## Before 1918, approximately 55% of the wolves in a region were male, and 45% were female. However, cattle ranchers in this area have made a d

Question

Before 1918, approximately 55% of the wolves in a region were male, and 45% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 65% of wolves in the region are male, and 35% are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (Round your answers to three decimal places.)

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2021-09-23T22:27:36+00:00
2021-09-23T22:27:36+00:00 1 Answer
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## Answers ( )

Answer:

(a) 0.1911

(b) 0.0610

(c) 0.3971

(c) 0.4255

(d) 0.6683

Step-by-step 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