7.3: Problems on Distribution and Density Functions
 Page ID
 10863
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise \(\PageIndex{1}\)
(See Exercises 3 and 4 from "Problems on Random Variables and Probabilities"). The class \(\{C_j: 1 \le j \le 10\}\) is a partition. Random variable \(X\) has values {1, 3, 2, 3, 4, 2, 1, 3, 5, 2} on \(C_1\) through \(C_{10}\), respectively, with probabilities 0.08, 0.13, 0.06, 0.09, 0.14, 0.11, 0.12, 0.07, 0.11, 0.09. Determine and plot the distribution function \(F_X\).
 Answer

T = [1 3 2 3 4 2 1 3 5 2]; pc = 0.01*[8 13 6 9 14 11 12 7 11 9]; [X,PX] = csort(T,pc); ddbn Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX % See MATLAB plot
Exercise \(\PageIndex{2}\)
(See Exercise 6 from "Problems on Random Variables and Probabilities"). A store has eight items for sale. The prices are $3.50, $5.00, $3.50, $7.50, $5.00, $5.00, $3.50, and $7.50, respectively. A customer comes in. She purchases one of the items with probabilities 0.10, 0.15, 0.15, 0.20, 0.10 0.05, 0.10 0.15. The random variable expressing the amount of her purchase may be written
\(X = 3.5 I_{C_1} + 5.0 I_{C_2} + 3.5 I_{C_3} + 7.5 I_{C_4} + 5.0 I_{C_5} + 5.0 I_{C_6} + 3.5 I_{C_7} + 7.5 I_{C_8}\)
Determine and plot the distribution function for \(X\).
 Answer

T = [3.5 5 3.5 7.5 5 5 3.5 7.5]; pc = 0.01*[10 15 15 20 10 5 10 15]; [X,PX] = csort(T,pc); ddbn Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX % See MATLAB plot
Exercise \(\PageIndex{3}\)
(See Exercise 12 from "Problems on Random Variables and Probabilities"). The class \(\{A, B, C, D\}\) has minterm probabilities
\(pm = 0.001 *\) [5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302]
Determine and plot the distribution function for the random variable \(X = I_A + I_B + I_C + I_D\), which counts the number of the events which occur on a trial.
 Answer

npr06_12 Minterm probabilities in pm, coefficients in c T = sum(mintable(4)); % Alternate solution. See Exercise 6.2.12 from "Problems on Random Variables and Probabilities" [X,PX] = csort(T,pm); ddbn Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX % See MATLAB plot
Exercise \(\PageIndex{4}\)
Suppose a is a ten digit number. A wheel turns up the digits 0 through 9 with equal probability on each spin. On ten spins what is the probability of matching, in order, k or more of the ten digits in \(a\), \(0 \le k \le 10\)? Assume the initial digit may be zero.
 Answer

\(P =\) cbinom(10, 0.1, 0 : 10).
Exercise \(\PageIndex{5}\)
In a thunderstorm in a national park there are 127 lightning strikes. Experience shows that the probability of of a lightning strike starting a fire is about 0.0083. What is the probability that \(k\) fires are started, \(k =\) 0,1,2,3?
 Answer

P = ibinom(127,0.0083,0:3) P = 0.3470 0.3688 0.1945 0.0678
Exercise \(\PageIndex{6}\)
A manufacturing plant has 350 special lamps on its production lines. On any day, each lamp could fail with probability \(p = \) 0.0017. These lamps are critical, and must be replaced as quickly as possible. It takes about one hour to replace a lamp, once it has failed. What is the probability that on any day the loss of production time due to lamp failaures is \(k\) or fewer hours, \(k = 0, 1, 2, 3, 4, 5\)?
 Answer

P = 1  chinom(350, 0.0017, 1:6)
= 0.5513 0.8799 0.9775 0.9968 0.9996 1.0000
Exercise \(\PageIndex{7}\)
Two hundred persons buy tickets for a drawing. Each ticket has probability 0.008 of winning. What is the probability of \(k\) or fewer winners, \(k = 2, 3, 4\)?
 Answer

