# 7.2 Problems on distribution and density functions  (Page 4/4)

Customers arrive at a service center with independent interarrival times in hours, which have exponential (3) distribution. The time X for the third arrival is thus gamma $\left(3,\phantom{\rule{0.277778em}{0ex}}3\right)$ . Without using tables or m-programs, determine $P\left(X\le 2\right)$ .

$P\left(X\le 2\right)=P\left(Y\ge 3\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}Y\sim \phantom{\rule{0.277778em}{0ex}}\text{poisson}\left(3·2=6\right)$
$P\left(Y\ge 3\right)=1-P\left(Y\le 2\right)=1-{e}^{-6}\left(1+6+36/2\right)=0.9380$

Five people wait to use a telephone, currently in use by a sixth person. Suppose time for the six calls (in minutes) are iid, exponential (1/3).What is the distribution for the total time Z from the present for the six calls? Use an appropriate Poisson distribution to determine $P\left(Z\le 20\right)$ .

$Z\sim$ gamma (6,1/3).

$P\left(Z\le 20\right)=P\left(Y\ge 6\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}Y\sim \phantom{\rule{0.277778em}{0ex}}\text{poisson}\left(1/3·20\right)$
$P\left(Y\ge 6\right)=\mathtt{c}\mathtt{p}\mathtt{o}\mathtt{i}\mathtt{s}\mathtt{s}\mathtt{o}\mathtt{n}\left(\mathtt{20}/\mathtt{3},\mathtt{6}\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\mathtt{0}.\mathtt{6547}$

A random number generator produces a sequence of numbers between 0 and 1. Each of these can be considered an observed value of a random variableuniformly distributed on the interval [0, 1]. They assume their values independently.A sequence of 35 numbers is generated. What is the probability 25 or more are less than or equal to 0.71? (Assume continuity. Do not make a discrete adjustment.)

p = cbinom(35,0.71,25) = 0.5620

Five “identical” electronic devices are installed at one time. The units fail independently, and the time to failure, in days, of each is a randomvariable exponential (1/30). A maintenance check is made each fifteen days. What is the probability that at least four are still operating at the maintenance check?

p = exp(-15/30) = 0.6065 P = cbinom(5,p,4) = 0.3483

Suppose $X\sim N\left(4,81\right)$ . That is, X has gaussian distribution with mean $\mu =4$ and variance ${\sigma }^{2}=81$ .

1. Use a table of standardized normal distribution to determine $P\left(2 and $P\left(|X-4|\le 5\right)$ .
2. Calculate the probabilities in part (a) with the m-function gaussian.
1. $P\left(2
$\Phi \left(4/9\right)+\Phi \left(2/9\right)-1=0.6712+0.5875-1=0.2587$
$P\left(|X-4|\le 5\right)=2\Phi \left(5/9\right)-1=1.4212-1=0.4212$
2. P1 = gaussian(4,81,8) - gaussian(4,81,2) P1 = 0.2596P2 = gaussian(4,81,9) - gaussian(4,84,-1) P2 = 0.4181

Suppose $X\sim N\left(5,\phantom{\rule{0.277778em}{0ex}}81\right)$ . That is, X has gaussian distribution with $\mu =5$ and ${\sigma }^{2}=81$ . Use a table of standardized normal distribution to determine $P\left(3 and $P\left(|X-5|\le 5\right)$ . Check your results using the m-function gaussian.

$P\left(3
$P\left(|X-5|\le 5\right)=2\Phi \left(5/9\right)-1=1.4212-1=0.4212$
P1 = gaussian(5,81,9) - gaussian(5,81,3) P1 = 0.2596P2 = gaussian(5,81,10) - gaussian(5,84,0) P2 = 0.4181

Suppose $X\sim N\left(3,\phantom{\rule{0.277778em}{0ex}}64\right)$ . That is, X has gaussian distribution with $\mu =3$ and ${\sigma }^{2}=64$ . Use a table of standardized normal distribution to determine $P\left(1 and $P\left(|X-3|\le 4\right)$ . Check your results with the m-function gaussian.

$P\left(1
$\Phi \left(0.75\right)+\Phi \left(0.25\right)-1=0.7734+0.5987-1=0.3721$
$P\left(|X-3|\le 4\right)=2\Phi \left(4/8\right)-1=1.3829-1=0.3829$
P1 = gaussian(3,64,9) - gaussian(3,64,1) P1 = 0.3721P2 = gaussian(3,64,7) - gaussian(3,64,-1) P2 = 0.3829

Items coming off an assembly line have a critical dimension which is represented by a random variable $\sim$ N(10, 0.01). Ten items are selected at random. What is the probability that three or more are within 0.05 of themean value μ .

p = gaussian(10,0.01,10.05) - gaussian(10,0.01,9.95) p = 0.3829P = cbinom(10,p,3) P = 0.8036

The result of extensive quality control sampling shows that a certain model of digital watches coming off a production line have accuracy, in seconds per month,that is normally distributed with $\mu =5$ and ${\sigma }^{2}=300$ . To achieve a top grade, a watch must have an accuracy within the range of -5 to +10 secondsper month. What is the probability a watch taken from the production line to be tested will achieve top grade? Calculate, using a standardized normal table. Checkwith the m-function gaussian.

