Answer to Question #214024 in Statistics and Probability for tasa

Question

The lifetime T in tens of hours of a battery has a cumulative distribution function

F(t)={0,t<1

{k(t²+2t-3),1<t<or equal to 1.5

{1,t>5

Find k and hence find p(T>1.2) and find the median of T

Answer 1

$$F(t) = \begin{cases} 0, & t<1 \\ k(t^2+2t-3), &1\leq t\leq 1.5 \\ 1, & t>1.5 \end{cases}$$

1. Differentiate with respect to $t$

$$f(t) = \begin{cases} k(2t+2), &1\leq t\leq 1.5 \\ 0, & otherwise \end{cases}$$

$$\displaystyle\int_{-\infin}^{\infin}f(t)dt=\displaystyle\int_{1}^{1.5}k(2t+2)dt$$

$$=k[t^2+2t]\begin{matrix} 1.5 \\ 1 \end{matrix}=k(2.25+3-1-2)=2.25k=1$$

$$k=\dfrac{4}{9}$$

2.

$$P(T>1.2)=1-P(T\leq1.2)=1-F(1.2)$$

$$=1-\dfrac{4}{9}(1.2^2+2(1.2)-3)=\dfrac{1.88}{3}\approx0.6267$$

3.

$$\dfrac{4}{9}(m^2+2m-3)=0.5$$

$$m^2+2m-3=1.125$$

$$m^2+2m-4.125=0$$

$$m=-1\pm\sqrt{5.125}$$

SInce median $>0,$ we take

$$median=-1+\sqrt{5.125}\approx1.264$$