Question

Let X have a standard gamma distribution with
α=7
. Compute P(X<4 or X>6)
0.912
0.625
0.812
0.713

10 Let X = the time between two successive arrivals at the drive –up window of a bank. If X has a exponential distribution with h=1 ( which is identical to a standard gamma distribution with a=1). Compute the standard deviation of the time between successive arrivals
2
1
3
4

Answer 1

Answer on Question #62107 – Math – Statistics and Probability

Question

Let $X$ have a standard gamma distribution with $\alpha = 7$.

Compute $P(X < 4 \text{ or } X > 6)$

0.912

0.625

0.812

0.713

Solution

For $\alpha > 0$ the gamma function is defined as follows:

$$\Gamma(\alpha) = \int_{0}^{\infty} x^{\alpha - 1} e^{-x} dx.$$

For integer $n$: $\Gamma(n) = (n - 1)!$

If $X$ is a continuous random variable, then $X$ is said to have a gamma distribution if the pdf of $X$ is

$$f(x) = \frac{\left(\frac{x - \mu}{\beta}\right)^{\alpha - 1} \cdot \exp\left(-\frac{x - \mu}{\beta}\right)}{\beta \Gamma(\alpha)}, \quad x \geq \mu; \alpha, \beta > 0,$$ $$f(x) = 0, \text{otherwise};$$

where $\alpha$ is the shape parameter, $\mu$ is the location parameter, $\beta$ is the scale parameter, and $\Gamma$ is the gamma function.

If $\beta = 1$ and $\mu = 0$ then we have the standard gamma distribution.

$$f(x) = \frac{x^{\alpha - 1} e^{-x}}{\Gamma(\alpha)}, \quad x \geq 0; \alpha > 0$$

When $X$ follows the standard gamma distribution then its cdf is

$$F(x; \alpha) = \int_{0}^{x} \frac{y^{\alpha - 1} e^{-y}}{\Gamma(\alpha)} dy, \quad x > 0$$

This is also called the incomplete gamma function. The cumulative distribution function of the gamma distribution can be calculated using the function GAMMA.DIST in Microsoft Excel.

$$\begin{array}{l} P(X < 4 \text{ or } X > 6) = P(X < 4) + P(X > 6) = P(X < 4) + 1 - P(X < 6) = \\ = F(4; 7) + 1 - F(6; 7) \approx 0.110674 + 0.606303 = 0.716977 \approx \\ \approx 0.717. \end{array}$$

Answer: 0.717.

Question

Let $X =$ the time between two successive arrivals at the drive-up window of a bank. If $X$ has an exponential distribution with $h = 1$ (which is identical to a standard gamma distribution with $a = 1$). Compute the standard deviation of the time between successive arrivals.

2

1

3

4

Solution

The probability density function (pdf) of an exponential distribution is

$$f(x) = \begin{cases} he^{-hx}, & x \geq 0, \\ 0, & \text{otherwise.} \end{cases}$$

In our case $h = 1$ and $f(x) = e^{-x}$, $x \geq 0$.

\begin{aligned} E(X) &= \int_{-\infty}^{+\infty} x f(x) \, dx \\ &= \int_{0}^{+\infty} x e^{-x} \, dx \\ &= -\int_{0}^{+\infty} x \, de^{-x} = -\left(x e^{-x} \right|_{0}^{+\infty} - \int_{0}^{+\infty} e^{-x} \, dx \\ &= -\left(x e^{-x} \right|_{0}^{+\infty} + e^{-x} \right|_{0}^{+\infty}) = -\lim_{A \to +\infty} \left(x e^{-x} \right|_{0}^{A} + e^{-x} \right|_{0}^{A}) = \\ &= -\lim_{A \to +\infty} (A e^{-A} - 0 + e^{-A} - 1) = 1; \end{aligned} $$V(X) = \int_{0}^{\infty} (x - 1)^2 e^{-x} \, dx = \left| \begin{array}{cc} u = x - 1 & \text{du} = dx \\ (x - 1)^2 = u^2 & e^{-x} = \frac{1}{e} e^{-u} \end{array} \right| = \frac{1}{e} \int_{-1}^{\infty} u^2 e^{-u} \, du;$$ $$\begin{aligned} \int u^2 e^{-u} \, du &= -\int u^2 d(e^{-u}) = -u^2 e^{-u} + 2 \int u e^{-u} \, du = -u^2 e^{-u} - \\ -2 \int u d(e^{-u}) &= -u^2 e^{-u} - 2 u e^{-u} + 2 \int e^{-u} \, du = -u^2 e^{-u} - 2 u e^{-u} - 2 e^{-u} \\ V(x) &= \frac{1}{e} \cdot \left[-u^2 e^{-u} - 2 u e^{-u} - 2 e^{-u} \Big|_{-1}^{\infty}\right] = \frac{1}{e} \cdot [-0 - 0 - 0 + e - 2 e + 2 e] = 1. \end{aligned}$$

The standard deviation is

$$\sigma = \sqrt{V(X)} = \sqrt{1} = 1.$$

Answer: 1.

www.AssignmentExpert.com