$P\left(-5

$\mathtt{P}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\mathtt{g}\mathtt{a}\mathtt{u}\mathtt{s}\mathtt{s}\mathtt{i}\mathtt{a}\mathtt{n}\left(\mathtt{5},\mathtt{300},\mathtt{10}\right)\phantom{\rule{3.33333pt}{0ex}}-\phantom{\rule{3.33333pt}{0ex}}\mathtt{g}\mathtt{a}\mathtt{u}\mathtt{s}\mathtt{s}\mathtt{i}\mathtt{a}\mathtt{n}\left(\mathtt{5},\mathtt{300},-\mathtt{5}\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\mathtt{0}.\mathtt{3317}$

Use the m-procedure bincomp with various values of n from 10 to 500 and p from 0.01 to 0.7, to observe the approximation of the binomial distribution by the Poisson.

Experiment with the m-procedure bincomp.

Use the m-procedure poissapp to compare the Poisson and gaussian distributions. Use various values of μ from 10 to 500.

Experiment with the m-procedure poissapp.

Random variable X has density ${f}_{X}\left(t\right)=\frac{3}{2}{t}^{2},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}-1\le t\le 1$ (and zero elsewhere).

1. Determine $P\left(-0.5\le X<0,8\right)$ , $P\left(|X|>0.5\right)$ , $P\left(|X-0.25|\le 0.5\right)$ .
2. Determine an expression for the distribution function.
3. Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.
$\frac{3}{2}\int {t}^{2}={t}^{3}/2$
1. $P1=0.5*\left(0.{8}^{3}-{\left(-0.5\right)}^{3}\right)=0.3185\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P2=2{\int }_{0.5}^{1}\frac{3}{2}{t}^{2}=\left(1-{\left(-0.5\right)}^{3}\right)=7/8$
$P3=P\left(|X-0.25|\le 0.5\right)=P\left(-0.25\le X\le 0.75\right)=\frac{1}{2}\left[{\left(3/4\right)}^{3}-{\left(-1/4\right)}^{3}\right]=7/32$
2. ${F}_{X}\left(t\right)={\int }_{-1}^{t}{f}_{X}=\frac{1}{2}\left({t}^{3}+1\right)$
3. tappr Enter matrix [a b]of x-range endpoints [-1 1] Enter number of x approximation points 200Enter density as a function of t 1.5*t.^2 Use row matrices X and PX as in the simple casecdbn Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % See MATLAB plot

Random variable X has density function ${f}_{X}\left(t\right)=t-\frac{3}{8}{t}^{2},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le t\le 2$ (and zero elsewhere).

1. Determine $P\left(X\le 0.5\right)$ , $P\left(0.5\le X<1.5\right)$ , $P\left(|X-1|<1/4\right)$ .
2. Determine an expression for the distribution function.
3. Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.
$\int \left(t-\frac{3}{8}{t}^{2}\right)=\frac{{t}^{2}}{2}-\frac{{t}^{3}}{8}$
1. $P1=0.{5}^{2}/2-0.{5}^{3}/8=7/64\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P2=1.{5}^{2}/2-1.{5}^{3}/8-7/64=19/32\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P3=79/256$
2. ${F}_{X}\left(t\right)=\frac{{t}^{2}}{2}-\frac{{t}^{3}}{8},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le t\le 2$
3. tappr Enter matrix [a b]of x-range endpoints [0 2] Enter number of x approximation points 200Enter density as a function of t t - (3/8)*t.^2 Use row matrices X and PX as in the simple casecdbn Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % See MATLAB plot

Random variable X has density function

${f}_{X}\left(t\right)=\left\{\begin{array}{cc}\left(6/5\right){t}^{2}& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le t\le 1\hfill \\ \left(6/5\right)\left(2-t\right)& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1
1. Determine $P\left(X\le 0.5\right)$ , $P\left(0.5\le X<1.5\right)$ , $P\left(|X-1|<1/4\right)$ .
2. Determine an expression for the distribution function.
3. Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.
1. $P1=\frac{6}{5}{\int }_{0}^{1/2}{t}^{2}=1/20\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P2=\frac{6}{5}{\int }_{1/2}^{1}{t}^{2}+\frac{6}{5}{\int }_{1}^{3/2}\left(2-t\right)=4/5$
$P3=\frac{6}{5}{\int }_{3/4}^{1}{t}^{2}+\frac{6}{5}{\int }_{1}^{5/4}\left(2-t\right)=79/160$
2. ${F}_{X}\left(t\right)={\int }_{0}^{t}{f}_{X}={I}_{\left[0,1\right]}\left(t\right)\frac{2}{5}{t}^{3}+{I}_{\left(1.2\right]}\left(t\right)\left[-\frac{7}{5}+\frac{6}{5}\left(2t-\frac{{t}^{2}}{2}\right)\right]$
3. tappr Enter matrix [a b]of x-range endpoints [0 2] Enter number of x approximation points 400Enter density as a function of t (6/5)*(t<=1).*t.^2 + ... (6/5)*(t>1).*(2 - t) Use row matrices X and PX as in the simple casecdbn Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % See MATLAB plot

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