91影视

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

(a)

(The graph of\({\rm{f(x)}}\)is shown below. The location of the roots of\({\rm{f(x)}}\)will stay same at\({\rm{x = 2,4}}\). But the height and hence the area under the graph of\({\rm{f(x)}}\)will change with\({\rm{k}}\). Hence\({\rm{f(x)}}\)will be a pdf for only one value of\({\rm{k}}\)鈥�

Therefore, the graph is obtained.

(b)

Since\({\rm{f(x)}}\)is a legitimate pdf, hence it must satisfy following condition 鈥�

\(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(x)}} \cdot {\rm{dx = 1}}} \)

For the given pdf 鈥�

\(\begin{aligned}\int_{ - \infty }^\infty f (x) \cdot dx &= \int_2^4 k \left( {1 - {{(x - 3)}^2}} \right) \cdot dx = 1 \hfill \\k\int_2^4 {\left( {1 - \left( {{x^2} - 6x + 9} \right) \cdot dx } \right.} &= 1 \hfill \\k\int_2^4 {\left( {1 - {x^2} + 6x - 9} \right)} \cdot dx &= 1 \hfill \\k\int_2^4 {\left( { - {x^2} + 6x - 8} \right)} \cdot dx &= 1 \hfill \\k\left[ { - \frac{{{x^3}}}{3} + 3{x^2} - 8x} \right]_2^4 &= 1 \hfill \\k\left[ {\left( { - \frac{{{4^3}}}{3} + 3 \cdot {4^2} - 8 \cdot 4} \right) - \left( { - \frac{{{2^3}}}{3} + 3 \cdot {2^2} - 8 \cdot 2} \right)} \right] &= 1 \hfill \\k\left[ {\left( {\frac{{ - 16}}{3}} \right) - \left( {\frac{{ - 20}}{3}} \right)} \right] &= 1 \hfill \\k \cdot \frac{4}{3} &= 1 \hfill \\k &= \frac{3}{4} \hfill \\ \end{aligned} \)

Hence the pdf can finally be written as 鈥�

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{1 - (x - 3}}{{\rm{)}}^{\rm{2}}}} \right)}&{{\rm{2}} \le {\rm{X}} \le {\rm{4}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

Conditions satisfied by a pdf: For a pdf to be a legitimate pdf, it must satisfy the following two conditions 鈥�

1. \({\rm{f(x)}} \ge {\rm{0}}\)for all\({\rm{x}}\)
2. \(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(x)}} \cdot {\rm{dx = 1}}} \)

Therefore, the value is obtained as \({\rm{k = }}\frac{{\rm{3}}}{{\rm{4}}}\).

(c)

The prescribed weight is \({\rm{3 g}}\). Hence the probability that the actual tracking weight is greater than the prescribed weight is \({\rm{P(X > 3)}}\). From the pdf it can be written 鈥�

\(\begin{array}{c}{\rm{P(X > 3) = }}\int_{\rm{3}}^{\rm{4}} {\frac{{\rm{3}}}{{\rm{4}}}} \left( {{\rm{1 - (x - 3}}{{\rm{)}}^{\rm{2}}}} \right) \cdot {\rm{dx}}\\{\rm{ = }}\frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{x - }}\frac{{{{{\rm{(x - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{3}}^{\rm{4}}\\{\rm{ = }}\frac{{\rm{3}}}{{\rm{4}}}\left( {\left( {{\rm{4 - }}\frac{{{{{\rm{(4 - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right){\rm{ - }}\left( {{\rm{3 - }}\frac{{{{{\rm{(3 - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\{\rm{ = }}\frac{{\rm{3}}}{{\rm{4}}}\left( {\left( {{\rm{4 - }}\frac{{\rm{1}}}{{\rm{3}}}} \right){\rm{ - (3)}}} \right)\\{\rm{ = }}\frac{{\rm{3}}}{{\rm{4}}}\left( {\left( {\frac{{{\rm{11}}}}{{\rm{3}}}} \right){\rm{ - (3)}}} \right)\\{\rm{ = }}\frac{{\rm{3}}}{{\rm{4}}}\left( {\frac{{\rm{2}}}{{\rm{3}}}} \right)\\{\rm{P(X > 3) = }}\frac{{\rm{1}}}{{\rm{2}}}\end{array}\)

Therefore, the value is obtained as \(\frac{1}{2}\).

