# 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?

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 $$0.5-0.05=0.45$$

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 $a=30$.