P = 1  cbinom(200,0.008,3:5) = 0.7838 0.9220 0.9768
Exercise \(\PageIndex{8}\)
Two coins are flipped twenty times. What is the probability the results match (both heads or both tails) \(k\) times, \(0 \le k \le 20\)?
 Answer

P = ibinom(20,1/2,0:20)
Exercise \(\PageIndex{9}\)
Thirty members of a class each flip a coin ten times. What is the probability that at least five of them get seven or more heads?
 Answer

p = cbinom(10,0.5,7) = 0.1719
P = cbinom(30,p,5) = 0.6052
Exercise \(\PageIndex{10}\)
For the system in Exercise 6, call a day in which one or more failures occur among the 350 lamps a “service day.” Since a Bernoulli sequence “starts over” at any time, the sequence of service/nonservice days may be considered a Bernoulli sequence with probability p_{1}, the probability of one or more lamp failures in a day.
 Beginning on a Monday morning, what is the probability the first service day is the first, second, third, fourth, fifth day of the week?
 What is the probability of no service days in a seven day week?
 Answer

p1 = 1  (1  0.0017)^350 = 0.4487 k = 1:5; (prob given day is a service day)

P = p1*(1  p1).^(k1) = 0.4487 0.2474 0.1364 0.0752 0.0414

P0 = (1  p1)^7 = 0.0155

Exercise \(\PageIndex{11}\)
For the system in Exercise 6 and Exercise 10 assume the plant works seven days a week. What is the probability the third service day occurs by the end of 10 days? Solve using the negative binomial distribution; repeat using the binomial distribution.
 Answer

p1 = 1  (1  0.0017)^350 = 0.4487
P = sum(nbinom(3,p1,3:10)) = 0.8990
Pa = cbinom(10,p1,3) = 0.8990
Exercise \(\PageIndex{12}\)
A residential College plans to raise money by selling “chances” on a board. Fifty chances are sold. A player pays $10 to play; he or she wins $30 with probability \(p =\) 0.2. The profit to the College is
\(X = 50 \cdot 10  30 N\), where \(N\) is the number of winners
Determine the distribution for \(X\) and calculate \(P(X > 0)\), \(P(X \ge 200)\), and \(P(X \ge 300)\)
 Answer

N = 0:50; PN = ibinom(50,0.2,0:50); X = 500  30*N; Ppos = (X>0)*PN' Ppos = 0.9856 P200 = (X>=200)*PN' P200 = 0.5836 P300 = (X>=300)*PN' P300 = 0.1034
Exercise \(\PageIndex{13}\)
A single sixsided die is rolled repeatedly until either a one or a six turns up. What is the probability that the first appearance of either of these numbers is achieved by the fifth trial or sooner?
 Answer

P = 1  (2/3)^5 = 0.8683
Exercise \(\PageIndex{14}\)
Consider a Bernoulli sequence with probability \(p =\) 0.53 of success on any component trial.
 The probability the fourth success will occur no later than the tenth trial is determined by the negative binomial distribution. Use the procedure nbinom to calculate this probability .
 Calculate this probability using the binomial distribution.
 Answer

P = sum(nbinom(4,0.53,4:10)) = 0.8729
Pa = cbinom(10,0.53,4) = 0.8729
Exercise \(\PageIndex{15}\)
Fifty percent of the components coming off an assembly line fail to meet specifications for a special job. It is desired to select three units which meet the stringent specifications. Items are selected and tested in succession. Under the usual assumptions for Bernoulli trials, what is the probability the third satisfactory unit will be found on six or fewer trials?
 Answer

P = cbinom(6,0.5,3) = 0.6562
Exercise \(\PageIndex{16}\)
The number of cars passing a certain traffic count position in an hour has Poisson (53) distribution. What is the probability the number of cars passing in an hour lies between 45 and 55 (inclusive)? What is the probability of more than 55?
 Answer

