Quantile Function of a Normal RV

Question

I am wondering if I can solve this with Quantile function.

Suppose LCD screens have lifetimes that are normally distributed with a mean lifetime of 18000 hours with a variance of 1000000 hours. (c) The top 10% of lifetimes will be at least how long?

This is my solution P(Z≥z)=1-P(Z≤z)=0.10 but can I find by using the inverse of the CDF? If so help me understand when I do need use inverse of the CDF because I do not see anything in my problem - find quantile.

Answer 1

1. the variance cannot be expressed in hours....the standard deviation can.

2. To calculate the inverse of the CDF you can use tha gaussian tables.

Thus your solution (you did it correctly) is

$$\mathbb{P}[Z \geq z]=0.1$$

$$\Phi\Bigg[\frac{x-18,000}{1,000}\Bigg]=0.9$$

$$\frac{x-18,000}{1,000}=1.28$$

$$x\approx 19,280$$

using a calculator you can have a more precise result $x=19,281.55$