Finding µ and σ using N (µ, σ^2)

Question

Suppose that 10% of the probability for a certain distribution that is N (µ, σ^2 ) is below 60 and that 5% is above 90. What are the values of µ and σ?

This is what I have so far:

P(x < 60) = 0.1 // since 10% of the probability is below 60

P(x > 90) = 0.05 // since  5% of the probability is above 90

$${P({(x-µ)\overσ} < {(60-µ)\overσ} = 0.1)} $$

$${P({(x-µ)\overσ} < {(90-µ)\overσ} = 0.95)} $$

Step 2:

$$(1) {P(Z < {(60-µ)\overσ} = 0.1)} $$ $$(2) {P(Z < {(90-µ)\overσ} = 0.95)}$$ // Why is (2) = 0.95 and not 0.9?

Step 3:

$${{(60-µ)\overσ} = -1.282)} $$ $${{(90-µ)\overσ} = 1.645)} $$

Step 4:

$${{(60-µ)} = -1.282σ)} $$ $${{(90-µ)} = 1.645σ)} $$

The final answer is provided but I am unsure on how to get to it. Any help on this would be appreciated

Solution µ = 73.14 and σ = 10.25

Answer 1

Hint: find values $z_1,\,z_2$ with$$\mu+z_1\sigma=60,\,\mu+z_2\sigma=90$$by solving$$\Phi(z_1)=0.1,\,\Phi(z_2)=0.95,$$with $\Phi$ the $N(0,\,1)$ CDF. Now solve simultaneous equations for $\mu,\,\sigma$.

Answer 2

$P(X<60) $

$= P(Z < \frac{60-\mu}{\sigma})=0.1=\Phi(-1.281552)$

$=P(Z < -1.281552)$

$\implies 60-\mu = -1.281552\sigma$

Similarly,

$P(X\leq 90)=1-P(X>90)=1-0.05=0.95$

$=P(Z \leq \frac{90-\mu}{\sigma}) =\Phi(1.644854)$

$=P(Z \leq 1.644854)$

$\implies 90-\mu = 1.644854\sigma$

Now, solving the $2 \times 2$ system of linear equations

$\mu -1.281552\sigma = 60$

$\mu + 1.644854\sigma = 90$

you have $\sigma = \frac{30}{1.281552 + 1.644854}=10.25148$ and

$\mu=60+1.281552*10.25148=73.1378$