4.
The main thing here is to realise that the area under a probability density function has to add to 1. So if we integrate it and let it equal 1, we can solve for k. Also that we need to use limits to evaluate this one as we get an
in one of the terminals.
a)
\right) & =1<br />\\ \underset{t\rightarrow\infty}{\lim}\left(k(1-e^{-t})\right) & =1<br />\end{alignedat}<br />)
Now what happens as t approaches infinity?
will become very small and approach
, so we are left with
(If you have trouble with this part, visualise it and draw it out
)
b) Now the
)
will be given by the area under the probability density function from negative infinity to 2, but in this case the function is 0 for y<0, so our terminals become 0 to 2.
<br />\\ & =1-e^{-2}<br />\\ & =0.865<br />\end{alignedat})
5) t-intercepts at
and
, since there is a negatie coefficeint on only one of the t's, it will be a parabola with a maximum. The function will need to be restricted to
as the probability density function has to be equal to or greater than 0 for all t.
a)
b) Now for it to appear within 7 days of contact we are looking for the area under the probability density function for
.
 & =\int_{5}^{7}\frac{1}{36}(t-5)(11-t)dt<br />\\ & =\frac{1}{36}\int_{5}^{7}-t^{2}+6t-55dt<br />\\ & =\frac{1}{36}\left[-\frac{1}{3}t^{3}+3t^{2}-55t\right]_{5}^{7}<br />\\ & =\frac{7}{27}<br />\\ & =0.259<br />\end{alignedat})
EDIT: Added q5 and graph.
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