Question

Before 1918, approximately 60% of the wolves in the New Mexico and Arizona region were male, and 40% 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 9 wolves spotted in the region, what is the probability that 6 or more were male?


b) What is the probability that 6 or more were female?


c) What is the probability that fewer than 3 were female?


d) For the period from 1918 to the present, in a random sample of 9 wolves spotted in the region, what is the probability that 6 or more were male?


e) What is the probability that 6 or more were female?


f) What is the probability that fewer than 3 were female?

Answer 1

(a)

Let X and Y be the random variable denoting the number of male and female wolves respectively in a sample.
Before 1918:
P(X) = 0.60
P(Y) = 0.40

$P(X\geq 6) = \sum_{k=6}^{9}P(X=k)$
$P(X=k) = _{k}^{9}\textrm{C}(0.60)^{k}(0.40)^{9-k}$
P(X=6) = 0.2508
P(X=7) = 0.1612
P(X=8) = 0.0605
P(X=9) = 0.0101
$P(X\geq 6) = 0.2508 + 0.1612 + 0.0605 + 0.0101 = 0.4826$

(b)

$P(Y\geq 6) = \sum_{k=6}^{9}P(Y=k)$
$P(Y=k) = _{k}^{9}\textrm{C}(0.40)^{k}(0.60)^{9-k}$
P(Y=6) = 0.0743
P(Y=7) = 0.0212
P(Y=8) = 0.0035
P(Y=9) = 0.0003
$P(Y\geq 6) = 0.0743 + 0.0212 + 0.0035 + 0.0003 = 0.0993$

(c)

P(Y=0) = 0.0101
P(Y=1) = 0.0605
P(Y=2) = 0.1612
P(Y < 3) = 0.2318

(d)
After 1918:
P(X) = 0.65
P(Y) = 0.35

$P(X\geq 6) = \sum_{k=6}^{9}P(X=k)$
$P(X=k) = _{k}^{9}\textrm{C}(0.65)^{k}(0.35)^{9-k}$
P(X=6) = 0.2716
P(X=7) = 0.2162
P(X=8) = 0.1004
P(X=9) = 0.0207
$P(X\geq 6) = 0.2716 + 0.2162 + 0.1004 + 0.0207 =0.6089$

(e)
$P(Y\geq 6) = \sum_{k=6}^{9}P(Y=k)$
$P(Y=k) = _{k}^{9}\textrm{C}(0.35)^{k}(0.65)^{9-k}$
P(Y=6) = 0.0424
P(Y=7) = 0.0098
P(Y=8) = 0.0013
P(Y=9) = 0.0000
$P(Y\geq 6) = 0.0424 + 0.0098 + 0.0013+ 0.0000 = 0.0535$

(f)


P(Y=0) = 0.0207
P(Y=1) = 0.1004
P(Y=2) = 0.2162
P(Y < 3) = 0.3373