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) \(\displaystyle{P}{\left({1}\le{T}\le{3}\right)}={P}{\left({T}\le{3}\right)}-{P}{\left({T}\le{1}\right)}\)

\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({3}\right)}^{{{2}}}-{\frac{{{1}}}{{{16}}}}{\left({1}\right)}^{{{2}}}\)

\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({9}-{1}\right)}\)

\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({8}\right)}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{P}{\left({1}\le{T}\le{3}\right)}={0.5}\)

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) \(\displaystyle{P}{\left({1}\le{T}\le{3}\right)}={P}{\left({T}\le{3}\right)}-{P}{\left({T}\le{1}\right)}\)

\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({3}\right)}^{{{2}}}-{\frac{{{1}}}{{{16}}}}{\left({1}\right)}^{{{2}}}\)

\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({9}-{1}\right)}\)

\(\displaystyle={\frac{{{1}}}{{{16}}}}{\left({8}\right)}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{P}{\left({1}\le{T}\le{3}\right)}={0.5}\)