A random variable X, the time taken by a garage to service a car, takes values between 0 and 10
hours with cumulative distribution function. F(x)= A + B In (3x + 2) for 0≤ x ≤ 10.
Find the values of A and B.
F(x)=A+Bln(3x+2)F(0)=A+Bln(2)=0F(10)=A+Bln(32)=1A=−Bln(2)−Bln(2)+Bln(32)=1B(ln(32)−ln(2))=1B=1ln(32)−ln(2)=13.465−0.693=12.772=0.361A=−0.361×0.693=−0.25F(x)= A + Bln(3x + 2) \\ F(0) = A + Bln(2) = 0 \\ F(10) = A +Bln(32) = 1 \\ A = -Bln(2) \\ -Bln(2) +Bln(32) = 1 \\ B(ln(32) -ln(2)) = 1 \\ B = \frac{1}{ln(32) -ln(2)} = \frac{1}{3.465 -0.693}= \frac{1}{2.772} = 0.361 \\ A = -0.361 \times 0.693 = -0.25F(x)=A+Bln(3x+2)F(0)=A+Bln(2)=0F(10)=A+Bln(32)=1A=−Bln(2)−Bln(2)+Bln(32)=1B(ln(32)−ln(2))=1B=ln(32)−ln(2)1=3.465−0.6931=2.7721=0.361A=−0.361×0.693=−0.25
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