P1 = cpoisson(53,45)  cpoisson(53,56) = 0.5224
P2 = cpoisson(53,56) = 0.3581
Exercise \(\PageIndex{17}\)
Compare \(P(X \le k)\) and \(P(Y \le k)\) for \(X~\) binomial(5000, 0.001) and \(Y~\) Poisson (5), for \(0 \le k \le 10\). Do this directly with ibinom and ipoisson. Then use the mprocedure bincomp to obtain graphical results (including a comparison with the normal distribution).
 Answer

k = 0:10; Pb = 1  cbinom(5000,0.001,k+1); Pp = 1  cpoisson(5,k+1); disp([k;Pb;Pp]') 0 0.0067 0.0067 1.0000 0.0404 0.0404 2.0000 0.1245 0.1247 3.0000 0.2649 0.2650 4.0000 0.4404 0.4405 5.0000 0.6160 0.6160 6.0000 0.7623 0.7622 7.0000 0.8667 0.8666 8.0000 0.9320 0.9319 9.0000 0.9682 0.9682 10.0000 0.9864 0.9863
bincomp Enter the parameter n 5000 Enter the parameter p 0.001 Binomial stairs Poisson .. Adjusted Gaussian o o o gtext('Exercise 17')
Exercise \(\PageIndex{18}\)
Suppose \(X~\) binomial (12, 0.375), \(Y~\) Poisson (4.5), and \(Z~\) exponential (1/4.5). For each random variable, calculate and tabulate the probability of a value at least \(k\), for integer values \(3 \le k \le 8\).
 Answer

k = 3:8; Px = cbinom(12,0.375,k); Py = cpoisson(4.5,k); Pz = exp(k/4.5); disp([k;Px;Py;Pz]') 3.0000 0.8865 0.8264 0.5134 4.0000 0.7176 0.6577 0.4111 5.0000 0.4897 0.4679 0.3292 6.0000 0.2709 0.2971 0.2636 7.0000 0.1178 0.1689 0.2111 8.0000 0.0390 0.0866 0.1690
Exercise \(\PageIndex{19}\)
The number of noise pulses arriving on a power circuit in an hour is a random quantity having Poisson (7) distribution. What is the probability of having at least 10 pulses in an hour? What is the probability of having at most 15 pulses in an hour?
 Answer

P1 = cpoisson(7,10) = 0.1695 P2 = 1  cpoisson(7,16) = 0.9976
Exercise \(\PageIndex{20}\)
The number of customers arriving in a small specialty store in an hour is a random quantity having Poisson (5) distribution. What is the probability the number arriving in an hour will be between three and seven, inclusive? What is the probability of no more than ten?
 Answer

P1 = cpoisson(5,3)  cpoisson(5,8) = 0.7420
P2 = 1  cpoisson(5,11) = 0.9863
Exercise \(\PageIndex{21}\)
Random variable \(X~\) binomial (1000, 0.1).
 Determine \(P(X \ge 80)\), \(P(X \ge 100)\), \(P(X \ge 120)\)
 Use the appropriate Poisson distribution to approximate these values.
 Answer

k = [80 100 120]; P = cbinom(1000,0.1,k) P = 0.9867 0.5154 0.0220 P1 = cpoisson(100,k) P1 = 0.9825 0.5133 0.0282
Exercise \(\PageIndex{22}\)
The time to failure, in hours of operating time, of a televesion set subject to random voltage surges has the exponential (0.002) distribution. Suppose the unit has operated successfully for 500 hours. What is the (conditional) probability it will operate for another 500 hours?
 Answer

\(P(X > 500 + 500X > 500) = P(X > 500) = e^{0.002 \cdot 500} = 0.3679\)
Exercise \(\PageIndex{23}\)
For \(X~\) exponential (\(\lambda\)), determine \(P(X \ge 1/\lambda)\), \(P(X \ge 2/\lambda)\).
 Answer

