



Customers arrive at a service center with independent interarrival
times in hours, which have exponential (3) distribution. The time
X for the
third arrival is thus gamma
$(3,\phantom{\rule{0.277778em}{0ex}}3)$ . Without using tables or mprograms,
determine
$P(X\le 2)$ .
$$P(X\le 2)=P(Y\ge 3),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}Y\sim \phantom{\rule{0.277778em}{0ex}}\text{poisson}(3\xb72=6)$$
$$P(Y\ge 3)=1P(Y\le 2)=1{e}^{6}(1+6+36/2)=0.9380$$
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Five people wait to use a telephone, currently in use by a
sixth person. Suppose time for the six calls (in minutes) are iid, exponential (1/3).What is the distribution for the total time
Z from the present for the six calls? Use an
appropriate Poisson distribution to determine
$P(Z\le 20)$ .
$Z\sim $ gamma (6,1/3).
$$P(Z\le 20)=P(Y\ge 6),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}Y\sim \phantom{\rule{0.277778em}{0ex}}\text{poisson}(1/3\xb720)$$
$$P(Y\ge 6)=\mathtt{c}\mathtt{p}\mathtt{o}\mathtt{i}\mathtt{s}\mathtt{s}\mathtt{o}\mathtt{n}(\mathtt{20}/\mathtt{3},\mathtt{6})\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\mathtt{0}.\mathtt{6547}$$
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A random number generator produces a sequence of numbers between
0 and 1. Each of these can be considered an observed value of a random variableuniformly distributed on the interval [0, 1]. They assume their values independently.A sequence of 35 numbers is generated. What is the probability 25 or more are
less than or equal to 0.71? (Assume continuity. Do not make a discrete adjustment.)
p = cbinom(35,0.71,25) = 0.5620
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Five “identical” electronic devices are installed at one time. The
units fail independently, and the time to failure, in days, of each is a randomvariable exponential (1/30). A maintenance check is made each fifteen days. What is
the probability that at least four are still operating at the maintenance check?
p = exp(15/30) = 0.6065 P = cbinom(5,p,4) = 0.3483
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Suppose
$X\sim N(4,81)$ . That is,
X has gaussian distribution with mean
$\mu =4$ and variance
${\sigma}^{2}=81$ .
 Use a table of standardized normal distribution to
determine
$P(2<X<8)$ and
$P\left(\rightX4\le 5)$ .
 Calculate the probabilities in part (a) with the mfunction gaussian.

$$P(2<X<8)=\Phi \left(\right(84)/9)\Phi \left(\right(24)/9)=$$
$$\Phi (4/9)+\Phi (2/9)1=0.6712+0.58751=0.2587$$
$$P\left(\rightX4\le 5)=2\Phi (5/9)1=1.42121=0.4212$$

P1 = gaussian(4,81,8)  gaussian(4,81,2)
P1 = 0.2596P2 = gaussian(4,81,9)  gaussian(4,84,1)
P2 = 0.4181
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Suppose
$X\sim N(5,\phantom{\rule{0.277778em}{0ex}}81)$ . That is,
X has gaussian distribution with
$\mu =5$ and
${\sigma}^{2}=81$ . Use a table of standardized normal distribution to
determine
$P(3<X<9)$ and
$P\left(\rightX5\le 5)$ . Check your results using the
mfunction gaussian.
$$P(3<X<9)=\Phi \left(\right(95)/9)\Phi \left(\right(35)/9)=\Phi (4/9)+\Phi (2/9)1=0.6712+0.58751=0.2587$$
$$P\left(\rightX5\le 5)=2\Phi (5/9)1=1.42121=0.4212$$
P1 = gaussian(5,81,9)  gaussian(5,81,3)
P1 = 0.2596P2 = gaussian(5,81,10)  gaussian(5,84,0)
P2 = 0.4181
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Suppose
$X\sim N(3,\phantom{\rule{0.277778em}{0ex}}64)$ . That is,
X has gaussian distribution with
$\mu =3$ and
${\sigma}^{2}=64$ . Use a table of standardized normal distribution to
determine
$P(1<X<9)$ and
$P\left(\rightX3\le 4)$ . Check your results with the
mfunction gaussian.
$$P(1<X<9)=\Phi \left(\right(93)/8)\Phi \left(\right(13)/9)=$$
$$\Phi (0.75)+\Phi (0.25)1=0.7734+0.59871=0.3721$$
$$P\left(\rightX3\le 4)=2\Phi (4/8)1=1.38291=0.3829$$
P1 = gaussian(3,64,9)  gaussian(3,64,1)
P1 = 0.3721P2 = gaussian(3,64,7)  gaussian(3,64,1)
P2 = 0.3829
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Items coming off an assembly line have a critical dimension which
is represented by a random variable
$\sim $ N(10, 0.01). Ten items are selected
at random. What is the probability that three or more are within 0.05 of themean value
μ .
p = gaussian(10,0.01,10.05)  gaussian(10,0.01,9.95)
p = 0.3829P = cbinom(10,p,3)
P = 0.8036
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The result of extensive quality control sampling shows that a certain model
of digital watches coming off a production line have accuracy, in seconds per month,that is normally distributed with
$\mu =5$ and
${\sigma}^{2}=300$ . To achieve
a top grade, a watch must have an accuracy within the range of 5 to +10 secondsper month. What is the probability a watch taken from the production line to be
tested will achieve top grade? Calculate, using a standardized normal table. Checkwith the mfunction gaussian.
$P(5<X<10)=\Phi (5/\sqrt{300})+\Phi (10/\sqrt{300})1=\Phi (0.289)+\Phi (0.577)1=0.614+0.7171=0.331$
$$\mathtt{P}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\mathtt{g}\mathtt{a}\mathtt{u}\mathtt{s}\mathtt{s}\mathtt{i}\mathtt{a}\mathtt{n}(\mathtt{5},\mathtt{300},\mathtt{10})\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathtt{g}\mathtt{a}\mathtt{u}\mathtt{s}\mathtt{s}\mathtt{i}\mathtt{a}\mathtt{n}(\mathtt{5},\mathtt{300},\mathtt{5})\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\mathtt{0}.\mathtt{3317}$$
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Use the mprocedure bincomp with various values of
n from 10 to 500 and
p from 0.01 to 0.7, to observe the approximation of the binomial distribution by
the Poisson.
Experiment with the mprocedure bincomp.
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Use the mprocedure poissapp to compare the Poisson and gaussian distributions.
Use various values of
μ from 10 to 500.
Experiment with the mprocedure poissapp.
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Random variable
X has density
${f}_{X}\left(t\right)=\frac{3}{2}{t}^{2},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1\le t\le 1$ (and zero elsewhere).
 Determine
$P(0.5\le X<0,8)$ ,
$P\left(\rightX>0.5)$ ,
$P\left(\rightX0.25\le 0.5)$ .
 Determine an expression for the distribution function.
 Use the mprocedures tappr and cdbn to plot an approximation to the
distribution function.
$$\frac{3}{2}\int {t}^{2}={t}^{3}/2$$

