91Ó°ÊÓ

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 probability density curve is given below. The total area under this curve can be given as –

\(\begin{aligned} Area &= \int_{\rm{3}}^{\rm{5}} {{\rm{(0}}{\rm{.075x + 0}}{\rm{.2)}}} \cdot {\rm{dx}}\\ &= \int_{\rm{3}}^{\rm{5}} {{\rm{(0}}{\rm{.075x)}}} \cdot {\rm{dx + }}\int_{\rm{3}}^{\rm{5}} {{\rm{(0}}{\rm{.2)}}} \cdot {\rm{dx}}\\ &= 0 {\rm{.075}}\int_{\rm{3}}^{\rm{5}} {\rm{x}} \cdot {\rm{dx + 0}}{\rm{.2}}\int_{\rm{3}}^{\rm{5}} \cdot {\rm{dx}}\\ &= 0 {\rm{.075}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{3}}^{\rm{5}}{\rm{ + 0}}{\rm{.2(x)}}_{\rm{3}}^{\rm{5}}\\ & = 0 {\rm{.075}}\left( {\frac{{{{\rm{5}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{3}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ + 0}}{\rm{.2(5 - 3)}}\\ &= 0 {\rm{.075(12}}{\rm{.5 - 4}}{\rm{.5) + 0}}{\rm{.2(2)}}\\ &= 0 {\rm{.075(8) + 0}}{\rm{.4}}\\& = 0 {\rm{.6 + 0}}{\rm{.5 = 1}}\end{aligned}\)

Hence the area under the density curve is indeed equal to\({\rm{1}}\).

Therefore, the graph is obtained.

(b)

The probability \({\rm{P(X < 4)}}\) can be given as –

\(\begin{aligned} P(X < 4) &= \int_{\rm{3}}^{\rm{4}} {{\rm{(0}}{\rm{.075x + 0}}{\rm{.2)}}} \cdot {\rm{dx}}\\ &= \int_{\rm{3}}^{\rm{4}} {{\rm{(0}}{\rm{.075x)}}} \cdot {\rm{dx + }}\int_{\rm{3}}^{\rm{4}} {{\rm{(0}}{\rm{.2)}}} \cdot {\rm{dx}}\\ & = 0 {\rm{.075}}\int_{\rm{3}}^{\rm{4}} {\rm{x}} \cdot {\rm{dx + 0}}{\rm{.2}}\int_{\rm{3}}^{\rm{4}} \cdot {\rm{dx}}\\ & = 0 {\rm{.075}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{3}}^{\rm{4}}{\rm{ + 0}}{\rm{.2(x)}}_{\rm{3}}^{\rm{4}}\\ &= 0 {\rm{.075}}\left( {\frac{{{{\rm{4}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{3}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ + 0}}{\rm{.2(4 - 3)}}\\ &= 0 {\rm{.075(8 - 4}}{\rm{.5) + 0}}{\rm{.2(1)}}\\ & = 0 {\rm{.075(8) + 0}}{\rm{.2}}\\ & = 0 {\rm{.2625 + 0}}{\rm{.2 = 0}}{\rm{.4625}}\end{aligned}\)

Since the given pdf is continuous hence \({\rm{P(X < 4) = P(X}} \le {\rm{4)}}\). Hence –

\({\rm{P(X < 4) = P(X}} \le {\rm{4) = 0}}{\rm{.4625}}\)

Practical consequence for continuous random variable –

When \({\rm{X}}\) is continuous random variable, then the probability that \({\rm{X}}\) lies in some interval between \({\rm{a}}\) and \({\rm{b}}\) does not depend on whether the lower limit \({\rm{a}}\) or the upper limit \({\rm{b}}\) is included in the probability calculation –

\({\rm{P(a < X < b) = P(a}} \le {\rm{X < b) = P(a < X}} \le {\rm{b) = P(a}} \le {\rm{X}} \le {\rm{b)}}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.4625}}\).

