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)

Since the pdf\({\rm{f(x)}}\)is uniform inside the interval\({\rm{A = 0}}{\rm{.2}}\)and\({\rm{B = 4}}{\rm{.25}}\), hence\({\rm{f(x)}}\)inside this interval can be given as 鈥�

\(\begin{array}{c}{\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{4}}{\rm{.25 - 0}}{\rm{.2}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{4}}{\rm{.05}}}}{\rm{ = 0}}{\rm{.2469}}\end{array}\)

Outside this interval\({\rm{f(x)}}\)is zero. Hence overall\({\rm{f(x)}}\)can be written as 鈥�

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.2469}}}&{{\rm{0}}{\rm{.2}} \le {\rm{X}} \le {\rm{4}}{\rm{.25}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

The graph of\({\rm{f(x)}}\)is shown below.

Therefore, the pdf is \({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.2469}}}&{{\rm{0}}{\rm{.2}} \le {\rm{X}} \le {\rm{4}}{\rm{.25}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\) and the graph is obtained.

(b)

The probability that diameter exceeds\({\rm{3 mm}}\)can be written as\({\rm{P(X > 3)}}\). Using\({\rm{f(x)}}\)from part (a), it can be written as 鈥�

\(\begin{array}{l}{\rm{P(X > 3) = }}\int_{\rm{3}}^{{\rm{4}}{\rm{.25}}} {\rm{f}} {\rm{(x) \times dx}}\\{\rm{ = }}\int_{\rm{3}}^{{\rm{4}}{\rm{.25}}} {{\rm{(0}}{\rm{.2469)}}} {\rm{dx}}\\{\rm{ = 0}}{\rm{.2469}}\int_{\rm{3}}^{{\rm{4}}{\rm{.25}}} {\rm{d}} {\rm{x}}\\{\rm{ = 0}}{\rm{.2469(x)}}_{\rm{3}}^{{\rm{4}}{\rm{.25}}}\\{\rm{ = 0}}{\rm{.2469(4}}{\rm{.25 - 3)}}\\{\rm{ = 0}}{\rm{.2469(1}}{\rm{.25)}}\\{\rm{ = 0}}{\rm{.3086}}\end{array}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.3086}}\).

(c)

For uniform distribution, the mean\({\rm{(\mu )}}\)is equal to the mean of the values between which it is nonzero. Which are denoted as\({\rm{A}}\)and\({\rm{B}}\)in the exercise.

\({\rm{\mu = }}\frac{{{\rm{4}}{\rm{.25 + 0}}{\rm{.2}}}}{{\rm{2}}}{\rm{ = 2}}{\rm{.225}}\)

The probability that diameter is within\({\rm{1 mm}}\)of the mean diameter can be written as\({\rm{P(1}}{\rm{.225}} \le {\rm{X}} \le {\rm{3}}{\rm{.225)}}\). Using\({\rm{f(x)}}\)from part (a), it can be written as 鈥�

\(\begin{array}{c}{\rm{P(1}}{\rm{.225}} \le {\rm{X}} \le {\rm{3}}{\rm{.225) = }}\int_{{\rm{1}}{\rm{.225}}}^{{\rm{3}}{\rm{.225}}} {\rm{f}} {\rm{(x)}} \cdot {\rm{dx}}\\{\rm{ = }}\int_{{\rm{1}}{\rm{.225}}}^{{\rm{3}}{\rm{.225}}} {{\rm{(0}}{\rm{.2469)}}} {\rm{dx}}\\{\rm{ = 0}}{\rm{.2469}}\int_{{\rm{1}}{\rm{.225}}}^{{\rm{3}}{\rm{.225}}} {\rm{d}} {\rm{x}}\\{\rm{ = 0}}{\rm{.2469(x)}}_{{\rm{1}}{\rm{.225}}}^{{\rm{3}}{\rm{.225}}}\\{\rm{ = 0}}{\rm{.2469(3}}{\rm{.225 - 1}}{\rm{.225)}}\\{\rm{ = 0}}{\rm{.2469(2)}}\\{\rm{ = 0}}{\rm{.4938}}\end{array}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.4938}}\).

(d)

It is given that\({\rm{0}}{\rm{.20 < a < (a + 1) < 4}}{\rm{.25}}\).

Using\({\rm{f(x)}}\)from part(a), we can write\({\rm{P(a < X < a + 1)}}\)as 鈥�

\(\begin{array}{c}{\rm{P(a < X < a + 1) = }}\int_{\rm{a}}^{{\rm{a + 1}}} {\rm{f}} {\rm{(x)}} \cdot {\rm{dx}}\\{\rm{ = }}\int_{\rm{a}}^{{\rm{a + 1}}} {{\rm{(0}}{\rm{.2469)}}} {\rm{dx}}\\{\rm{ = 0}}{\rm{.2469}}\int_{\rm{a}}^{{\rm{a + 1}}} {\rm{d}} {\rm{x}}\\{\rm{ = 0}}{\rm{.2469(x)}}_{\rm{a}}^{{\rm{a + 1}}}\\{\rm{ = 0}}{\rm{.2469((a + 1) - a)}}\\{\rm{ = 0}}{\rm{.2469(1)}}\\{\rm{ = 0}}{\rm{.2469}}\end{array}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.2469}}\).