Find the value you are 90% sure the price will be above

Question

I have been stuck on a problem and am not even sure if I am going the right direction.

Problem is as follows: "Tomorrow's prices of items in a shop (assume normally distributed random variable P) are given by $\mu = 10$ and standard deviation $= 2$" I am to find the value I am $90\%$ sure the price will be above.

I applied Chebyshev's inequality (which may be completely wrong), and did:

$$P(|E - E[X]| \geq k) \leq \frac{\sigma^2}{k^2}$$

We want to be $90\%$ sure, so $\frac{\sigma^2}{k^2}$ has to equal $\frac{1}{10}$, therefore $\frac{4}{k^2} = \frac{1}{10}$, $k^2 = 40$ and therefore $k = \sqrt{40}$.

Putting this all together gives $P(|X - 10| \geq \sqrt{40} \leq \frac{1}{10}$

However this does not actually give me the value. Any tips?

Answer 1

Let $P \sim N(10,4)$.

We want $P(P \geq k) = 0.9$, which is the same as $1 - P(P \leq k) = 0.9$ or $P(P\leq k) = 0.1$

To find $k$, first we standardize $P$:

$$Z = \frac{P-\mu}{\sigma} = \frac{P-10}{2} \sim N(0,1)$$

So

$$P(P \leq k) = 0.1 \iff P\left(\frac{P-10}{2} \leq \frac{k-10}{2}\right) = 0.1$$

$$\iff P(Z \leq z) = 0.1 \qquad \left(z = \frac{k-10}{2}\right)$$

Using a standard normal distribution table we have $z = -1.3$ so

$$-1.3 = \frac{k-10}{2} \implies k = 7.4$$

Answer 2

You want to find the value $q$ such that for a random variable $X$ which has $N(10,4)$ distribution you have $$0.9=\mathbb{P}(X>q)=P(\frac{X-10}{2}>\frac{q-10}{2})$$

Now notice that $Z=\frac{X-10}{2}$ has a $N(0,1)$ distribution, hence you want to find $q$ such that

$$\mathbb{P}(Z>\frac{q-10}{2})=0.9$$

A statistical table gives the value $\frac{q-10}{2}=-1.28$, hence you get $q=7.44$