Question

8 Let X have a uniform distribution on the interval [A,B]. compute V(X)

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

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

Answer 1

Answer on Question #59915 – Math – Statistics and Probability

Question

8 Let X have a uniform distribution on the interval [A,B]. Compute V(X)

Solution

Since x is uniformly distributed in [A,B], the probability density function is

$$f(x) = \begin{cases} \dfrac{1}{B - A}, & A \leq x \leq B, \\ 0, & \text{otherwise.} \end{cases}$$

The mean is

$$\mu = \int_{-\infty}^{+\infty} f(x) \, dx = \int_{A}^{B} \dfrac{x \, dx}{B - A} = \dfrac{1}{B - A} \left(\dfrac{x^2}{2}\right)_A^B = \dfrac{1}{2} \dfrac{1}{B - A} (B^2 - A^2) = \dfrac{A + B}{2}.$$

The variance is

$$V(X) = \int_{-\infty}^{+\infty} (x - \mu)^2 f(x) \, dx = \int_{A}^{B} \dfrac{(x - \mu)^2 \, dx}{B - A} = \dfrac{1}{B - A} \int_{A}^{B} (x - \mu)^2 d(x - \mu) = \dfrac{1}{3} \dfrac{1}{B - A} \left[ \left(B - \dfrac{A + B}{2}\right)^3 - \left(A - \dfrac{A + B}{2}\right)^3 \right] = \dfrac{(B - A)^2}{12}.$$

Answer: $V(X) = \dfrac{(B - A)^2}{12}$.

Question

9 Let X have a standard gamma distribution with $\alpha = 7$. Compute P(X<4 or X>6)

Solution

Let the cumulative distribution function of the standard gamma distribution with shape parameter $\alpha$ be $F(x)$.

Then

$$P(X < 4) = F(4); \quad P(X > 6) = 1 - F(6).$$ $$P(X < 4 \text{ or } X > 6) = P(X < 4) + P(X > 6) = F(4) + 1 - F(6) = 0.1107 + 1 - 0.3937 = 0.7170.$$

We used Excel function GAMMA.DIST to calculate the values of the cumulative distribution function:

$F(6) = \text{GAMMA.DIST}(6,7,1,\text{TRUE})$ and $F(4) = \text{GAMMA.DIST}(4,7,1,\text{TRUE})$.

Answer: $P(X < 4 \text{ or } X > 6) = 0.7170$.

Question

10 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.

Solution

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

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

where $\frac{1}{h}$ is the mean.

In our case $f(x) = \begin{cases} e^{-x}, & x \geq 0, \\ 0, & otherwise. \end{cases}$

The variance is

$$\begin{aligned} V(X) &= \int_{0}^{\infty} (x - 1)^2 e^{-x} dx = |y = x - 1| = \int_{-1}^{\infty} y^2 e^{-y - 1} dy = \frac{1}{e} \int_{-1}^{\infty} y^2 e^{-y} dy \\ &\quad \int_{-1}^{\infty} y^2 e^{-y} dy = - \int_{-1}^{\infty} y^2 d e^{-y} = (-y^2 e^{-y})_{-1}^{\infty} + 2 \int_{-1}^{\infty} y e^{-y} dy \\ &\quad (-y^2 e^{-y})_{-1}^{\infty} = e \\ &\quad \int_{-1}^{\infty} y e^{-y} dy = (-y e^{-y})_{-1}^{\infty} + \int_{-1}^{\infty} e^{-y} dy \\ &\quad (-y e^{-y})_{-1}^{\infty} = -e \\ &\quad \int_{-1}^{\infty} e^{-y} dy = (-e^{-y})_{-1}^{\infty} = e \\ V(X) &= \frac{1}{e} [e + 2(e - e)] = 1 \end{aligned}$$

The standard deviation is

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

Answer: $\sigma = 1$.

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