Let T be the time needed to complete a job at a certain factory. By us

nemired9 2021-12-11 Answered
Let T be the time needed to complete a job at a certain factory. By using the historical data, we know that
\[P(T \le t)=\begin{cases}\frac{1}{16}t^{2}& for\ 0 \le t \le 4\\1& for\ t \geq 4 \end{cases}\]
a. Find the probability that the job is completed in less than one hour, i.e., find \(\displaystyle{P}{\left({T}\le{1}\right)}\).
b. Find the probability that the job needs more than 2 hours.
c. Find the probability that \(\displaystyle{1}\le{T}\le{3}\).

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Expert Answer

Lynne Trussell
Answered 2021-12-12 Author has 5247 answers
Step 1
Given:
Let T be the time needed to complete a job at a certain factory.
We know that,
\[P(T \le t)=\begin{cases}\frac{1}{16}t^{2}& for\ 0 \le t \le 4\\1& for\ t \geq 4 \end{cases}\]
We want to find,
(a) \(\displaystyle{P}{\left({T}\le{1}\right)}=?\)
(b) \(\displaystyle{P}{\left({T}\geq{2}\right)}=?\)
(c) \(\displaystyle{P}{\left({1}\le{T}\le{3}\right)}=?\)
Step 2
Solution:
(a) From the given information
\(\displaystyle{P}{\left({T}\le{1}\right)}={\frac{{{1}}}{{{16}}}}{\left({1}\right)}^{{{2}}}={\frac{{{1}}}{{{16}}}}\)
\(\displaystyle{P}{\left({T}\le{1}\right)}={0.0625}\)
(b) Let \(\displaystyle{P}{\left({T}\geq{2}\right)}={1}-{P}{\left({T}\le{2}\right)}\)
\(\displaystyle={1}-{\frac{{{1}}}{{{16}}}}{\left({2}\right)}^{{{2}}}\)
\(\displaystyle={1}-{\frac{{{1}}}{{{16}}}}{\left({4}\right)}\)
\(\displaystyle={1}-{\frac{{{1}}}{{{4}}}}\)
\(\displaystyle={\frac{{{3}}}{{{4}}}}\)
\(\displaystyle{P}{\left({T}\geq{2}\right)}={0.75}\)
(c) \(\displaystyle{P}{\left({1}\le{T}\le{3}\right)}={P}{\left({T}\le{3}\right)}-{P}{\left({T}\le{1}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({3}\right)}^{{{2}}}-{\frac{{{1}}}{{{16}}}}{\left({1}\right)}^{{{2}}}\)
\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({9}-{1}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({8}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{P}{\left({1}\le{T}\le{3}\right)}={0.5}\)
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