Before 1918, approximately 55% of the wolves in the New Mexico and Arizona regio

Question

Before 1918, approximately 55% of the wolves in the New Mexico and Arizona 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.)

(a) Before 1918, in a random sample of 10 wolves spotted in the region, what is the probability that 7 or more were male?

What is the probability that 7 or more were female?

What is the probability that fewer than 4 were female?

(b) For the period from 1918 to the present, in a random sample of 10 wolves spotted in the region, what is the probability that 7 or more were male?

What is the probability that 7 or more were female?

What is the probability that fewer than 4 were female?

please circle anser.

Explanation / Answer

before 1918 , probability of male = 0.55

probability of female = 0.45

from 1918 to present, probability of male = 0.65

probability of female = 0.35

a)p(7 or more are male) = p(7)+p(8)+p(9)+p(10)

p(7) = 10C7*(0.55)^7*(0.45)^3 = 0.166

P(8) = 10C8*(0.55)^8*(0.45)^2 = 0.076

P(9) = 10C9*(0.55)^9*0.45 = 0.020

P(10) = (0.55)^10 = 0.002

ADDING ALL WE WILL GET THE ANSWER = 0.262

2)P(7 OR MORE ARE FEMALE ) = P(7)+P(8)+P(9)+P(10)

SIMILARLY WE WILL FIND AS IN ABOVE PART

P(7) = 10C7*(0.55)^3*(0.45)^7 = 0.074

SIMILARLY P(8) = 0.066

P(9)= 0.014

P(10) = 0.001

ADDING ALL WE WILL GET = 0.155

3)P(LESS THEN 4 FEMALES) = P(0)+P(1)+P(2)+P(3)

P(0)= 0.55^10 = 0.001

P(1) = 10C1*(0.55)^9*(0.45) = 0.020

P(2) = 10C2*(0.55)^8*(0.45)^8 = 0.076

P(3) = 10C3*(0.55)^7*(0.45)^3 = 0.166

ADDING WE WILL GET THE ANSWER = 0.262

B)PRESENT

P(7 OR MORE MALE) = P(7)+P(8)+P(9)+P(10)

P(7) = 10C7*(0.65)^7*(0.35)^3 = 0.252

P(8) = 10C8*(0.65)^8*(0.35)^2 = 0.175

P(9) = 10C9*(0.65)^9*(0.35) = 0.072

P(10) - 0.65^10 = 0.013

ANSWER WILL BE = 0.452

2) P(7 OR MORE FEMALE) = P(7)+P(8)+P(9)+P(10)

P(7) = 10C7*(0.65)^3*(0.35)^7 = 0.021

SIMILARLY P(8) = 0.016

P(9) = 0.010

P(10)= 0.0.0002

ADDDING WE GET THE ANSWER = 0.0472

3) (LESS THEN 4 FEMALES) = P(0)+P(1)+P(2)+P(3)

P(0)= 0.013

P(1)=0.072

P(2)=0.175

P(3) = 0.252

ADDING WE GET THE ANSWER = 0.452

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