Question

. Let X denote the wing length in millimeters of a male gallinule and Y the wing length in millimeters of a female gallinule. Assume that X is N(184.09, 39.37) and Y is N(171.93, 50.88) and that X and Y are independent. If a male and a female gallinule are captured, what is the probability that X is greater than Y

Answer 1

Answer:

The probability that X is greater than Y is 0.5754.

Step-by-step explanation:

It is provided that,

$X\sim N(184.09, 39.37)\\\\Y\sim N(171.93, 50.88)$

It is also provided that X and Y are independent.

We need to compute the value of:

P (X > Y) = P (X - Y > 0)

Compute the mean and standard deviation of (X - Y) as follows:

$E(X-Y)=E(X)-E(Y)=184.09-171.93=12.16\\\\SD(X-Y)=\sqrt{V(X-Y)}=\sqrt{V(X)+V(Y)}=\sqrt{39.37^{2}+50.88^{2}}=64.33$

Compute the the value of P (X > Y) as follows:

$P(X>Y)=P(X-Y>0)\\\\=P(\frac{(X-Y)-E(X-Y)}{SD(X-Y)}>\frac{0-12.16}{64.33})\\\\=P(Z>-0.19)\\\\=P(Z<0.19)\\\\=0.57535\\\\\approx 0.5754$

Thus, the probability that X is greater than Y is 0.5754.