Answer:
a)0.259
b)0.407
c)693.1hrs
Step-by-step explanation:
This is an exponential density function with a constant c and mean of 1000:
[tex]f(t)={0,t<0\choose ce^{-ct,t\geq }0[/tex]
The constant c is calculated as:
[tex]\mu=\frac{1}{c}\\\\c=\frac{1}{\mu}=\frac{1}{1000}\\\\c=0.001[/tex]
#To find the probability that i falls with the first 300 hrs;
[tex]P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0{0.001e^{-0.001t}dt\\\\\\[/tex]
[tex]=[-e^{-0.001t}]\limits^{300}_{0}\\\\=-0.7408+1\\\\=0.2592[/tex]
Hence, the probability that a bulb fails within the first 300 hours is 0.259
b. For more than 900hrs, integrate between 900 and infinity.
#Use the limit method for dealing with improper integrals:
[tex]P(X\geq 900)=\int\limits^{\infty}_{900} {f(t)} \, dt\\ \\\\=\int\limits^{\infty}_{900}{0.001e^{-0.001t}dt\\\\[/tex]
[tex]=\int\limits^{b}_{900}{0.001e^{-0.001t}dt, b->\infty\\\\[/tex]
[tex]=[-e^{-0.001t}]\limits^{b}_{900}\\\\=[-e^{-0.001t}]\limits^{b}_{900}\\\\=0--0.4066\\\\=0.407[/tex]
Hence, the probability that a bulb burns for more than 900 hours is 0.407
c. To find the median lifetime of these bulbs:
[tex]P(X\leq median)=P(X\geq median)=0.5\\\\P(X\geq median)=e^{-median/\mu}=0.5\\\\-median/1000=In(0.5)\\\\-median=1000\ In(0.5)\\\\median=693.1[/tex]
Hence, the median lifetime of these bulbs is 693.1hrs
