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  ## Continuous Probability Distributions
  
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  ### Continuous Probability Distributions
  
  Chapter 8
  
  0
  
  1/3
  
  1/2
  
  1
  
  2/3
  
  8.1 Continuous Probability Distributions
  • Unlike a discrete random variable which we studied in Chapter 7, a continuous random variable（連續型隨機變數）is one that can assume an uncountable number of values in the interval (a,b).
  • We cannot list the possible values because there is an infinite number of them.
  • Because there is an infinite number of values, the probability that a continuous variable X will assume any particular value is 0. Why?
  
  The probability of each value
  
  1/4 + 1/4 + 1/4 + 1/4 = 1
  
  1/3 + 1/3 + 1/3 = 1
  
  1/2 + 1/2 = 1
  
  0
  
  1/3
  
  1/2
  
  1
  
  2/3
  
  8.1 Continuous Probability Distributions
  
  As the number of values increases the probability of each
  
  value decreases. This is so because the sum of all the
  
  probabilities remains 1.
  
  When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0.
  
  The probability of each value
  
  1/4 + 1/4 + 1/4 + 1/4 = 1
  
  1/3 + 1/3 + 1/3 = 1
  
  1/2 + 1/2 = 1
  
  Point Probabilities are Zero
  • Because there is an infinite number of values, the probability of each individual value is virtually 0.
  • Thus, we can determine the probability of a range of values only.
  • E.g. with a discrete random variable like tossing a die, it is meaningful to talk about P(X=5), say.
  • In a continuous setting (e.g. with time as a random variable), the probability the random variable of interest, say task length, takes exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.
  • It is meaningful to talk about P(X ≤ 5).
  
  Area = 1
  
  x1
  
  x2
  
  P(x1≤X≤x2)
  
  Probability Density Function
  • To calculate probabilities we define a probability density function f(x).
  • A function f(x) is called a probability density function（機率密度函數）(over the range a ≤ x ≤ b if it meets the following requirements:
  
  1) f(x) ≥ 0 for all x between a and b, and
  
  2) The total areaunder the curve between a and b is 1.0
  
  • The probability that X falls between x1 and x2 is found by calculating the area under the graph of f(x) between x1 and x2.
  Uniform Distribution（均等分配）
  • Consider the uniform probability distribution (sometimes called the rectangular probability distribution（矩形分配）).
  • It is described by the function:
  
  f(x)
  
  a
  
  b
  
  x
  
  Uniform Distribution
  • A random variable X is said to be uniformly distributed if its density function is
  • The expected value and the variance are
  Example 8.1
  • The amount of gasoline sold daily at a service station is uniformly distributed with a minimum of 2,000 gallons and a maximum of 5,000 gallons. Find the probability that sales are:
  • Between 2,500 and 3,000 gallons
  • More than 4,000 gallons
  • Exactly 2,500 gallons
  Example 8.1
  • The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons.
  • Find the probability that sales are between 2,500 and 3,000 gallons.
  
  f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
  
  P(2500£X£3000) = (3000-2500)(1/3000) = .1667
  
  There is about a 17% chance that between 2,500 and 3,000 gallons of gas will be sold on a given day.
  
  1/3000
  
  x
  
  2000
  
  2500
  
  3000
  
  5000
  
  Example 8.1
  • The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons.
  • Find the probability that sales are more than 4,000 gallons.
  
  f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
  
  There is a 33% chance the gas station will sell more than 4,000 gallons on any given day.
  
  P(X³4000) = (5000-4000)(1/3000) = .333
  
  1/3000
  
  x
  
  2000
  
  4000
  
  5000
  
  Example 8.1
  • The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons.
  • Find the probability that sales are exactly 2,500 gallons.
  
  f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
  
  The probability that the gas station will sell exactly 2,500 gallons is zero.
  
  P(X=2500) = (2500-2500)(1/3000) = 0
  
  1/3000
  
  x
  
  2000
  
  2500
  
  5000
  
  8.2 Normal Distribution（常態分配）
  • This is the most important continuous distribution.
  • Many distributions can be approximated by a normal distribution.
  • The normal distribution is the cornerstone distribution of statistical inference.
  Normal Distribution
  • A random variable X with mean m and variance s2is normally distributed if its probability density function is given by
  
  Unlike the range of the uniform distribution (a ≤ x ≤ b),
  
  Normal distributions range from minus infinity to plus infinity.
  
  The Normal Distribution
  • Important things to note:
  
  The normal distribution is fully defined by two parameters: its standard deviation andmean
  
  The normal distribution is bell shaped and symmetrical about the mean.
  
  The Shape of the Normal Distribution
  
  The normal distribution is bell shaped, and symmetrical around m.
  
  m
  
  90
  
  110
  
  Why symmetrical? Let m = 100. Suppose x = 110.
  