\(P(X > k\lambda) = e^{\lambda k/ \lambda} = e^{k}\)
Exercise \(\PageIndex{24}\)
Twenty “identical” units are put into operation. They fail independently. The times to failure (in hours) form an iid class, exponential (0.0002). This means the “expected” life is 5000 hours. Determine the probabilities that at least \(k\), for \(k\) = 5,8,10,12,15, will survive for 5000 hours.
 Answer

p = exp(0.0002*5000) p = 0.3679 k = [5 8 10 12 15]; P = cbinom(20,p,k) P = 0.9110 0.4655 0.1601 0.0294 0.0006
Exercise \(\PageIndex{25}\)
Let \(T~\) gamma (20, 0.0002) be the total operating time for the units described in Exercise 24.
 Use the mfunction for the gamma distribution to determine \(P(T \le 100,000)\).
 Use the Poisson distribution to determine \(P(T \le 100,000)\).
 Answer

P1 = gammadbn(20,0.0002,100000) = 0.5297 P2 = cpoisson(0.0002*100000,20) = 0.5297
Exercise \(\PageIndex{26}\)
The sum of the times to failure for five independent units is a random variable \(X~\) gamma (5, 0.15). Without using tables or mprograms, determine \(P(X le 25)\).
 Answer

\(P(X \le 25) = P(Y \ge 5)\), \(Y~\) Poisson \((0.15 \cdot 25 = 3.75)\)
\(P(Y \ge 5) = 1  P(Y \le 4) = 1  e^{3.35} (1 + 3.75 + \dfrac{3.75^2}{2} + \dfrac{3.75^3}{3!} + \dfrac{3.75^4}{24}) = 0.3225\)
Exercise \(\PageIndex{27}\)
Interarrival times (in minutes) for fax messages on a terminal are independent, exponential (\(\lambda =\) 0.1). This means the time \(X\) for the arrival of the fourth message is gamma(4, 0.1). Without using tables or mprograms, utilize the relation of the gamma to the Poisson distribution to determine \(P \le 30\).
 Answer

\(P(X \le 30) = P(Y \ge 4)\), \(Y~\) poisson (\(0.2 \cdot 30 = 3\))
\(P(Y \ge 4) = 1  P(Y \le 3) = 1  e^{3} (1 + 3 + \dfrac{3^2}{2} + \dfrac{3^3}{3!}) = 0.3528\)
Exercise \(\PageIndex{28}\)
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 (3, 3). Without using tables or mprograms, determine \(P(X \le 2)\).
 Answer

\(P(X \le 2) = P(Y \ge 3)\), \(Y ~\) poisson (\(3 \cdot 2 = 6\))
\(P(Y \ge 3) = 1  P(Y \le 2) = 1  e^{6} (1 + 6 + 36/2) = 0.9380\)
Exercise \(\PageIndex{29}\)
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 \(P(Z \le 20)\).
 Answer

\(Z~\) gamma (6, 1/3).
\(P(Z \le 20) = P(Y \ge 6)\), \(Y~\) poisson\((1/3 \cdot 20)\)
\(P(Y \ge 6)\) = cpoisson(20/3, 6) = 0.6547
Exercise \(\PageIndex{30}\)
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 variable uniformly 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.)
 Answer

p = cbinom(35,0.71,25) = 0.5620
Exercise \(\PageIndex{31}\)
Five “identical” electronic devices are installed at one time. The units fail independently, and the time to failure, in days, of each is a random variable 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?
 Answer

p = exp(15/30) = 0.6065 P = cbinom(5,p,4) = 0.3483
Exercise \(\PageIndex{32}\)
Suppose \(X ~ N\)(4, 81). That is, \(X\) has gaussian distribution with mean \(\mu\) = 4 and variance \(\sigma^2\) = 81.
 Use a table of standardized normal distribution to determine \(P(2 < X < 8)\) and \(P(X  4 \le 5)\).
 Calculate the probabilities in part (a) with the mfunction gaussian.
 Answer

