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

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

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