Answer :
Answer: The required time is 0.743.
Step-by-step explanation:
Since we have given that
[tex]\lambda=0.02[/tex]
We need to find the time above which is 68% of the cameras have to be reset.
So, exponential distribution would be
[tex]f(x)=0.02e^{-0.02}[/tex]
So, it becomes,
[tex]P(X<68)=\int\limits^{68}_0 {0.02e^{-0.20x}} \, dx \\\\=(-e^{-0.02x})|_0^{68}\\\\=1-e^{-0.02\times 68}\\\\=1-e^{-1.36}\\\\=0.743[/tex]
Hence, the required time is 0.743.
Answer:
After 19 days, 68% of the cameras have to be reset.
Step-by-step explanation:
The probability density function of an exponential distribution is:
[tex]f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0[/tex]
The value of λ is 0.02.
It is given that, P (X > x) = 0.68.
Compute the value of x as follows:
[tex]P(X>x)=0.68\\\int\limits^{\infty}_{x} {\lambda e^{-\lambda x}} \, dx =0.68\\\lambda\int\limits^{\infty}_{x} { e^{-\lambda x}} \, dx =0.68\\\lambda|\frac{e^{-\lambda x}}{-\lambda} |^{\infty}_{x}=0.68\\|e^{-\lambda x} |^{\infty}_{x}=0.68\\e^{-0.02x}=0.68[/tex]
Take natural log on both sides.
[tex]ln(e^{-0.02x})=ln(0.68)\\-0.02x=-0.386\\x=\frac{0.386}{0.02}\\ =19.30\approx19[/tex]
Thus, for times more than 19 days, 68% of the cameras have to be reset.