  Now suppose x = 90
  
  The effects of m and s
  
  How does the standard deviation affect the shape of f(x)?
  
  s= 2
  
  Increasing the standard deviation “flattens” the curve.
  
  s =3
  
  s =4
  
  How does the expected value affect the location of f(x)?
  
  m = 10
  
  m = 11
  
  m = 12
  
  Increasing the mean shifts the curve to the right.
  
  0
  
  1
  
  1
  
  Standard Normal Distribution
  • A normal distribution whose mean is 0 and standard deviation is 1 is called the standard normal distribution（標準常態分配）.
  • As we shall see shortly, any normal distribution can be converted to a standard normal distribution with simple algebra. This makes calculations much easier.
  Finding Normal Probabilities
  • Two facts help calculate normal probabilities:
  • The normal distribution is symmetrical.
  • Any normal distribution can be transformed into a specific normal distribution called
  
  “STANDARD NORMAL DISTRIBUTION”
  
  • Example
  
  The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?
  
  Finding Normal Probabilities
  • Solution
  • If Xdenotes the assembly time of a computer, we seek the probability P(45<X<60).
  • This probability can be calculated by creating a new normal variable the standard normal variable.
  
  This shifts the mean of X to zero…
  
  Every normal variable
  
  with some m and s, can be transformed into this Z.
  
  Therefore, once probabilities for Z are calculated, probabilities of any normal variable can be found.
  
  E(Z) = m = 0, V(Z) = s2 = 1
  
  This changes the shape of the curve…
  
  Finding Normal Probabilities
  • Example - continued
  
  - m
  
  45
  
  - 50
  
  X
  
  60
  
  - 50
  
  P(45<X<60) = P( < < )
  
  s
  
  10
  
  10
  
  = P(-0.5 < Z < 1)
  
  To complete the calculation we need to compute
  
  the probability under the standard normal distribution
  
  P(0<Z<z0)
  
  Using the Standard Normal Table
  
  Standard normal probabilities have been
  
  calculated and are provided in a table .
  
  The tabulated probabilities correspond
  
  to the area between Z=0 and some Z = z0 >0
  
  Z = z0
  
  Z = 0
  
  - m
  
  45
  
  - 50
  
  X
  
  60
  
  - 50
  
  P(45<X<60) = P( < < )
  
  s
  
  10
  
  10
  
  z0 = 1
  
  z0 = -.5
  
  Finding Normal Probabilities
  • Example - continued
  
  = P(-.5 < Z < 1)
  
  We need to find the shaded area
  
  - m
  
  45
  
  - 50
  
  X
  
  60
  
  - 50
  
  P(45<X<60) = P( < < )
  
  s
  
  10
  
  10
  
  z0 = 1
  
  z0 =-.5
  
  Finding Normal Probabilities
  • Example - continued
  
  = P(-.5<Z<1) =
  
  P(-.5<Z<0)+ P(0<Z<1)
  
  P(0<Z<1
  
  .3413
  
  z=0
  
  Finding Normal Probabilities
  • The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows:
  
  -z0
  
  +z0
  
  0
  
  P(-z0<Z<0) = P(0<Z<z0)
  
  Finding Normal Probabilities
  • Example - continued
  
  .3413
  
  .1915
  
  -.5
  
  .5
  
  .1915
  
  .1915
  
  .1915
  
  .1915
  
  Finding Normal Probabilities
  • Example - continued
  
  .3413
  
  1.0
  
  -.5
  
  .5
  
  P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328
  
  “Just over half the time, 53% or so, a computer will have an assembly time between 45 minutes and 1 hour”
  
  Using the Normal Table (Table 3)
  • What is P(Z > 1.6) ?
  
  P(0 < Z < 1.6) = .4452
  
  z
  
  0
  
  1.6
  
  P(Z > 1.6) = .5 – P(0 < Z < 1.6)
  
  = .5 – .4452
  
  = .0548
  
  Using the Normal Table (Table 3)
  • What is P(Z < -2.23) ?
  
  P(0 < Z < 2.23)
  
  P(Z < -2.23)
  
  P(Z > 2.23)
  
  z
  
  0
  
  -2.23
  
  2.23
  
  P(Z < -2.23) = P(Z > 2.23)
  
  = .5 – P(0 < Z < 2.23)
  
  = .0129
  
  Using the Normal Table (Table 3)
  • What is P(Z < 1.52) ?
  
  P(0 < Z < 1.52)
  
  P(Z < 0) = .5
  
  z
  
  0
  
  1.52
  
  P(Z < 1.52) = .5 + P(0 < Z < 1.52)
  
  = .5 + .4357
  
  = .9357
  
  Using the Normal Table (Table 3)
  • What is P(0.9 < Z < 1.9) ?
  