a.
\(P(2 < X < 8) = \phi((8  4)/9)  \phi ((2  4)/9)\) =
\(\phi (4/9) + \phi (2/9)  1 = 0.6712 + 0.5875  1 = 0.2587\)
\(P(X  4 \le 5) = 2\phi(5/9)  1 = 1.4212  1 = 0.4212\)
b.
P1 = gaussian(4,81,8)  gaussian(4,81,2) P1 = 0.2596 P2 = gaussian(4,81,9)  gaussian(4,84,1) P2 = 0.4181
Exercise \(\PageIndex{33}\)
Suppose \(X ~ N(5, 81)\). That is, \(X\) has gaussian distribution with \(\mu\) = 5 and \(\sigma^2\) = 81. Use a table of standardized normal distribution to determine \(P(3 < X < 9)\) and \(P(X  5 le 5)\). Check your results using the mfunction gaussian.
 Answer

\(P(3 < X < 9) = \phi ((9  5)/9)  \phi ((3  5)/9) = \phi(4/9) + \phi(2/9)  1 = 0.6712 + 0.5875  1 = 0.2587\)
\(P(X  5 \le 5) = 2 \phi(5/9)  1 = 1.4212  1 = 0.4212\)
P1 = gaussian(5,81,9)  gaussian(5,81,3) P1 = 0.2596 P2 = gaussian(5,81,10)  gaussian(5,84,0) P2 = 0.4181
Exercise \(\PageIndex{34}\)
Suppose \(X ~ N(3, 64)\). That is, \(X\) has gaussian distribution with \(\mu\) = 3 and \(\sigma^2\) = 64. Use a table of standardized normal distribution to determine \(P(1 < X < 9)\) and \(P(X  3 le 4)\). Check your results with the mfunction gaussian.
 Answer

\(P(1 < X < 9) = \phi((9  3)/8)  \phi(1  3)/9) =\)
\(\phi(0.75) + \phi(0.25)  1 = 0.7734 + 0.5987  1 = 0.3721\)
\(P(X  3 \le 4) = 2 \phi(4/8)  1 = 1.3829  1 = 0.3829\)
P1 = gaussian(3,64,9)  gaussian(3,64,1) P1 = 0.3721 P2 = gaussian(3,64,7)  gaussian(3,64,1) P2 = 0.3829
Exercise \(\PageIndex{35}\)
Items coming off an assembly line have a critical dimension which is represented by a random variable \(~ N\) (10, 0.01). Ten items are selected at random. What is the probability that three or more are within 0.05 of the mean value \(\mu\).
 Answer

p = gaussian(10,0.01,10.05)  gaussian(10,0.01,9.95) p = 0.3829 P = cbinom(10,p,3) P = 0.8036
Exercise \(\PageIndex{36}\)
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 seconds per 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. Check with the mfunction gaussian.
 Answer

\(P(5 \le X \le 10) = \phi(5/ \sqrt{300}) + \phi(10/\sqrt{300})  1 = \phi(0.289) + \phi(0.577)  1 = 0.614 + 0.717  1 = 0.331\)
\(P =\) gaussian(5, 300, 10)  gaussian(5, 300, 5) = 0.3317
Exercise \(\PageIndex{37}\)
Use the mprocedure 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.
 Answer

Experiment with the mprocedure bincomp.
Exercise \(\PageIndex{38}\)
Use the mprocedure poissapp to compare the Poisson and gaussian distributions. Use various values of \(\mu\) from 10 to 500.
 Answer