(c)

The probability \({\rm{P(3}}{\rm{.5}} \le {\rm{X}} \le {\rm{4}}{\rm{.5)}}\) can be given as –

\(\begin{aligned}{\rm{P(3}}{\rm{.5}} \le {\rm{X}} \le 4) &= \int_{{\rm{3}}{\rm{.5}}}^{{\rm{4}}{\rm{.5}}} {{\rm{(0}}{\rm{.075x + 0}}{\rm{.2)}}} \cdot {\rm{dx}}\\ & = \int_{{\rm{3}}{\rm{.5}}}^{{\rm{4}}{\rm{.5}}} {{\rm{(0}}{\rm{.075x)}}} \cdot {\rm{dx + }}\int_{{\rm{3}}{\rm{.5}}}^{{\rm{4}}{\rm{.5}}} {{\rm{(0}}{\rm{.2)}}} \cdot {\rm{dx}}\\ &= 0 {\rm{.075}}\int_{{\rm{3}}{\rm{.5}}}^{{\rm{4}}{\rm{.5}}} {\rm{x}} \cdot {\rm{dx + 0}}{\rm{.2}}\int_{{\rm{3}}{\rm{.5}}}^{{\rm{4}}{\rm{.5}}} \cdot {\rm{dx}}\\ &= 0 {\rm{.075}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{{\rm{3}}{\rm{.5}}}^{{\rm{4}}{\rm{.5}}}{\rm{ + 0}}{\rm{.2(x)}}_{{\rm{3}}{\rm{.5}}}^{{\rm{4}}{\rm{.5}}}\\ & = 0 {\rm{.075}}\left( {\frac{{{{{\rm{(4}}{\rm{.5)}}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{{\rm{(3}}{\rm{.5)}}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ + 0}}{\rm{.2(4}}{\rm{.5 - 3}}{\rm{.5)}}\\ &= 0 {\rm{.075(10}}{\rm{.125 - 6}}{\rm{.125) + 0}}{\rm{.2(1)}}\\ & = 0 {\rm{.075(4) + 0}}{\rm{.2}}\\ &= 0 {\rm{.2625 + 0}}{\rm{.2 = 0}}{\rm{.5}}\end{aligned}\)

The probability \({\rm{P(4}}{\rm{.5 < X)}}\) can be given as –

\(\begin{aligned}{{\rm P(4}} .5 < X) &= \int_{{\rm{4}}{\rm{.5}}}^{\rm{5}} {{\rm{(0}}{\rm{.075x + 0}}{\rm{.2)}}} \cdot {\rm{dx}}\\ &= \int_{{\rm{4}}{\rm{.5}}}^{\rm{5}} {{\rm{(0}}{\rm{.075x)}}} \cdot {\rm{dx + }}\int_{{\rm{4}}{\rm{.5}}}^{\rm{5}} {{\rm{(0}}{\rm{.2)}}} \cdot {\rm{dx}}\\&= 0 {\rm{.075}}\int_{{\rm{4}}{\rm{.5}}}^{\rm{5}} {\rm{x}} \cdot {\rm{dx + 0}}{\rm{.2}}\int_{{\rm{4}}{\rm{.5}}}^{\rm{5}} \cdot {\rm{dx}}\\ &= 0 {\rm{.075}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{{\rm{4}}{\rm{.5}}}^{\rm{5}}{\rm{ + 0}}{\rm{.2(x)}}_{{\rm{4}}{\rm{.5}}}^{\rm{5}}\\ & = 0{\rm{.075}}\left( {\frac{{{{\rm{5}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{{\rm{(4}}{\rm{.5)}}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ + 0}}{\rm{.2(5 - 4}}{\rm{.5)}}\\ &= 0 {\rm{.075(12}}{\rm{.5 - 10}}{\rm{.125) + 0}}{\rm{.2(0}}{\rm{.5)}}\\ &= 0 {\rm{.075(2}}{\rm{.375) + 0}}{\rm{.1}}\\ &= 0 {\rm{.1781 + 0}}{\rm{.1 = 0}}{\rm{.2781}}\end{aligned}\)

Therefore, the values obtained are \({\rm{0}}{\rm{.5}}\) and \({\rm{0}}{\rm{.2781}}\).