[Probability Theory] Distribution Function and Mathematical Expectation of Continuous Random Variables (2)

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1. If the density function of X is
   $p(x)=\{\begin{array}{ll}x,& 0\le x<1;\\ 2-x,& 1\le x<2;\\ 0,& Other\end{array}$
   Try to find P{X≤1.5}=_____ 【Correct answer: 7/8 or 0.875】(express the result in fractional form)
   Solution:
   P { X ≤ 1.5 } = ∫ − ∞ 1.5 p ( x ) dx = ∫ 0 1 xdx + ∫ 1 1.5 ( 2 − x ) dx = x 2 2 ∣ 0 1 + ( 2 x − x 2 2 ) ∣ 1 1.5 = 1 2 − 0 + ( 3 − 1. 5 2 2 ) − 3 2 = 7 8 P \{ X \le 1.5 \} = \int _{- \infty }^{1.5}p(x)dx= \int _{0}^{1}xdx+ \int _{1}^{1.5}( 2-x)dx= \frac {x^{2}}{2}|_{0}^{1}+(2x- \frac {x^{2}}{2})|^{1.5}_1 =\frac{1}{2}-0+(3- \frac {1.5^{2}}{2})- \frac {3}{2}= \frac {7}{8}P{ X≤1.5}=∫−∞1.5​p(x)dx=∫01​x d x+∫11.5​(2−x)dx=2x2​∣01​+( 2x _−2x2​)∣11.5​=21​−0+(3−21.52​)−23​=87​

2. Let the density function of random variable X be
   $p(x)=\{\begin{array}{ll}a+b{x}^2,& 0\le x\le 1;\\ 0,& Other\end{array}$
   If $E\left(X\right)=\frac{2}{3}$,
   seek: a=_____ [correct answer: 1/3] and b=_____ [correct answer: 2]; (please express the result of a in approximate fractions) multiple answers
   (multiple blanks) can be filled in a question, Insert the solution where you fill in the answer
   : ∫ − ∞ + ∞ p ( x ) dx = ∫ 0 1 ( a + bx 2 ) dx = ( ax + b ⋅ x 3 3 ) ∣ 0 1 = a + b from
   regularity ${\int }_{-\infty }^{+\infty }p(x)dx={\int }_0^1(a+b{x}^2)dx=(ax+b\cdot \frac{x^3}{3}){|}_0^1=a+\frac{b}{3}=1$,又$E\left(X\right)={\int }_{-\infty }^{+\infty }xp(x)dx={\int }_0^{4}x(a+b{x}^2)dx=(a\cdot \frac{x^2}{2}+b\cdot \frac{x^4}{4}){|}_0^1=\frac{a}{2}+\frac{b}{4}=\frac{2}{3}$,故$a=\frac{1}{3},b=2$

3. Let the density function of random variable X be as follows,
   $p(x)=\{\begin{array}{ll}{e}^{-x},& x>0;\\ 0,& x\le 0.\end{array}$
   Try $E(2X+5)$=_____ 【正确答案: 7】;
   解:
   $E(2X+5)={\int }_0^{+\infty }(2x\_+5)e^{-x}dx={\int }_0^{+\infty }(2x\_+5)(-d{e}^{-x})=-(2x\_+5)e^{-x}{|}_0^{+\infty }+{\int }_0^{+\infty }2e{\_}^{-x}dx=5-2e\_\_{\_}^{-x}{|}_0^{+\infty }=7$

4. The market share X of a new product in the future is a random variable that only takes values ​​on the interval (0, 1), and its density function is
   $p(x)=\{\begin{array}{ll}4(1-x{)}^3,& 0<x<1;\\ 0,& Other\end{array}$
   Average market share =_____【Correct answer: 0.2 or 1/5】
   Solution:
   $E\left(X\right)={\int }_0^{1}x\cdot 4(1-x{)}^{3}dx={\int }_0^1(4x\_-12x^2+12x^3-4x{\_}^4)dx=(2x{\_}^2-4x{\_}^3+3x{\_}^4-\frac{4}{5}{x}^5){|}_0^1=\frac{1}{5}$

