let x=1 with p=0.56 and x=-1 with p=0.44
here from above given data
x | f(x) | xP(x) | x2P(x) |
1 | 0.56 | 0.560 | 0.560 |
-1 | 0.44 | -0.440 | 0.440 |
total | 0.120 | 1.000 | |
E(x) =μ= | ΣxP(x) = | 0.1200 | |
E(x2) = | Σx2P(x) = | 1.0000 | |
Var(x)=σ2 = | E(x2)-(E(x))2= | 0.9856 | |
std deviation= | σ= √σ2 = | 0.9928 |
therefore for 100 steps ; expected distance Y=100*E(X)=100*0.12=12
and std deviation =sqrt(100*Var(X))=sqrt(100*0.9856)=9.93
hence probability that the particle will be less than 6 units from the origin after 100 steps
=P(-5.5<X<5.5)=P((-5.5-12)/9.93<Z<(5.5-12)/9.93)=P(-1.76<Z<-0.65)=0.2578-0.0392=0.2186
0.2186 is approximately equal to 0.238
so option b is correct
A particle executes a simple unrestricted random walk on the line: a step to the right...
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