a)
for this to be valid:
P(x) =P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6) must be1
P(x) =P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=0.15+0.2+0.34+0.19+0.06+0.05+0.01
=1
since all individual sample point probabilities are between 0 and 1 and sum of probability of all sample point is 1, this is a valid probability distribution
b)
E(X)=xP(x) =0*0.15+1*0.2+2*0.34+3*0.19+4*0.06+5*0.05+6*0.01=2
c)
E(X2)=x2P(x) =0^2*0.15+1^2*0.2+2^2*0.34+3^2*0.19+4^2*0.06+5^2*0.05+6^2*0.01=5.84
Var(x)=σ2 = | E(x2)-(E(x))2= | 1.8400 |
std deviation= | σ= √σ2 = | 1.3565 |
d)
P(X=0) =0.15
e)\
P(X>=1) =1-P(X=0)=1-0.15 =0.85
I need help with question 11 part A,B, C, D, and E 11. The 4077th Mobile...
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