Question

In a recent study, the distances of certain raptor nests to the nearest marshland were studied. The study found that these distances are normally distributed with mean 4.42 km and standard deviation 0.86 km. Let Y be the distance of a randomly selected nest to the nearest marshland. Complete parts (a) and (b) Click here to view page 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table. a. Determine P(Y > 5) P(Y > 5)-| | (Round to four decimal places as needed.) b. Determine P(3 sYs6) P(3sYs6)(Round to four decimal places as needed.)

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 4.42 km

Standard deviation = 0.86 km

a) P(Y > 5) = 1 - P(Y < 5)

= 1 - P(Z < (5 - 4.42)/0.86)

= 1 - P(Z < 0.67)

= 1 - 0.7486

= 0.2514

b) P(3 $\small \leq$ Y $\small \leq$ 6) = P(Y < 6) - P(Y < 3)

= P(Z < (6 - 4.42)/0.86) - P(Z < (3 - 4.42)/0.86)

= P(Z < 1.84) - P(Z < -1.65)

= 0.9671 - 0.0495

= 0.9176

a) Area to the right of z = -0.92, P(z > -0.92)

= 1 - P(z < -0.92)

= 1 - 0.1788

= 0.8212

b) Area to the right of z = -1.19, P(z > -1.19)

= 1 - P(z < -1.19)

= 1 - 0.1170

= 0.8830

c) Area to the right of z = 0.15, P(z > 0.15)

= 1 - P(z < 0.15)

= 1 - 0.5596

= 0.4404

d) Area to the right of z = 1.55, P(z > 1.55)

= 1 - P(z < 1.55)

= 1 - 0.9394

= 0.0606