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

Applied probability 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 $(3, 3)$ . Without using tables or m-programs, determine $P (X \leq 2)$ .

P (X \leq 2) = P (Y \geq 3), Y \sim poisson (3 \cdot 2 = 6)

P (Y \geq 3) = 1 - P (Y \leq 2) = 1 - e^{- 6} (1 + 6 + 36 / 2) = 0.9380

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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 $P (Z \leq 20)$ .

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

P (Z \leq 20) = P (Y \geq 6), Y \sim poisson (1 / 3 \cdot 20)

P (Y \geq 6) = c p o i s s o n (20 / 3, 6) = 0.6547

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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

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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

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

Use a table of standardized normal distribution to determine $P (2 < X < 8)$ and $P (| X - 4 | \leq 5)$ .
Calculate the probabilities in part (a) with the m-function gaussian.

$P (2 < X < 8) = Φ ((8 - 4) / 9) - Φ ((2 - 4) / 9) =$

$Φ (4 / 9) + Φ (2 / 9) - 1 = 0.6712 + 0.5875 - 1 = 0.2587$

$P (| X - 4 | \leq 5) = 2 Φ (5 / 9) - 1 = 1.4212 - 1 = 0.4212$
P1 = gaussian(4,81,8) - gaussian(4,81,2) P1 = 0.2596P2 = gaussian(4,81,9) - gaussian(4,84,-1) P2 = 0.4181

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Suppose $X \sim N (5, 81)$ . That is, X has gaussian distribution with $μ = 5$ and $σ^{2} = 81$ . Use a table of standardized normal distribution to determine $P (3 < X < 9)$ and $P (| X - 5 | \leq 5)$ . Check your results using the m-function gaussian.

P (3 < X < 9) = Φ ((9 - 5) / 9) - Φ ((3 - 5) / 9) = Φ (4 / 9) + Φ (2 / 9) - 1 = 0.6712 + 0.5875 - 1 = 0.2587

P (| X - 5 | \leq 5) = 2 Φ (5 / 9) - 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

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Suppose $X \sim N (3, 64)$ . That is, X has gaussian distribution with $μ = 3$ and $σ^{2} = 64$ . Use a table of standardized normal distribution to determine $P (1 < X < 9)$ and $P (| X - 3 | \leq 4)$ . Check your results with the m-function gaussian.

P (1 < X < 9) = Φ ((9 - 3) / 8) - Φ ((1 - 3) / 9) =

Φ (0.75) + Φ (0.25) - 1 = 0.7734 + 0.5987 - 1 = 0.3721

P (| X - 3 | \leq 4) = 2 Φ (4 / 8) - 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

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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

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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 $μ = 5$ and $σ^{2} = 300$ . 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.

$P (- 5 < X < 10) = Φ (5 / \sqrt{300}) + Φ (10 / \sqrt{300}) - 1 = Φ (0.289) + Φ (0.577) - 1 = 0.614 + 0.717 - 1 = 0.331$

P = g a u s s i a n (5, 300, 10) - g a u s s i a n (5, 300, - 5) = 0.3317

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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.

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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.

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Random variable X has density $f_{X} (t) = \frac{3}{2} t^{2}, - 1 \leq t \leq 1$ (and zero elsewhere).

Determine $P (- 0.5 \leq X < 0, 8)$ , $P (| X | > 0.5)$ , $P (| X - 0.25 | \leq 0.5)$ .
Determine an expression for the distribution function.
Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.

\frac{3}{2} \int t^{2} = t^{3} / 2

$P 1 = 0.5 * (0 . 8^{3} - {(- 0.5)}^{3}) = 0.3185 P 2 = 2 \int_{0.5}^{1} \frac{3}{2} t^{2} = (1 - {(- 0.5)}^{3}) = 7 / 8$

$P 3 = P (| X - 0.25 | \leq 0.5) = P (- 0.25 \leq X \leq 0.75) = \frac{1}{2} [{(3 / 4)}^{3} - {(- 1 / 4)}^{3}] = 7 / 32$
$F_{X} (t) = \int_{- 1}^{t} f_{X} = \frac{1}{2} (t^{3} + 1)$
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

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Random variable X has density function $f_{X} (t) = t - \frac{3}{8} t^{2}, 0 \leq t \leq 2$ (and zero elsewhere).

Determine $P (X \leq 0.5)$ , $P (0.5 \leq X < 1.5)$ , $P (| X - 1 | < 1 / 4)$ .
Determine an expression for the distribution function.
Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.

\int (t - \frac{3}{8} t^{2}) = \frac{t^{2}}{2} - \frac{t^{3}}{8}

$P 1 = 0 . 5^{2} / 2 - 0 . 5^{3} / 8 = 7 / 64 P 2 = 1 . 5^{2} / 2 - 1 . 5^{3} / 8 - 7 / 64 = 19 / 32 P 3 = 79 / 256$
$F_{X} (t) = \frac{t^{2}}{2} - \frac{t^{3}}{8}, 0 \leq t \leq 2$
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

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Random variable X has density function

f_{X} (t) = \{\begin{matrix} (6 / 5) t^{2} & for 0 \leq t \leq 1 \\ (6 / 5) (2 - t) & for 1 < t \leq 2 \end{matrix}) = I [0, 1] (t) \frac{6}{5} t^{2} + I_{(1, 2]} (t) \frac{6}{5} (2 - t)

Determine $P (X \leq 0.5)$ , $P (0.5 \leq X < 1.5)$ , $P (| X - 1 | < 1 / 4)$ .
Determine an expression for the distribution function.
Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.

$P 1 = \frac{6}{5} \int_{0}^{1 / 2} t^{2} = 1 / 20 P 2 = \frac{6}{5} \int_{1 / 2}^{1} t^{2} + \frac{6}{5} \int_{1}^{3 / 2} (2 - t) = 4 / 5$

$P 3 = \frac{6}{5} \int_{3 / 4}^{1} t^{2} + \frac{6}{5} \int_{1}^{5 / 4} (2 - t) = 79 / 160$
$F_{X} (t) = \int_{0}^{t} f_{X} = I_{[0, 1]} (t) \frac{2}{5} t^{3} + I_{(1.2]} (t) [- \frac{7}{5} + \frac{6}{5} (2 t - \frac{t^{2}}{2})]$
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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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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