Understanding confidence intervals and percentiles 
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 $40$ mm?
b. If two sunflower plants are randomly selected, what
is the probability that both plants will have a basal
diameter of more than $40$ mm?
c. Within what limits would you expect the basal
diameters to lie, with probability $.95$ ?
d. What diameter represents the $90$th percentile of the
distribution of diameters?
$\mathbf{\text{MY ATTEMPT:}}$
a-) $P(X >40)=P(Z>1.66)=0.0475$
b-) for this part , i tried to use the formula for selection more than one elements such that $$P\bigg(Z > \frac{40-35}{3 / \sqrt{2}}\bigg)=P\bigg(Z > \frac{5}{2.12132}\bigg) =P\bigg(Z >2.357\bigg)=0.00921$$
However , the answer is $0.00226$ and it has such a solution : $$P(Z>1.66) \times P(Z>1.66)=(0.0475)^2 $$
Why didn't my formula work ? I always use it to find normal dist. probabilities when we choose more than one object.
c-) I could not do this with mathematical formula , the only way i could do is to seach for values to satisfy this condition in z table , so i am looking for a nice answer for it. By the way , the given answer is $29.12$ to $40.88$
d-) I have never solved a percentile problem , answer says that $P(Z \geq -z) =0.9$ and answr is $38.84$. I could not understand why they used "-z" insteaf of "+z" and why we used $"\geq"$ instead of $>$.
Thanks in advance..
 A: Since this is clearly homework, I won't do all the work for you. Instead I will try to guide you in the right direction.
Part b):  When two events $A$ and $B$ are assumed independent (such as choosing something randomly), then $P(A \text{ and } B)=P(A)P(B)$. Do you know the probability that one sunflower plant will have a basal diameter of more than 40 mm? Now the probability of both being greater than 40 mm should be this probability times itself.
Part c): For a Normal distribution, values will lie in the following interval with 95% probability: $$[\mu - z_{0.975} \cdot \sigma,\mu + z_{0.975} \cdot \sigma].$$You look in your "Z-table" the value of $z_{0.975}$. If you want to know more about this, try googling confidence intervals for the normal distribution.
Part d): Let $X$ denote the sunflower basal diameter. To find the 90% percentile for $X$ you are asking what value of $x$ satisfies that
$$
0.9 \overset{!}{=}P(X \leq x) = F(x)
$$
Hence taking the inverse of the cumulative distribution function $F$ on both sides you get that
$$
x = F^{-1}(0.9)
$$
This value can be calculated by most advanced calculators/stastical software or you can similarly look the corresponding value up for the standard normal distribution (which is what you call $Z$)

Since the density of a Normal distribution is symmetrical, we have that $P(Z \leq z) = P(Z \geq -z)$. Since the Normal distribution is (absolutely) continuous, we have that $P(Z=z)=0$ and hence $P(Z > -z) = P(Z \geq -z)$.
A: a) correct...even if the exact result is $\approx 0.0478$
b) incorrect: the result is $(0.0475)^2\approx 0.0023$. This because you are requested to calculate the probability of the intersection of two independent events
Your formula is not correct because it is related to the probability of the sample mean of two objects
c) the distribution of the diameter is $D\sim N(35;3^2)$. Thus we expect to find the 95% of the distribution in the range
$$\mu\pm \sigma z_{0.975}$$
that is
$$35\pm 3\cdot1.96$$
or, equivalently
$$[29.12;40.88]$$
d) using Gaussian CDF (and tables to calculate the requested percentile), you get
$$P(D\le d)=0.90$$
that is
$$\frac{d-35}{3}=1.28$$
$$d=38.84$$

$\ge$ or $>$ is the same given that the distribution is continuous and thus
$$P(D=d)=0$$
$\forall d$