(d)

The prescribed weight is \({\rm{3 g}}\). Hence the probability that the actual weight is within \({\rm{0}}{\rm{.25 g}}\)of the prescribed weight is \({\rm{P(2}}{\rm{.75 < X < 3}}{\rm{.25)}}\). From the pdf it can be written 鈥�

\(\begin{aligned}{\rm{P(2}}{\rm{.75 < X < 3}} .25) &= \int_{{\rm{2}}{\rm{.75}}}^{{\rm{3}}{\rm{.25}}} {\frac{{\rm{3}}}{{\rm{4}}}} \left( {{\rm{1 - (x - 3}}{{\rm{)}}^{\rm{2}}}} \right) \cdot {\rm{dx}}\\ &= \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{x - }}\frac{{{{{\rm{(x - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{{\rm{2}}{\rm{.75}}}^{{\rm{3}}{\rm{.25}}}\\ &= \frac{{\rm{3}}}{{\rm{4}}}\left( {\left( {{\rm{3}}{\rm{.25 - }}\frac{{{{{\rm{(3}}{\rm{.25 - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right){\rm{ - }}\left( {{\rm{2}}{\rm{.75 - }}\frac{{{{{\rm{(2}}{\rm{.75 - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\ & = \frac{{\rm{3}}}{{\rm{4}}}\left( {\left( {{\rm{3}}{\rm{.25 - }}\frac{{{\rm{0}}{\rm{.2}}{{\rm{5}}^{\rm{3}}}}}{{\rm{3}}}} \right){\rm{ - }}\left( {{\rm{2}}{\rm{.75 + }}\frac{{{\rm{0}}{\rm{.2}}{{\rm{5}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\ &= \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{3}}{\rm{.25 - }}\frac{{{\rm{0}}{\rm{.2}}{{\rm{5}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - 2}}{\rm{.75 - }}\frac{{{\rm{0}}{\rm{.2}}{{\rm{5}}^{\rm{3}}}}}{{\rm{3}}}}\right)\\\left. &= \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{(3}}{\rm{.25 - 2}}{\rm{.75) - 2 \times }}\frac{{{\rm{0}}{\rm{.2}}{{\rm{5}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\ &= \frac{{\rm{3}}}{{\rm{4}}}{\rm{(0}}{\rm{.5 - 0}}{\rm{.0104)}}\\{\rm{P(2}}{\rm{.75 < X < 3}}{\rm{.25) = 0}}{\rm{.3672}}\end{aligned}\)

Therefore, the value is obtained as \(0.3672\).

(e)

The prescribed weight is \({\rm{3 g}}\). Hence the probability that the actual weight differs from the prescribed weight by more than \({\rm{0}}{\rm{.5 g}}\) is \({\rm{P(2}}{\rm{.75 < X < 3}}{\rm{.25)}}\). From the pdf it can be written 鈥�

\({\rm{P(X < 2}}{\rm{.5 or X > 3}}{\rm{.5) = 1 - P(2}}{\rm{.5}} \le {\rm{X}} \le {\rm{3}}{\rm{.5)}}\)

\(\begin{aligned} &= 1 - \int_{{\rm{2}}{\rm{.5}}}^{{\rm{3}}{\rm{.5}}} {\frac{{\rm{3}}}{{\rm{4}}}} \left( {{\rm{1 - (x - 3}}{{\rm{)}}^{\rm{2}}}} \right) \cdot {\rm{dx}}\\ &= 1 - \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{x - }}\frac{{{{{\rm{(x - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{{\rm{2}}{\rm{.5}}}^{{\rm{3}}{\rm{.5}}}\\ &= 1 - \frac{{\rm{3}}}{{\rm{4}}}\left( {\left( {{\rm{3}}{\rm{.5 - }}\frac{{{{{\rm{(3}}{\rm{.5 - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right){\rm{ - }}\left( {{\rm{2}}{\rm{.5 - }}\frac{{{{{\rm{(2}}{\rm{.5 - 3)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\ &= 1 - \frac{{\rm{3}}}{{\rm{4}}}\left( {\left( {{\rm{3}}{\rm{.5 - }}\frac{{{\rm{0}}{\rm{.125}}}}{{\rm{3}}}} \right){\rm{ - }}\left( {{\rm{2}}{\rm{.5 + }}\frac{{{\rm{0}}{\rm{.125}}}}{{\rm{3}}}} \right)} \right)\\ &= 1 - \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{3}}{\rm{.5 - }}\frac{{{\rm{0}}{\rm{.125}}}}{{\rm{3}}}{\rm{ - 2}}{\rm{.5 - }}\frac{{{\rm{0}}{\rm{.125}}}}{{\rm{3}}}} \right)\\ &= 1 - \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{(3}}{\rm{.5 - 2}}{\rm{.5) - 2 \times }}\frac{{{\rm{0}}{\rm{.125}}}}{{\rm{3}}}} \right)\\ &= 1 - \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{1 - }}\frac{{{\rm{0}}{\rm{.25}}}}{{\rm{3}}}} \right)\\ &= 1 - \frac{{\rm{3}}}{{\rm{4}}}\left( {\frac{{{\rm{2}}{\rm{.75}}}}{{\rm{3}}}} \right)\\ &= 1 - \frac{{{\rm{2}}{\rm{.75}}}}{{\rm{4}}}\\{\rm{P(X > 3}} & .5 = 0 {\rm{.3125}}\end{aligned}\)

Therefore, the value is obtained as \(0.3125\)颅.