Question

A random number generator picks a number from one to nine in a uniform manner.

a. X ~ _________

b. Graph the probability distribution.

c. f(x) = _________

d. μ = _________

e. σ = _________

f. P(3.5 < x < 7.25) = _________

g. P(x > 5.67)

h. P(x > 5|x > 3) = _________

i. Find the 90th percentile

Short answer

a. $X~\upsilon \left(1,9\right)$

b.

c.

$f\left(x\right)=\left\{\begin{array}{l}\frac{1}{8}\\ 0\end{array},1<x<9\right.$

d. $\mu =5$

e.$\sigma =2.31$

f.$P(3.5<x<7.25)=0.4688$

g.$P(x>5.67)=0.41625$

h.$P(x>5|x>3)=0.67$

i. The 90th percentile is 8.2

Step-by-step solution

 Definition of measured variables 

The measurement variable means the expressions of some types of measurements which has an associated number.

Findings the results of measured variables 

a. Uniform of all data

$X~U(1,9)$

The height of probability distribution

b. the probability distribution function is

localid="1649324054718">f(x)=1b-aa=1b=53f(x)=19-1=18

The height of the distribution function= 1/8

Graph the probability distribution

c. The formulation of the

$\begin{array}{rcl}f\left(x\right)& =& \left\{\begin{array}{l}\frac{1}{b-a}a<X<b\\ 0\end{array}\right.\\ & & 0forotherwise\\ f\left(x\right)& =& \left\{\begin{array}{l}\frac{1}{8}1<X<8\\ 0\end{array}\right.\\ & & 0forotherwise\end{array}$

d. The mean value is

$\begin{array}{rcl}\mu & =& \frac{a+b}{2}\\ & =& \frac{1+9}{2}\\ & =& 5\end{array}$

Then, the required value is 5

e. The standard deviation

localid="1649328736962" $\begin{array}{rcl}\sigma & =& \frac{\sqrt{{\left(b-a\right)}^2}}{12}\\ \sigma & =& \frac{\sqrt{{\left(9-1\right)}^2}}{12}\\ & =& 2.31\end{array}$

Then, the required value is 2.31.

f. The required probability is as below:

localid="1649328744005" $\begin{array}{rcl}P(3.5& <& x<7.25)=base\times height\\ & =& (7.25-3.5)\times \frac{1}{8}\\ & =& 0.4688\end{array}$

Then, the required value is 0.4688

g. The required probability is as below:

$\begin{array}{rcl}P(x& >& 5.67)=base\times height\\ & =& (9-5.67)\times \frac{1}{8}\\ & =& 0.41625\end{array}$

Then, the required value is 0.41625

h. The required probability is calculated as below:

$\begin{array}{rcl}P(x& >& 5|x>3)=base\times height\\ & =& (9-5)\times \frac{1}{\left(9-3\right)}\\ & =& 0.67\end{array}$

Therefore, the required value is 0.67

i. The 19th percentile of the data can be calculated as below:

For 90 percentile

localid="1649328754478" $\begin{array}{rcl}P(x& <& k)=base\times height\\ 0.90& =& (k-1)\times \frac{1}{8}\\ k& =& 8.2\end{array}$

The value of the 90th percentile is 8.2