Find the parameters of this distribution and the probability The mean weight of a leopard is 190kg. We look at the leopards in the Zoo and assume their weight to be normally distributed. We know that 5% of all leopards weigh more than 220kg.
(a) Determine the parameters µ and σ for this distribution.
(b) What is the probability that a randomly chosen leopard weighs between 160kg and 190kg 
So lets say that W= weight in kg, E[W]=190kg
a)$0.95=P[W \le 220]= P[\frac{W-190}{\sigma \le \frac{30}{\sigma}}]=\phi \frac{30}{\sigma}≈ 1.64$
b)$P[160 ≤ G ≤ 190] = P[G ≤ 190]−P[G ≤ 160] = Φ(0/σ)−Φ(30/σ) ≈ Φ(2.87)−Φ(2.05)≈ 0.018$
Is this correct? I was unsure on how to calculate this. Are the values right?
 A: For $(a)$ we have
$$\begin{align*}
\mathsf P(W\gt220)
&=1-\mathsf P(W<220)\\\\
&=1-\Phi\left(\frac{30}{\sigma}\right)\\\\
&=0.05
\end{align*}$$
We have then that $\Phi\left(\frac{30}{\sigma}\right)=0.95$ which occurs when $$\frac{30}{\sigma}=1.645\Rightarrow \sigma\approx18.24$$
More accurately than a standard normal table,  R statistical software gives
> qnorm(.95)
[1] 1.644854

> 30/qnorm(.95)
[1] 18.2387

You now have the ingredients to solve for $(b)$ except you should have
$$\begin{align*}
\mathsf P(160<W<190)
&=\mathsf P(W<190)-\mathsf P(W<160)\\\\
&=\Phi\left(\frac{0}{18.24}\right)-\Phi\left(\frac{-30}{18.24}\right)\\\\
&\approx0.45
\end{align*}$$
Alternatively, you can note that we are given that $5$% of leopards weigh more that $220$ so by symmetry, $5$% of leopards weigh less than $160$ giving $$0.5-0.05=0.45$$
A: You can calculate b) without resorting to normal table look up.
Due to symmetry of the normal curve around its mean
$$ 
P(\mu-a < W < \mu) = P(\mu< W < \mu + a) = 0.5 - P(W > \mu+a) = 0.45
$$
here $a=30$.
