Question

Suppose the reaction temperature X (in °C) in a certain chemical process has a uniform distribution with

A = −9

and B = 9.(a) Compute P(X < 0).


(b) Compute

P(−4.5

< X < 4.5).


(c) Compute

P(−7

≤ X ≤ 8). (Round your answer to two decimal places.)


(d) For k satisfying

−9 < k < k + 4 < 9,

compute P(k < X < k + 4). (Round your answer to two decimal places.)

Answer 1

Concepts and reason

Uniform distribution: Uniform distribution is a continuous probability distribution. It is defined between two parameters A and B. Here, A is called minimum value and B is called maximum value. Let X be continuous random variable with uniform distribution$U\left( {A,B} \right)$. That is, $X \sim U\left( {A,B} \right)$. Moreover, the probability density function for X is,

${f_X}\left( x \right) = \frac{1}{{B - A}}{\rm{ }}A < x < B$.

Fundamentals

Formula for finding the value of $P\left( {X > x} \right)$ is$P\left( {X > x} \right) = \int\limits_x^B {f\left( x \right)dx}$.

Formula for finding the value of $P\left( {X \le x} \right)$ is$P\left( {X \le x} \right) = \int\limits_A^x {f\left( x \right)dx}$.

Formula for finding the value of $P\left( {X \ge x} \right)$ is $P\left( {X \ge x} \right) = \int\limits_x^B {f\left( x \right)dx}$ or $P\left( {X \ge x} \right) = 1 - P\left( {X < x} \right)$.

Formula for finding the value of $P\left( {a \le X \le b} \right)$ is $P\left( {a \le X \le b} \right) = \int\limits_a^b {f\left( x \right)\,\,\,dx}$

(a)

The probability of $X < 0$ is obtained below:

From the information given, reaction temperature X (in °C) in a certain chemical process has a uniform distribution with $A = - 9\,\,{\rm{and}}\,\,B = 9$.

The density function is,

$\begin{array}{c}\\{f_X}\left( x \right) = \frac{1}{{B - A}}{\rm{ ; }}A < x < B\\\\ = \frac{1}{{9 - \left( { - 9} \right)}}\,\,\,;\, - 9 < x < 9\\\\ = \frac{1}{{18}}\\\end{array}$

The required probability is,

$\begin{array}{c}\\P\left( {X < 0} \right) = \int\limits_{ - 9}^0 {f\left( x \right)dx} \\\\ = \int\limits_{ - 9}^0 {\frac{1}{{18}}dx} \\\\ = \frac{1}{{18}}\left[ x \right]_{ - 9}^0\\\\ = \frac{1}{{18}}\left[ {0 + 9} \right]\\\end{array}$

$\begin{array}{c}\\ = \frac{1}{2}\\\\ = 0.5\\\end{array}$

(b)

The probability of $- 4.5 < X < 4.5$ is obtained below:

$\begin{array}{c}\\P\left( { - 4.5 < X < 4.5} \right) = \int\limits_{ - 4.5}^{4.5} {f\left( x \right)\,\,\,dx} \\\\ = \int\limits_{ - 4.5}^{4.5} {\frac{1}{{18}}\,\,\,dx} \\\\ = \frac{1}{{18}}\int\limits_{ - 4.5}^{4.5} {1\,\,\,dx} \\\\ = \frac{1}{{18}}\left[ x \right]_{ - 4.5}^{4.5}\\\end{array}$

$\begin{array}{c}\\ = \frac{1}{{18}}\left[ {4.5 - \left( { - 4.5} \right)} \right]\\\\ = \frac{9}{{18}}\\\\ = 0.5\\\end{array}$

(c)

The probability of $- 7 \le X \le 8$ is obtained below:

$\begin{array}{c}\\P\left( { - 7 \le X \le 8} \right) = \int\limits_{ - 7}^8 {f\left( x \right)\,\,\,dx} \\\\ = \int\limits_{ - 7}^8 {\frac{1}{{18}}\,\,\,dx} \\\\ = \frac{1}{{18}}\int\limits_{ - 7}^8 {1\,\,\,dx} \\\\ = \frac{1}{{18}}\left[ x \right]_{ - 7}^8\\\end{array}$

$\begin{array}{c}\\ = \frac{1}{{18}}\left[ {8 - \left( { - 7} \right)} \right]\\\\ = \frac{{15}}{{18}}\\\\ = 0.83\\\end{array}$

(d)

The value of $P\left( {k < X < k + 4} \right)$ is obtained below:

The required probability is,

$\begin{array}{c}\\P\left( {k < X < k + 4} \right) = \int\limits_k^{k + 4} {f\left( x \right)} dx\\\\ = \int\limits_k^{k + 4} {\frac{1}{{18}}\,\,} dx\\\\ = \frac{1}{{18}}\int\limits_k^{k + 4} {1\,\,} dx\\\\ = \frac{1}{{18}}\left[ x \right]_k^{k + 4}\\\end{array}$

$\begin{array}{c}\\ = \frac{1}{{18}}\left[ {k + 4 - k} \right]\\\\ = \frac{4}{{18}}\\\\ = 0.22\\\end{array}$

Ans: Part a

The probability that the random variable X takes values less than 0 is 0.50.

Part b

The probability that the random variable X takes values between –4.5 and 4.5 is 0.50.

Part c

The probability that the random variable X takes values between –7 and 8 is 0.83.

Part d

The probability of the random variable X lies between values k and k+4 is 0.22.