Experiment with the mprocedure poissapp.
Exercise \(\PageIndex{39}\)
Random variable \(X\) has density \(f_X(t) = \dfrac{3}{2} t^2\), \(1 \le t \le 1\) (and zero elsewhere).
 Determine \(P(0.5 \le X < 0.8)\), \(P(X > 0.5)\), \(P(X  0.25) \le 0.5)\).
 Determine an expression for the distribution function.
 Use the mprocedures tappr and cdbn to plot an approximation to the distribution function.
 Answer


\(\dfrac{3}{2} \int t^2 = t^3/2\)
\(P1 = 0.5 * (0.8^3  (0.5)^3) = 0.3185\) \(P2 = 2 \int_{0.5}^{1} \dfrac{3}{2} t^2 = (1  (0.5)^3) = 7/8\)
\(P3 = P(X  0.25 \le 0.5) = P(0.25 \le X \le 0.75) = \dfrac{1}{2}[(3/4)^3  (1/4)^3] = 7/32\)  \(F_X (t) = \int_{1}^{1} f_X = \dfrac{1}{2} (t^3 + 1)\)

tappr Enter matrix [a b] of xrange endpoints [1 1] Enter number of x approximation points 200 Enter density as a function of t 1.5*t.^2 Use row matrices X and PX as in the simple case cdbn Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX % See MATLAB plot

Exercise \(\PageIndex{40}\)
Random variable \(X\) has density function \(f_X(t) = t  \dfrac{3}{8}t^3\), \(0 \le t \le 2\) (and zero elsewhere).
 Determine \(P(X \le 0.5)\), \(P(0.5 \le X < 1.5)\), \(P(X  1 < 1/4)\).
 Determine an expression for the distribution function.
 Use the mprocedures tappr and cdbn to plot an approximation to the distribution function.
 Answer


\(\int (t  \dfrac{3}{8} t^2) = \dfrac{t^2}{2}  \dfrac{t^3}{8}\)
\(P1 = 0.5^2/2  0.5^3/8 = 7/64\) \(P2 = 1.5^2 /2  1.5^3 /8  7/64 = 19/32\) \(P3 = 79/256)\)  \(F_X (t) = \dfrac{t^2}{2}  \dfrac{t^3}{8}\), \(0 \le t \le 2\)

tappr Enter matrix [a b] of xrange endpoints [0 2] Enter number of x approximation points 200 Enter density as a function of t t  (3/8)*t.^2 Use row matrices X and PX as in the simple case cdbn Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX % See MATLAB plot

Exercise \(\PageIndex{41}\)
Random variable \(X\) has density function
\(f_X (t) = \begin{cases} (6/5) t^2 & \text{for } 0 \le t \le 1 \\ (6/5) (2 t) & \text{for } 1 < t \le 2 \end{cases} = I[0, 1] (t) \dfrac{6}{5} t^2 + I_{(1, 2]} (t) \dfrac{6}{5} (2 t)\)
 Determine \(P(X \le 0.5)\), \(P(0.5 \le X < 1.5)\), \(P(X  1 < 1/4)\).
 Determine an expression for the distribution function.
 Use the mprocedures tappr and cdbn to plot an approximation to the distribution function.
 Answer


\(P1 = \dfrac{6}{5} \int_{0}^{1/2} t^2 = 1/20\) \(P2 = \dfrac{6}{5} \int_{1/2}^{1} t^2 + \dfrac{6}{5} \int_{1}^{3/2} (2  t) = 4/5\)
\(P3) = \dfrac{6}{5} \int_{3/4}^{1} t^2 + \dfrac{6}{5} \int_{1}^{5/4} (2  t) = 79/160\) 
\(F_X (t) = \int_{0}^{1} f_X = I_{[0, 1]} (t) \dfrac{2}{5}t^3 + I_{(1, 2]} (t) [\dfrac{7}{5} + \dfrac{6}{5} (2t  \dfrac{t^2}{2})]\)

tappr Enter matrix [a b] of xrange endpoints [0 2] Enter number of x approximation points 400 Enter 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 case cdbn Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX % See MATLAB plot
