find the value of the missing variate for the following distribution whose mean is 10:
Variate(xi) 5, 7, 9, 11,--- 15, 20
frequency fi 4, 4, 4, 7, 3, 2, 1
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Hey
Here's your answer
The frequencies are 4, 4, 4, 7, 3, 2, 1
The variates are 5, 7, 9, 11,x, 15, 20
So number of terms are 4+ 4+ 4+ 7+ 3+ 2+ 1
= 25.
The sum of all the observations are
= (5*4)+(7*4)+(9*4)+(11*7)+(3*x)+(2*15)+(1*20)
= 20+28+36+77+3x+30+20.
= 211+3x
Mean of the observations is 10.
So total of the observations is = 25*10 = 250
ATQ,
250 = 211 + 3x
=> 39 = 3x
=> 11 = x (Answer)
Hope it helps!!
Here's your answer
The frequencies are 4, 4, 4, 7, 3, 2, 1
The variates are 5, 7, 9, 11,x, 15, 20
So number of terms are 4+ 4+ 4+ 7+ 3+ 2+ 1
= 25.
The sum of all the observations are
= (5*4)+(7*4)+(9*4)+(11*7)+(3*x)+(2*15)+(1*20)
= 20+28+36+77+3x+30+20.
= 211+3x
Mean of the observations is 10.
So total of the observations is = 25*10 = 250
ATQ,
250 = 211 + 3x
=> 39 = 3x
=> 11 = x (Answer)
Hope it helps!!
Hybridization
sp
sp2
sp3
sp3d
sp3d2
sp3d3
Formula
I=GA+1/2(V:E-V-C)
GA=Number of group atom attached to the central atom
VE= number of valence electron present in central atom
V= valency of central atom
C= charge present on central atom
1 .Question.[Ni(CN)4]--
2.Question [Ni(NH3)4]2+
3.Question [Cu(CN)4]--
4.Question [Fe(CN)6]----
5.Question [Ptcl4]--
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Hybridization
sp
sp2
sp3
sp3d
sp3d2
sp3d3
Formula
I=GA+1/2(V:E-V-C)
GA=Number of group atom attached to the central atom
VE= number of valence electron present in central atom
V= valency of central atom
C= charge present on central atom
1 .Question.[Ni(CN)4]--
2.Question [Ni(NH3)4]2+
3.Question [Cu(CN)4]--
4.Question [Fe(CN)6]----
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