Answer:
(a) 0.1911
(b) 0.0610
(c) 0.3971
(c) 0.4255
(d) 0.6683
Explanation:
Let p = 0.55, q = 1 - 0.55 = 0.45, n = 11
(a) the required Probability is Pr(X> or = 8)
PR(X>=8) = Pr(X= 8)+ Pr(X=9) +Pr( 10) + Pr(11)
Pr(X>= 8) = 11C8(.55)^8(0.45)³ + 11C9(0.55)^9(0.45)² + 11C10(0.55)^10(0.45) + (0.55)^11
P(X>= 8) = 0.1259 + 0.0513 + 0.0125 + 0.00139
~= 0.1911
(b) p = 0.45, q = 0.55
Pr(X>= 8) = 11C8 × (.45)^8 × (0.55)³ + 11C9 × (0.45)^9 × (0.55)² + 11C10 × (0.45)^10 × (0.55) + (0.45)^11
P(X>= 8) =0.04616 + 0.01259 +0.002060 + 0.000153
~= 0.0610
(c )the required probability is Pr( X<5).
Where p= 0.45, q = 0.55
Pr(X<5) = Pr(X= 4)+ Pr(X=3) +Pr(=2) + Pr(=1) + P(X=0)
Pr(X<5) = 11C4 × (0.45)⁴× (0.55)^7 + 11C3 ×(0.45)³ × (0.55)^8 + 11C2 × (0.45)² ×(0.55)^9 +11C1 × (0.45)(0.55)^10 + 11C0 × (0.55)^11
= 0.2060 + 0.1259 + 0.0513 + 0.0125 + 0.0014
= 0.3971
(c) the probability changes from 1918 to p= 0.65, q = 0.35
The required probability that 8 or more were male
PR(X>=8) = Pr(X= 8)+ Pr(X=9) +Pr( 10) + Pr(11)
Pr(X>= 8) = 11C8 × (.65)^8 × (0.35)³ + 11C9 × (0.65)^9 × (0.35)² + 11C10 × (0.65)^10 × (0.35) + (0.65)^11
P(X>= 8) = 0.2254 + 0.1395 + 0.05183 + 0.00875
= 0.4255
(d) The required probability that 8 or more were female is given by
p = 0.35, q= 0.65
PR(X>=8) = Pr(X= 8)+ Pr(X=9) +Pr( 10) + Pr(11)
Pr(X>= 8) = 11C8 × (.35)^8 × (0.65)³ + 11C9 × (0.35)^9 × (0.65)² + 11C10 × (0.35)^10 × (0.65) + (0.35)^11
P(X>= 8) = 0.0102 + 0.00183+ 0.000197 +0.00000965
= 0.0129
(e) (b )the required probability is Pr( X<5).
p =0.35, q = 0.65
Pr(X<5) = Pr(X= 4)+ Pr(X=3) +Pr(=2) + Pr(=1) + P(X=0)
Pr(X<5) = 11C4 × (0.35)⁴× (0.65)^7 + 11C3 ×(0.35)³ × (0.65)^8 + 11C2 × (0.35)² ×(0.65)^9 +11C1 × (0.35)(0.65)^10 + 11C0 × (0.65)^11
= 0.2428 + 0.2254 + 0.1395 + 0.05183 + 0.00875
= 0.6683