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.

Since\({\rm{f(x)}}\)is uniform inside interval\({\rm{ - 5}} \le {\rm{X}} \le {\rm{5}}\), hence inside this interval its value is –

\({\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{5 - ( - 5)}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ = 0}}{\rm{.1}}\)

And outside this interval its value is zero. Hence, finally write\({\rm{f(x)}}\)as –

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.1}}}&{{\rm{ - 5}} \le {\rm{X}} \le {\rm{5}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

(a)

Derive\({\rm{P(a}} \le {\rm{X}} \le {\rm{b)}}\)for this pdf such that lower limit\({\rm{a}}\)and upper limit\({\rm{b}}\)belong in the interval\({\rm{ - 5}} \le {\rm{X}} \le {\rm{5}}\).

\(\begin{aligned}{\rm{P(a}} \le {\rm{X}} \le b) &= \int_{\rm{a}}^{\rm{b}} {{\rm{(0}}{\rm{.1)}}} \cdot {\rm{dx}}\\&= 0 {\rm{.1}}\int_{\rm{a}}^{\rm{b}} {\rm{d}} {\rm{x}}\\ &= 0 {\rm{.1(x)}}_{\rm{a}}^{\rm{b}}\\ &= 0 {\rm{.1(b - a)}}\\{\rm{P(a}} \le {\rm{X}} \le b) &= 0 {\rm{.1(b - a)}}\end{aligned}\)

Now for this part we have to calculate\({\rm{P( - 5}} \le {\rm{X < 0)}}\). Since the given distribution is continuous hence –

\({\rm{P( - 5}} \le {\rm{X < 0) = P( - 5}} \le {\rm{X}} \le {\rm{0)}}\)

Now taking \({\rm{a = - 5}}\) and \({\rm{b = 0}}\) in the derived relation, it is obtained –

\(\begin{aligned}{\rm{P( - 5}} \le X < 0) &= 0 {\rm{.1(0 - ( - 5))}}\\{\rm{P( - 5}} \le X < 0) &= 0 {\rm{.5}}\end{aligned}\)

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 obtained is \({\rm{0}}{\rm{.5}}\).

(b)

For this part calculate \({\rm{P( - 2}}{\rm{.5 < X < 2}}{\rm{.5)}}\). Since the given distribution is continuous hence –

\({\rm{P( - 2}}{\rm{.5 < X < 2}}{\rm{.5) = P( - 2}}{\rm{.5}} \le {\rm{X}} \le {\rm{2}}{\rm{.5)}}\)

Now taking \({\rm{a = - 2}}{\rm{.5}}\) and \({\rm{b = 2}}{\rm{.5}}\) in the derived relation, it is obtained –

\(\begin{aligned}{\rm{P( - 2}}{\rm{.5 < X < 2}}.5) &= 0{\rm{.1(2}}{\rm{.5 - ( - 2}}{\rm{.5))}}\\{\rm{P( - 2}}{\rm{.5}} \le {\rm{X < 2}} .5) &= 0 {\rm{.5}}\end{aligned}\)

Therefore, the value obtained is \({\rm{0}}{\rm{.5}}\).

(c)

For this part calculate \({\rm{P( - 2}} \le {\rm{X}} \le {\rm{3)}}\).

Now taking \({\rm{a = - 2}}\) and \({\rm{b = 3}}\) in the derived relation, it is obtained –

\(\begin{aligned}{\rm{P( - 2}} \le {\rm{X}} \le 3) &= 0 {\rm{.1(3 - ( - 2))}}\\{\rm{P( - 2}} \le {\rm{X}} \le 3) &= 0 {\rm{.5}}\end{aligned}\)

Therefore, the value obtained is \({\rm{0}}{\rm{.5}}\).

(d)

For this part calculate \({\rm{P(k < X < k + 4)}}\). Since the given condition is \({\rm{ - 5 < k < X < (k + 4) < 5}}\) the relation derived in part (a) can be used –

Now taking\({\rm{a = k}}\)and\({\rm{b = k + 4}}\)in the derived relation, it is obtained –

\(\begin{aligned} P(k < X < k + 4) &= 0 {\rm{.1((k + 4) - k)}}\\ &= 0{\rm{.1(4)}}\\P(k < X < k + 4) &= 0{\rm{.4}}\end{aligned}\)

Therefore, the value obtained is \({\rm{0}}{\rm{.4}}\).