119
For the standard deviation ,
σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
Σ[tex]\sqrt{x-m} ^{2}[/tex] = [tex](106-119)^{2} + (119-119)^{2} +(136-119)^{2} +(125-119)^{2} +(120-119)^{2} +(103-119)^{2} +(112-119)^{2} +(131-119)^{2}[/tex]
Σ[tex]\sqrt{x-m} ^{2}[/tex] = [tex](-13)^{2}+0+17^{2} +6^{2} +1^{2} +(-16)^{2} +(-7)^{2} +12^{2}[/tex]
Σ[tex]\sqrt{x-m} ^{2}[/tex] = 169+289+36+1+256+49+144
Σ[tex]\sqrt{x-m} ^{2}[/tex] = 944
Substitute this formula,
σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
σ = [tex]\sqrt{\frac{944}{8} }[/tex]
σ = [tex]\sqrt{118}[/tex]
σ = 10.86
For Team 2 :
Mean m = sum of the terms / number of the terms
sum of the terms = 122+128+130+151+126+115+132+144
sum of the terms = 1048
number of the terms N = 8
So,
Mean m = 1048/8
m = 131
For the standard deviation ,
σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
Σ[tex]\sqrt{x-m} ^{2}[/tex] = [tex](122-131)^{2} +(128-131)^{2} +(130-131)^{2}+(151-131)^{2} +(126-131)^{2} +(115-131)^{2} +(132-131)^{2} +(144-131)^{2}[/tex]
Σ[tex]\sqrt{x-m} ^{2}[/tex] = [tex](-9)^{2} +(-3)^{2} +(-1)^{2} +20^{2} +(-5)^{2} +(-16)^{2} +1^{2} +13^{2}[/tex]
Σ[tex]\sqrt{x-m} ^{2}[/tex] = 81+9+1+400+25+256+1+169
Σ[tex]\sqrt{x-m} ^{2}[/tex] = 942
Substitute this formula,
σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
σ = [tex]\sqrt{\frac{942}{8} }[/tex]
σ = [tex]\sqrt{117.75}[/tex]
σ = 10.85
Therefore,
The difference is,
10.86-10.85 = 0.1
We can use this expression,
We can write,
Then the difference of approximate value is,
4 and 5 years are,
Team 1 Mean m = (125+120) /2
m = 245 / 2 = 185
Team 2 Mean m = (151+126)/2
m = 277/2 = 214
Deviation of Team 1 σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
= [tex](125-185)^{2}+(120-185)^{2}[/tex]
= 60*60 + 65*65
= 3600+4225
= 7825
σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
= [tex]\sqrt{\frac{7825}{2} }[/tex]
= [tex]\sqrt{6025}[/tex]
= 77.6
Deviation of Team 1 σ = 77.6
Deviation of Team 2 σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
= [tex](151-214)^{2} +(126-214)^{2}[/tex]
= 63*63 +88*88
= 3969 + 7744
= 11713
σ = Σ [tex]\frac{\sqrt{x-m}^{2} }{N}[/tex]
= [tex]\sqrt{\frac{11713}{2} }[/tex]
= [tex]\sqrt{5856.5}[/tex]
= 76.52
Deviation of Team 2 σ = 76.52
Therefore,
The approximate result is = 77.6 - 76.52 = 1.06
The mean absolute deviation of the data for team = 1.06
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