5. The maintenance time T of a bulldozer in a factory is a random variable (unit: h), and its density function is
   $p(t)=\{\begin{array}{ll}0.02e^{-0.02t},& t>0;\\ 0,& t\le 0.\end{array}$
   Average maintenance time=______【Correct answer: 50】
   Solution:
   Average maintenance time $E(T)={\int }_0^{+\infty }t\cdot 0.02e^{-0.02t}dt={\int }_0^{+\infty }t(-d{e}^{-0.02t}){}^{\_}=-t{e}^{-0.02t}{|}_0^{+\infty }+{\int }_0^{+\infty }{e}^{-0.02t}dt=\frac{e^{-0.02t}}{-0.02}{|}_0^{+\infty }=50$

6. The time X for a student to complete an assignment is a random variable in hours. Its density function is
   $p(x)=\{\begin{array}{ll}c{x}^2+x,& 0\le x\le 0.5;\\ 0,& Other\end{array}$
   (1) Constant c=_____ [Correct answer: 21];
   (2) Find the probability of completing an assignment within 20 minutes =_____ [Correct answer: 17/54]; (Please express the result in approximate fractions) (
   3) Find 10 minutes Probability of completing an assignment above=_____ 【Correct answer: 103/108】; (Please express the result in approximate fractions)
   Solution:
   (1) According to the regularity of the density function, ${\int }_{-\infty }^{+\infty }p(x)dx={\int }_0^{0.5}(c{x}^2+x)dx=(c\frac{x^3}{3}+\frac{x^2}{2}){|}_0^{\infty }=\frac{c}{24}+\frac{1}{8}=1.$ So c=21;
   (2) Distribution function F(x)=P{X≤x}, the subsection point is x=0, 0.5,
   when x<0,F ( x ) = P { X ≤ x } = P ( ϕ ) = 0 F(x)=P\{X≤x\}=P(\phi)=0F(x)=P{ X≤x}=P ( ϕ )=0
   当0≤x<0.5时,F(x)= ${\int }_{-\infty }^{x}p\left(u\right)du={\int }_0^x(21u\_{}^2+u)du=(7u^3+\frac{u^2}{2}){|}_0^x=7x{\_}^3+\frac{x^2}{2}$
   When x≥0.5, F ( x ) = P { X ≤ x } = P ( Ω ) = 1 F(x)=P\{X≤x\}=P(Ω)=1F(x)=P{ X≤x}=P ( Ω )=1 ,
   so the distribution function of X
   $F\left(x\right)=\{\begin{array}{ll}0,& x<0;\\ 7x{\_}^3+\frac{x^2}{2},& 0\le x<0.5;\\ 1,& x\ge 0.5;\end{array}$
   (3) 所求概率为 P { X ≤ 20 60 = 1 3 } = F ( 1 3 ) = 7 × ( 1 3 ) 3 + 1 2 × ( 1 3 ) 2 = 7 27 + 1 18 = 17 54 P \{ X \le \frac {20}{60}= \frac {1}{3} \} =F \left ( \frac {1}{3} \right )=7 \times \left ( \frac {1}{3} \right )^{3}+ \frac {1}{2} \times \left ( \frac {1}{3} \right )^{2}= \frac {7}{27}+ \frac {1}{18}= \frac {17}{54} P{ X≤6020​=31​}=F(31​)=7×(31​)3+21​×(31​)2=277​+181​=5417​;
   (4) 所求概率为 P { X ≥ 10 60 = 1 6 } = 1 − F ( 1 6 ) = 1 − 7 × ( 1 6 ) 3 − 1 2 × ( 1 6 ) 2 = 1 − 7 216 − 1 72 = 103 108 P \{ X \ge \frac {10}{60}= \frac {1}{6} \} =1-F \left ( \frac {1}{6} \right )=1-7 \times \left ( \frac {1}{6} \right )^{3}- \frac {1}{2} \times \left ( \frac {1}{6} \right )^{2}=1- \frac {7}{216}- \frac {1}{72}= \frac {103}{108} P{ X≥6010​=61​}=1−F(61​)=1−7×(61​)3−21​×(61​)2=1−2167​−721​=108103​