Exercises: Probability

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Various exercises relating to probability.

If it exists, find the value of the constant $c$ below which makes the function a PDF on the interval $[e,e^2]$ . If no such constant exists, enter N/A. $f(x) = c \ln x$ $c = \answer {e^{-2}}.$
You’ll need to integrate by parts.
For the same function as above, find the value of the constant $c$ which makes it a PDF on the interval $[e^{-2},e^2]$ . If no such constant exists, enter N/A. $c = \answer {N/A}.$
The function $\ln x$ is negative on $[e^{-2},1)$ and positive on $(1,e^2]$ . What does that imply for the possibility of it being a PDF?

Suppose $X$ is a random variable which represents the length of time your customers remain on hold before reaching a customer service agent. A reasonable model for such a random variable is to use an exponential distribution, i.e., to have a PDF given by $f(x) = c e^{-cx}$ for some positive constant $c$ . If the mean hold time is $2$ minutes, what is the standard deviation? $\sigma = \answer {2}.$ What is the probability that a customer will have to wait more than one standard deviation beyond the mean before their call is answered? $P = \answer {e^{-2}} \approx 13.5\%.$

First you need to solve for $c$ .

The equation will be $2 = \int c x e^{-cx} dx.$

This gives $c = 1/2$ .

The easiest formula to use here is $\sigma ^2 = \int _0^\infty x^2 f(x) dx - \mu ^2,$ where $\mu$ is the mean.

Suppose $X$ is a random variable on the interval $[1,10]$ whose PDF is given by $f(x) = \frac {c}{x}$ for some positive constant $c$ (note: this is the PDF that arises in the phenomenon known as Benford’s Law). Compute the mean $\mu$ and the median $m$ associated to this PDF. Which is larger? $\mu = \answer {\frac {9}{\ln 10}} \ \ \text { and } m = \answer {\sqrt {10}}$ The meanmedian is more sensitive to tails. Since this distribution has a tail extending to the left/for smaller $x$ to the right/for larger $x$ , it is expected (and in fact, true) that the mean is largersmaller than the median.

First you need to solve for the constant $c$ .

$c = \frac {1}{\ln 10}$

Sample Quiz Questions

Find the value of $c$ which makes the function $f(x) = \frac {1}{2}e^{-x} - ce^{-2x}$ a probability density function on the interval $[0, \infty )$ . What is the value of the mean $\mu$ of the corresponding random variable? (Hints will not be revealed until after you choose a response.)

$\displaystyle c = 1, ~ \mu = \frac {1}{2}$ $\displaystyle c = -1, ~ \mu = \frac {3}{4}$ $\displaystyle c = 1, ~ \mu = 1$ $\displaystyle c = -1, ~ \mu = \frac {5}{4}$ $\displaystyle c = 1, ~ \mu = \frac {3}{2}$ $\displaystyle c = -1, ~ \mu = \frac {7}{4}$

To compute the constant $c$ , we use the fact that the integral of a probability density function must equal $1$ , so $\mu = \int _{0}^{\infty } \left ( \frac {1}{2}e^{-x} - ce^{-2x} \right ) ~ dx = 1.$ This gives the equation $\frac {1}{2} - \frac {1}{2}c = 1,$ which then implies that $c = -1$ .

To compute the mean $\mu$ , we use the formula $\mu = \int _{0}^{\infty } x \left ( \frac {1}{2}e^{-x} + e^{-2x} \right ) ~ dx.$ Calculating the integral gives $\mu = 3/4.$

A certain random variable $X$ takes values in the interval $\left [2 \pi , \frac {5}{2} \pi \right ]$ . If the probability density function is given by $A \sin {x}$ for some appropriate value of the constant $A$ , compute the expected value $\mu$ of $X$ . (Hints will not be revealed until after you choose a response.)

$\displaystyle \mu = -1 + 2 \pi$ $\displaystyle \mu = 1 + \frac {3}{2} \pi$ $\displaystyle \mu = 2 \pi$ $\displaystyle \mu = -1 + \frac {5}{2} \pi$ $\displaystyle \mu = 1 + 2 \pi$ $\displaystyle \mu = \frac {5}{2} \pi$

The constant $A$ will be the reciprocal of the integral $\int _{2 \pi }^{\frac {5}{2} \pi } \sin {x}\, dx$

One can check that $\begin {aligned} \int _{2 \pi }^{\frac {5}{2} \pi } \sin {x}\, dx & = 1. \end {aligned}$

To compute the expected value $\mu$ we also need to compute the integral $\int _{2 \pi }^{\frac {5}{2} \pi } x \sin {x}\, dx$ To compute the integral, we can use integration by parts. A reasonable strategy is to integrate $\sin {x}$ and differentiate $x$ .

