Question

what is the solution of 99th percentile?

Answer 1

When we say that an individual’s test score was at the 99th percentile of the population, we mean that 99% of all population scores were below that score and 1% were above. 

Let $p$ be a number between 0 and 1. The $(100p)$th percentile of the distribution of a continuous rv $X,$ denoted by $\eta(p)$, is defined by

$$p=F(\eta(p))=\displaystyle\int_{-\infin}^{\eta(p)}f(y)dy$$

Acoording the expression $\eta(0.99),$ the 99th percentile, is that value on the measurement axis such that $100(0.99)\%$ of the area under the graph of $f(x)$ lies to the left of $\eta(p)$ and $100(1-0.99)\%$ lies to the right.

For a example for the normal distribution the formula below is used to compute percentiles of a normal distribution.

$$X=\mu+Z\sigma$$

where $\mu$ is the mean and $\sigma$ is the standard deviation of the variable $X,$and $Z$ is the value from the standard normal distribution for the desired percentile.

For the 99th percentile $\alpha=0.01$ and $z_{\alpha}=2.3263.$