Question

The amount of time that a surveillance camera will run without having to be reset is a random variable having the exponential distribution with λ = .02 per day. Find the time above which 68% of the cameras have to be reset.

Answer 1

Answer: The required time is 0.743.

Step-by-step explanation:

Since we have given that

$\lambda=0.02$

We need to find the time above which is 68% of the cameras have to be reset.

So, exponential distribution would be

$f(x)=0.02e^{-0.02}$

So, it becomes,

$P(X$

Hence, the required time is 0.743.

Answer 2

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:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of λ is 0.02.

It is given that, P (X > x) = 0.68.

Compute the value of x as follows:

$P(X>x)=0.68\\\int\limits^{\infty}_{x} {\lambda e^{-\lambda x}} \, dx =0.68\\\lambda\int\limits^{\infty}_{x} { e^{-\lambda x}} \, dx =0.68\\\lambda|\frac{e^{-\lambda x}}{-\lambda} |^{\infty}_{x}=0.68\\|e^{-\lambda x} |^{\infty}_{x}=0.68\\e^{-0.02x}=0.68$

Take natural log on both sides.

$ln(e^{-0.02x})=ln(0.68)\\-0.02x=-0.386\\x=\frac{0.386}{0.02}\\ =19.30\approx19$

Thus, for times more than 19 days, 68% of the cameras have to be reset.