P(x)=1 which is not true in a and d. In c one P(x)= -0.1 but we know 0≤P(x)≤1 )
2) it should be approximately normal with mean 2.8 and standard deviation 0.4 (Since sd =σ /n
). However i don't find this in any option.
3) X: for lateness in hours
On time means X =0
E(X)= ∑x P(x) = 0(4/5) +1(1/10)+2(1/20)+3(1/20) = 7/20
4) a) 7C2 = 7!/(2! 5!) = 21
b) 7×6 =42
c) 7×7 =49
1) 1) (4/13)(4/13) =16/169 = 0.09467455621 (with replacement)
2) (4/13)(3/12)=12/156 = 0.076923 (without replacement)
6) The table is as follows:
F M T
S 80 55 135
N 120 XX 165
X XXX 100 XXX
a) X(S) = 135/300
b) P(F or N) = P(F∪N) = P(X)+P(N)-X(F∩ N) = 200/XXX + 165/300 - 120/XXX = XXX/XXX
c) X(S|M)= P(S∩ M)/P(X) =(XX/XXX)/(100/300)=XX/100
Note: All the XXXXXXXXXXXXX are XXXXXXXXXX from table
7) X =69.X
X =2.8
a) X(X&XX;XX)= 1-X(X≤XX) = 1-X [(XX-69.X)/2.X] = X-Φ(0.XXXXXXXXXXX) = X-X.7939=X.XXXX
b) X(68≤X≤72) = X [(72-XX.7)/X.X] - X [(XX-XX.7)/X.8] = X(0.82142857142) -Φ(-X.60714285714)
= Φ(0.XXXXXXXXXXX)-[X-Φ(0.XXXXXXXXXXX)]
= Φ(0.82) -1+ X(0.XX)
=X.XXXX -1+X.7291
=X.XXX
8) μ =101.5
X =X.X
n=XX
P(Xˉ&XX; 102.X)= X [(XXX.X-101.5)/(1.6/36
)]
= Φ [(X.6)/(X.XXXXX)]
= X (2.25)
=X.9878
X) p=X/X, q=X- X/3 =2/3, n=XX
X(X=X) = XXXX (X/X)8 (2/3)30-8 (XXX XXXXXXXX X(X=x)=f(x)=nCx pxqn-x)
= 5852925 (0.00015241579)(0.00013365718)
=X.XXXXXXXXX
b) X(X≥8)= 1-P(X≤X) =f(0)+f(X)+f(X)+f(X)+f(4)+f(X)+f(X)+f(X) where f(x)=nCx pxqn-x with n=XX, x=X,X,2,...,7
c) p=X.XX
X=0.XX
n=XX
np = 10
npq=20/3 =6.XXX
Mean=np ,μ =XX
XXXXXXXX deviation σ = npq
=XX/3
=X.XX
P(XX≥8)=1-X(XX < X)=1-Φ [(X-10)/(X.XX/3X
)] =X-X (-2/X.471)=1-X (-4.XX) =1- [X-Φ (4.24)] =X-1+1 =1
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