$$P1=0.5*(0.{8}^{3}{(0.5)}^{3})=0.3185\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P2=2{\int}_{0.5}^{1}\frac{3}{2}{t}^{2}=(1{(0.5)}^{3})=7/8$$
$$P3=P\left(\rightX0.25\le 0.5)=P(0.25\le X\le 0.75)=\frac{1}{2}[{(3/4)}^{3}{(1/4)}^{3}]=7/32$$

${F}_{X}\left(t\right)={\int}_{1}^{t}{f}_{X}=\frac{1}{2}({t}^{3}+1)$

tappr
Enter matrix [a b]of xrange endpoints [1 1]
Enter number of x approximation points 200Enter density as a function of t 1.5*t.^2
Use row matrices X and PX as in the simple casecdbn
Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % See MATLAB plot
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Random variable
X has density function
${f}_{X}\left(t\right)=t\frac{3}{8}{t}^{2},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le t\le 2$ (and zero elsewhere).
 Determine
$P(X\le 0.5)$ ,
$P(0.5\le X<1.5)$ ,
$P\left(\rightX1<1/4)$ .
 Determine an expression for the distribution function.
 Use the mprocedures tappr and cdbn to plot an approximation to the
distribution function.
$$\int (t\frac{3}{8}{t}^{2})=\frac{{t}^{2}}{2}\frac{{t}^{3}}{8}$$

$$P1=0.{5}^{2}/20.{5}^{3}/8=7/64\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P2=1.{5}^{2}/21.{5}^{3}/87/64=19/32\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P3=79/256$$

${F}_{X}\left(t\right)=\frac{{t}^{2}}{2}\frac{{t}^{3}}{8},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le t\le 2$

tappr
Enter matrix [a b]of xrange endpoints [0 2]
Enter number of x approximation points 200Enter density as a function of t t  (3/8)*t.^2
Use row matrices X and PX as in the simple casecdbn
Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % See MATLAB plot
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Random variable
X has density function
$${f}_{X}\left(t\right)=\left\{\begin{array}{cc}(6/5){t}^{2}& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le t\le 1\hfill \\ (6/5)(2t)& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1<t\le 2\hfill \end{array}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}=I[0,1]\left(t\right)\frac{6}{5}{t}^{2}+{I}_{(1,2]}\left(t\right)\frac{6}{5}(2t)$$
 Determine
$P(X\le 0.5)$ ,
$P(0.5\le X<1.5)$ ,
$P\left(\rightX1<1/4)$ .
 Determine an expression for the distribution function.
 Use the mprocedures tappr and cdbn to plot an approximation to the
distribution function.

$$P1=\frac{6}{5}{\int}_{0}^{1/2}{t}^{2}=1/20\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P2=\frac{6}{5}{\int}_{1/2}^{1}{t}^{2}+\frac{6}{5}{\int}_{1}^{3/2}(2t)=4/5$$
$$P3=\frac{6}{5}{\int}_{3/4}^{1}{t}^{2}+\frac{6}{5}{\int}_{1}^{5/4}(2t)=79/160$$

$${F}_{X}\left(t\right)={\int}_{0}^{t}{f}_{X}={I}_{[0,1]}\left(t\right)\frac{2}{5}{t}^{3}+{I}_{(1.2]}\left(t\right)[\frac{7}{5}+\frac{6}{5}(2t\frac{{t}^{2}}{2})]$$

tappr
Enter matrix [a b]of xrange endpoints [0 2]
Enter number of x approximation points 400Enter density as a function of t (6/5)*(t<=1).*t.^2 + ...
(6/5)*(t>1).*(2  t)
Use row matrices X and PX as in the simple casecdbn
Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % See MATLAB plot
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Questions & Answers
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
is there industrial application of fullrenes.
What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.?
How this robot is carried to required site of body cell.?
what will be the carrier material and how can be detected that correct delivery of drug is done
Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
nano basically means 10^(9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
Source:
OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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