How do you solve this statistics question

An experimenter publishing in the Annals of Botany investigated whether the stem diameters of the dicot sunflower would change depending on whether the plant was left to sway freely in the wind or was artificially supported. Suppose that the unsupported stem diameters at the base of a particular species of sunflower plant have a normal distribution with an average diameter of 35 millimeters (mm) and a standard deviation of 3 mm.

(a) What is the probability that a sunflower plant will have a basal diameter of more than 37 mm? (Round your answer to four decimal places.)
=.2514

(b) If two sunflower plants are randomly selected, what is the probability that both plants will have a basal diameter of more than 37 mm? (Round your answer to four decimal places.)
=.0632

(c) Within what limits would you expect the basal diameters to lie, with probability 0.95? (Round your answers to two decimal places.)
lower limit 29.12 mm
upper limit 40.88 mm

(d) What diameter represents the 60th percentile of the distribution of diameters? (Round your answer to two decimal places.)
=????
mm

Let X be the unsupported stem diameters at the base of a particular species of sunflower plant.

X ∼ n(35; 3)

a)

P(X ≥ 37) = P((X - 35)/3 ≥ (37 - 35)/3) = P(Z ≥ 0.67) = 0.251428

b)

n = 2

By Limit Central Theorem,

Xmean ∼ n(35; 3/sqrt(2))

3/sqrt(2) = 2.121320344

P(X ≥ 37) = P((X - 35)/2.121320344 ≥ (37 - 35)/2.121320344) = P(Z ≥ 0.94) = 0.173609

c)

P(x1 < X < x2 ) = 0.95

P((x1 - 35)/3 < (X - 35)/3 < (x2 - 35)/3 ) = 0.95

P( - z1 < Z < z2) = 0.95

P( - 1.96 < Z <1.96) = 0.95

- 1.96 = (x1 - 35)/3

x1 = 29.12mm, lower limit

1.96 = (x2 - 35)/3

x2 = 40.88mm, upper limit

d)

P(X ≥ x) = 0.6

P((X - 35)/3 ≥ (x - 35)/3) = 0.6

P(Z ≥ - z) = 0.6

z = - 0.253347

- 0.253347 = (x - 35)/3

x = 34mm