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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 2 ) .

P ( X 2 ) = P ( Y 3 ) , Y poisson ( 3 · 2 = 6 )
P ( Y 3 ) = 1 - P ( Y 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 20 ) .

Z gamma (6,1/3).

P ( Z 20 ) = P ( Y 6 ) , Y poisson ( 1 / 3 · 20 )
P ( Y 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 N ( 4 , 81 ) . That is, X has gaussian distribution with mean μ = 4 and variance σ 2 = 81 .

  1. Use a table of standardized normal distribution to determine P ( 2 < X < 8 ) and P ( | X - 4 | 5 ) .
  2. Calculate the probabilities in part (a) with the m-function gaussian.
  1. 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 | 5 ) = 2 Φ ( 5 / 9 ) - 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
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Suppose X 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 | 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 | 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 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 | 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 | 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 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 / 300 ) + Φ ( 10 / 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 ) = 3 2 t 2 , - 1 t 1 (and zero elsewhere).

  1. Determine P ( - 0 . 5 X < 0 , 8 ) , P ( | X | > 0 . 5 ) , P ( | X - 0 . 25 | 0 . 5 ) .
  2. Determine an expression for the distribution function.
  3. Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.
3 2 t 2 = t 3 / 2
  1. P 1 = 0 . 5 * ( 0 . 8 3 - ( - 0 . 5 ) 3 ) = 0 . 3185 P 2 = 2 0 . 5 1 3 2 t 2 = ( 1 - ( - 0 . 5 ) 3 ) = 7 / 8
    P 3 = P ( | X - 0 . 25 | 0 . 5 ) = P ( - 0 . 25 X 0 . 75 ) = 1 2 [ ( 3 / 4 ) 3 - ( - 1 / 4 ) 3 ] = 7 / 32
  2. F X ( t ) = - 1 t f X = 1 2 ( t 3 + 1 )
  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
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Random variable X has density function f X ( t ) = t - 3 8 t 2 , 0 t 2 (and zero elsewhere).

  1. Determine P ( X 0 . 5 ) , P ( 0 . 5 X < 1 . 5 ) , P ( | X - 1 | < 1 / 4 ) .
  2. Determine an expression for the distribution function.
  3. Use the m-procedures tappr and cdbn to plot an approximation to the distribution function.
( t - 3 8 t 2 ) = t 2 2 - t 3 8
  1. 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
  2. F X ( t ) = t 2 2 - t 3 8 , 0 t 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
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Random variable X has density function

f X ( t ) = ( 6 / 5 ) t 2 for 0 t 1 ( 6 / 5 ) ( 2 - t ) for 1 < t 2 = I [ 0 , 1 ] ( t ) 6 5 t 2 + I ( 1 , 2 ] ( t ) 6 5 ( 2 - t )
  1. Determine P ( X 0 . 5 ) , P ( 0 . 5 X < 1 . 5 ) , P ( | X - 1 | < 1 / 4 ) .
  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. P 1 = 6 5 0 1 / 2 t 2 = 1 / 20 P 2 = 6 5 1 / 2 1 t 2 + 6 5 1 3 / 2 ( 2 - t ) = 4 / 5
    P 3 = 6 5 3 / 4 1 t 2 + 6 5 1 5 / 4 ( 2 - t ) = 79 / 160
  2. F X ( t ) = 0 t f X = I [ 0 , 1 ] ( t ) 2 5 t 3 + I ( 1 . 2 ] ( t ) [ - 7 5 + 6 5 ( 2 t - t 2 2 ) ]
  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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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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