a)
P(1<X<4) =P(X=2))+P(X=3)=0.2+0.25=0.45
P(X=2)=0.2
b)
x | P(x) | xP(x) | x2P(x) |
0 | 0.1 | 0.000 | 0.000 |
1 | 0.15 | 0.150 | 0.150 |
2 | 0.2 | 0.400 | 0.800 |
3 | 0.25 | 0.750 | 2.250 |
4 | 0.2 | 0.800 | 3.200 |
5 | 0.06 | 0.300 | 1.500 |
6 | 0.04 | 0.240 | 1.440 |
total | 2.640 | 9.340 | |
E(x) =μ= | ΣxP(x) = | 2.6400 | |
E(x2) = | Σx2P(x) = | 9.3400 | |
Var(x)=σ2 = | E(x2)-(E(x))2= | 2.370 |
from above E(X)=2.64
Var(X)=2.370
c)
Y follows binomial distribution with parameter n=10 and p=0.2
hence P(Y<=2) =P(Y=0)+P(Y=1)+P(Y=2)
=10C0(0.2)0(0.8)10+10C1(0.2)1(0.8)9+10C2(0.2)2(0.8)8=0.6778
d)
n= | 64 | p= | 0.2000 | |
here mean of distribution=μ=np= | 12.8 | |||
and standard deviation σ=sqrt(np(1-p))= | 3.2000 | |||
for normal distribution z score =(X-μ)/σx | ||||
therefore from normal approximation of binomial distribution and continuity correction: |
probability = | P(11.5<X<12.5) | = | P(-0.41<Z<-0.09)= | 0.4641-0.3409= | 0.1232 |
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