Question

A distribution is given as $X~U(0,20)$. What is $P(2<x<18)$? Find the${90}^{\mathrm{th}}$percentile.

Short answer

The value of $P(2<x<18)$is $0.8$.

The value of the $90{}^{th}$percentile is$18$.

Step-by-step solution

Given Information 

Given in the question that, $X~U(0,20)$

We need to find what is$P(2<x<18)$and${90}^{th}$percentile.

Calculate the probability density function 

Here, $X$remains a random variable follows uniform distribution as $X~U(\mathrm{a},\mathrm{b})$

From the question $X~U(0,20)$

Where,

$a=0$

$b=20$

Hence, the probability density function or height of it will be

$f\left(x\right)=\frac{1}{b-a}$

$=\frac{1}{20-0}$

$=\frac{1}{20}$

Calculate the value of P(2<x<18)

Let's find the required probability,

$P(2<x<18)=\text{base}\times \text{height}$

$=(18-2)\times \left(\frac{1}{20}\right)$

$=16\times \frac{1}{20}$

$=0.8$

Calculate the 90th percentile

Let's calculate the ${90}^{th}$percentile for $0<x<20$as below

$P(x<k)=\text{base}\times \text{height}$

$0.90=(k-0)\left(\frac{1}{20}\right)$

$\text{or}k-0=0.90\times 20$

$k=18$