Question

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c. P(X < 30) d. Find the value of x such that P(X < x) = 0.95.

Answer 1

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

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 E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$

Thus, the value of P (X > 20) is 0.1353.

(c)

Compute the value of P (X < 30) as follows:

$P(X$

Thus, the value of P (X < 30) is 0.9502.

(d)

It is given that, P (X < x) = 0.95.

Compute the value of x as follows:

$P(X$

Take natural log on both sides.

$ln(e^{-0.10x})=ln(0.05)\\-0.10x=-2.996\\x=\frac{2.996}{0.10}\\ =29.96\\\approx30$

Thus, the value of x is 30.