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A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 5: Q. 5.1 (page 215)

The speed of a molecule in a uniform gas at equilibrium is a random variable whose probability density function is given by
$f (x) = \{\begin{cases} a x^{2} e^{鈭� b x^{2}} & x 鈮� 0 \\ 0 & x < 0 \end{cases}$
where $b = m / 2 k T$ and $m$ denote, respectively,
Boltzmann鈥檚 constant, the absolute temperature of the gas,
and the mass of the molecule. Evaluate $a$ in terms of $b$ .

Short Answer

Expert verified

Therefore $a$ in terms of $b$ then $a = \frac{4 b \frac{3}{2}}{\sqrt{蟺}}$ .

Step by step solution

01

Given Information.

$f (x) = \{\begin{cases} a x^{2} e^{鈭� b x^{2}} & x 鈮� 0 \\ 0 & x < 0 \end{cases}$

Where,

$b = m / 2 k T$

02

Explanation.

${鈭�}_{0}^{鈭�} f (x) d x = 1 {鈭�}_{0}^{鈭�} a x^{2} e^{- b x^{2}} d x = 1$

localid="1649615141474" $b x^{2} = t 2 x d x = d t {鈭�}_{0}^{鈭�} a x^{2} e^{- t} 脳 \frac{d t}{2 b x} = 1$

03

Explanation.

$b x^{2} = t 鈬� x^{2} = \frac{t}{b} 鈬� x = \sqrt{\frac{t}{b}} \frac{a}{2 b^{\frac{3}{2}}} {鈭�}_{0}^{鈭�} t^{\frac{1}{2}} e^{- t} d t = 1 \frac{a}{2 b \frac{3}{2}} 脳 \sqrt{\frac{3}{2}} = 1 \frac{a \sqrt{蟺}}{4 b \frac{3}{2}} = 1 a = \frac{4 b \frac{3}{2}}{\sqrt{蟺}}$

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Most popular questions from this chapter

Consider the function

$f (x) = \{\begin{cases} C (2 x 鈭� x^{3}) & 0 < x < \frac{5}{2} \\ 0 & o t h e r w i s e \end{cases}$

Could $f$ be a probability density function? If so, determine

$C$ . Repeat if $f (x)$ were given by

localid="1646632841793" $f (x) = \{\begin{cases} C (2 x 鈭� x 3) & 0 < x < \frac{5}{2} \\ 0 & o t h e r w i s e \end{cases}$

If $X$ is an exponential random variable with a parameter $位 = 1$ , compute the probability density function of the random variable $Y$ defined by $Y = \log (X)$

If X is an exponential random variable with parameter 位, and c > 0, show that cX is exponential with parameter 位/c

Jones figures that the total number of thousands of miles that an auto can be driven before it would need to be junked is an exponential random variable with a parameter $\frac{1}{20}$ . Smith has a used car that he claims has been driven only $10, 000$ miles. If Jones purchases the car, what is the

probability that she would get at least $20, 000$ additional miles out of it? Repeat under the assumption that the life-

time mileage of the car is not exponentially distributed, but rather is (in thousands of miles) uniformly distributed over $(0, 40)$ .

Let $Y = \frac{X - 饾湀}{饾浖}$ .

Show that if X is a Weibull random variable with parameters 谓, 伪, and 尾, then Y is an exponential random variable with parameter 位 = 1 and vice versa.

See all solutions

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