This gives the equality $\begin {aligned} \int x \sin {x}\, dx & = - x \cos {x} - \int \left (- \cos {x}\right )\, dx \\ & = - x \cos {x} + \sin {x}. \end {aligned}$ Therefore $\begin {aligned} \int _{2 \pi }^{\frac {5}{2} \pi } x \sin {x}\, dx & = \left . \left [- x \cos {x} + \sin {x} \right ] \right |_{2 \pi }^{\frac {5}{2} \pi }\\ & = 1 - \left (- 2 \pi \right ) = 1 + 2 \pi . \end {aligned}$ Therefore the expected value is the ratio of the integrals, i.e., $\begin {aligned} \mu & = \frac {1 + 2 \pi }{1} = 1 + 2 \pi . \end {aligned}$

Sample Exam Questions

A certain random variable $X$ has values in $(1,\infty )$ and has the property that there is some constant $C$ such that $P( X > a) = C \ln \frac {a^3+1}{a^3}$ for every $a > 1$ . Compute the value of $C$ and determine whether the expected value $\mu$ of $X$ is finite or infinite. [Hint: There is enough information given to compute $C$ without calculating any integrals.]

$C = \ln 2$ and $\mu < \infty$ $C = 1$ and $\mu < \infty$ $C = (\ln 2)^{-1}$ and $\mu < \infty$ $C = \ln 2$ and $\mu = \infty$ $C = 1$ and $\mu = \infty$ $C = (\ln 2)^{-1}$ and $\mu = \infty$

We know that $X$ is always greater than one, so $1 = P(X > 1) = C \ln \frac {1+1}{1},$ which gives $C = (\ln 2)^{-1}$ . If we let $f(x)$ denote the probability density function of $X$ , then $\frac {1}{\ln 2} \ln \frac {a^3+1}{a^3} = P(X > a) = \int _a^\infty f(x) dx.$ Differentiating both sides with respect to $a$ gives $\frac {1}{\ln 2} \left [ \frac {3 a^2}{a^3+1} - \frac {3}{a} \right ] = - f(a)$ so $f(a) = \frac {1}{\ln 2} \left [ \frac {3}{a} - \frac {3a^2}{a^3+1} \right ] = \frac {3}{a(a^3+1) \ln 2}.$ The expected value of $X$ must equal $\int _1^\infty \frac {3a}{a(a^3+1) \ln 2} da = \frac {3}{\ln 2} \int _1^\infty \frac {da}{a^3+1}.$ This integral will be finite by direct comparison to the convergent integral $\int _1^\infty a^{-3} da$ .

The function $f(x) = \begin {cases} \displaystyle \frac {k}{x^3} & 1 \leq x < \infty \\ 0 & \text {otherwise} \end {cases}$ is a probability density function for a certain value of $k$ . For that probability density function, find the probability that $x > 2$ .

$\displaystyle \frac {1}{2}$ $\displaystyle \frac {1}{3}$ $\displaystyle \frac {1}{4}$ $\displaystyle \frac {2}{3}$ $\displaystyle \frac {1}{5}$ $\displaystyle \frac {1}{6}$

For a certain real number $k$ , the function $f(X) = \begin {cases} \displaystyle \frac {k}{X^2+1} & \text {if } X \geq 0 \\ 0 & \text {otherwise} \end {cases}$ is a probability density function for a continuous random variable $X$ . For this value of $k$ , find the probability that $X > 1$ .

$0$ $\displaystyle \frac {1}{3}$ $\displaystyle \frac {2}{3}$ $1$ $\displaystyle \frac {1}{2}$ $\displaystyle \frac {1}{4}$

Let $f(r) = \begin {cases} C r^2 e^{-2r/b} & r \geq 0 \\ 0 & r < 0 \end {cases}.$ Find $C$ so that this is a probability density function (pdf) for the random variable $r$ . Here $b$ is a positive constant. This function is used to model the distance between the nucleus and the electron in a hydrogen atom. The constant $b$ is called the Bohr length. Find the mean of the pdf.

$\displaystyle C = \frac {b^3}{4}$ , mean $\displaystyle = b$ $\displaystyle C = \frac {4}{b^2}$ , mean $\displaystyle = b$ $\displaystyle C = \frac {4}{b}$ , mean $\displaystyle = b^2$ $\displaystyle C = \frac {4}{b^3}$ , mean $\displaystyle = \frac {3}{2}b$ $\displaystyle C = \frac {4}{b^2}$ , mean $\displaystyle = \frac {3}{2}b^2$ $\displaystyle C = \frac {4}{b}$ , mean $\displaystyle = \frac {3}{2}b^3$