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dolphi86 [110]
2 years ago
13

Suppose the time it takes a certain printer to print a document is an Exponential random variable with an average time of 15 sec

onds. You send a document to the printer at 1:00pm, and it is the third document in the queue. What is the probability that your printout will be ready by 1:01pm
Mathematics
1 answer:
dybincka [34]2 years ago
3 0

Answer:

0.8046

Step-by-step explanation:

We're asked to calculate the probability that the job will be ready by 1.01 pm

From the question, we can note that the parameter of an Exponential is E(X)= 15

Next, to calculate the third job probability, we will have to use Poisson Distribution with parameter 1/λ

Therefore, E(Y) = 1/15

Then, The third job will be ready for 1:01 pm, and thus E(Y) = 61/15

Therefore, the required probability is

P(X ≥ 3) = 1 - P(X < 3)

= 1 - Poisson(3,4, true)

= 1 - 0.1954

= 0.8046

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x=-7

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Step-by-step explanation:

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Jessie spent $10 on 3 hair accessories. Which point represents this relationship? On a coordinate plane, a graph titled Jessie's
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  A.  (3, 10)

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You are told that ...

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so the ordered pair is ...

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Let X denote the temperature (degree C) and let Y denote thetime in minutes that it takes for the diesel engine on anautomobile
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Step-by-step explanation:

Given f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y

a)

we know that \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1

therefore \int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1

on integrating we get

c=(1/6640)

b)

P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

on doing the integration we get

                        =0.37349

c)

marginal density of X is

f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy

on doing integration we get

f(x)=(4x+3)/3320 ; 0<x<40

marginal density of Y is

f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx

on doing integration we get

f(y)=\frac{(y+40.5)}{83}

d)

P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

solve the above integration we get the answer

e)

P(X>20, 0

solve the above integration we get the answer

f)

Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.

we know f(x,y)

In the (c) bit we got f(x) and f(y)

f(x,y)cramster-equation-2006112927536330036287f(x).f(y)

therefore X and Y are not independent

4 0
2 years ago
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