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Q. 77

Expert-verifiedFound in: Page 353

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

A subway train arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution. a. Define the random variable. X = _______ b. X ~ _______ c. Graph the probability distribution. d. f(x) = _______ e. μ = _______ f. σ = _______ g. Find the probability that the commuter waits less than one minute. h. Find the probability that the commuter waits between three and four minutes. i. Sixty percent of commuters wait more than how long for the train? State this in a probability question, similarly to parts g and h, draw the picture, and find the probabilit

All data has been provided below

a.

As per basis of provided information , X is the time length commuter that wait for a train

b.

Uniform distribution of random variable X is

$X=U(0,8)$

c. The probability distribution is

_{$\begin{array}{rcl}f\left(x\right)& =& \frac{1}{b-a}\\ & =& \frac{1}{8-0}\\ & =& \frac{1}{8}\end{array}$}

So, the graph is

d.

The calculation of part c is

$f\left(x\right)=\left\{\begin{array}{l}\frac{1}{8}\\ 0\end{array}\right.0<x<8\phantom{\rule{0ex}{0ex}}oisforotherwise$

e.

The mean value is

$\begin{array}{rcl}\mu & =& \frac{a+b}{2}\\ & =& \frac{0+18}{2}\\ & =& 8/2\\ & =& 4\end{array}$

f.

The value of standard deviation

$\begin{array}{rcl}\sigma & =& \frac{\sqrt{{\left(b-a\right)}^{2}}}{12}\\ \sigma & =& \frac{\sqrt{{\left(8-0\right)}^{2}}}{12}\\ & =& 2.31\end{array}$

g.

$P(x<1)=base\times height\phantom{\rule{0ex}{0ex}}=(1-0)\times \frac{1}{8}\phantom{\rule{0ex}{0ex}}=0.125$

h. The calculation

$P(3<x<4)=base\times height\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & (4-3)\times \frac{1}{8}\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 0\end{array}\begin{array}{rcl}& & .\end{array}\begin{array}{rcl}& & 125\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}$

i. The calculation of probability

$\begin{array}{rcl}& & P(x>k)=base\times height\\ 0.60& =& (8-k)\times \frac{1}{8}\\ k& =& 3.2\\ & & \end{array}$

The curve of the probability

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