  P(0 < Z < 0.9)
  
  P(0.9 < Z < 1.9)
  
  z
  
  0
  
  0.9
  
  1.9
  
  P(0.9 < Z < 1.9) = P(0 < Z < 1.9) – P(0 < Z < 0.9)
  
  =.4713 – .3159
  
  = .1554
  
  Finding Normal Probabilities
  • Determine the following probability:
  
  P(Z>1.47)
  
  = P(Z>1.47)
  
  0.5
  
  - P(0<Z<1.47)
  
  0
  
  1.47
  
  P(Z>1.47) = 0.5 - 0.4292 = 0.0708
  
  Finding Normal Probabilities
  
  • Determine the following probability:
  
  P(-2.25<Z<1.85)
  
  P(-2.25<Z<0) = ?
  
  =.4878
  
  P(0<Z<1.85) = .4678
  
  P(0<Z<2.25) = .4878
  
  0
  
  2.25
  
  -2.25
  
  1.85
  
  P(-2.25<Z<1.85) = 0.4878 + 0.4678 = 0.9556
  
  Finding Normal Probabilities
  
  • Determine the following probability:
  
  P(.65<Z<1.36)
  
  P(0<Z<1.36) = .4131
  
  P(0<Z<.65) = .2422
  
  1.36
  
  0
  
  .65
  
  P(.65<Z<1.36) = .4131 - .2422 = .1709
  
  0%
  
  10%
  
  Example 8.2
  • The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%.
  • What is the probability of losing money?
  
  Some advice: always draw a picture!
  
  X
  
  0 - 10
  
  5
  
  .4772
  
  (i) P(X< 0 ) = P(Z< ) = P(Z< - 2)
  
  Z
  
  2
  
  -2
  
  0
  
  =P(Z>2) =
  
  0.5 - P(0<Z<2) = 0.5 - .4772 = .0228
  
  0%
  
  10%
  
  1
  
  Example 8.2
  • The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%.
  • What is the probability of losing money?
  
  X
  
  0 - 10
  
  10
  
  .3413
  
  (ii) P(X< 0 ) = P(Z< )
  
  Z
  
  -1
  
  = P(Z< - 1) = P(Z>1) =
  
  0.5 - P(0<Z<1) = 0.5 - .3413 = .1587（>.0228）
  
  Increasing the standard deviation increases the probability of losing money, which is why the standard deviation (or the variance) is a measure of risk.
  
  Finding Values of Z
  • Sometimes we need to find the value of Z for a given probability
  • We use the notation zA to express a Z value for which P(Z > zA) = A
  
  A
  
  zA
  
  0.05
  
  Finding Values of Z
  • Example 8.3 & 8.4
  • Determine z exceeded by 5% of the population
  • Determine z such that 5% of the population is below
  • Solution
  
  z.05is defined as the z value for which the area on its right under the standard normal curve is .05.
  
  0.45
  
  0.05
  
  1.645
  
  Z0.05
  
  -Z0.05
  
  0
  
  Finding Values of Z
  • Other Z values are
  • Z.05 = 1.645
  • Z.01 = 2.33
  Using the values of Z
  • Because z.025 = 1.96 and - z.025= -1.96, it follows that we can state
  
  P(-1.96 < Z < 1.96) = .95
  
  • Similarly
  
  P(-1.645 < Z < 1.645) = .90
  
  8.3 Exponential Distribution（指數分配）
  • The exponential distribution can be used to model
  • the length of time between telephone calls
  • the length of time between arrivals at a service station
  • the life-time of electronic components.
  • When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.
  Exponential Distribution
  • A random variable is exponentially distributed if its probability density function is given by
  • Note that x ≥ 0. Time (for example) is a non-negative quantity; the exponential distribution is often used for time related phenomena such as the length of time between phone calls or between parts arriving at an assembly station. Note also that the mean and standard deviation are equal to each other and to the inverse of the parameter of the distribution (lambda)
  Exponential Distribution
  • The exponential distribution depends upon the value of .
  • Smaller values of “flatten” the curve.
  • E.g. exponential distributions for  = .5, 1, 2
  Exponential Distribution
  • If X is an exponential random variable, then we can calculate probabilities by:
  • P(X > a) = e–la
  • P(X < a) = 1 – e –la
  • P(a1 < X < a2) = e – l(a1) – e – l(a2)
  Exponential Distribution
  • Example 8.6
  • The lifetime of an alkaline battery is exponentially distributed with l = .05 per hour.
  • What is the mean and standard deviation of the battery’s lifetime?
  • Find the following probabilities:
  • The battery will last between 10 and 15 hours.
  • The battery will last for more than 20 hours.
  Exponential Distribution
  • Solution
  • The mean = standard deviation = 1/l = 1/.05 = 20 hours.
  • Let X denote the lifetime.
  • P(X > 20) = e-.05(20) = .3679
  
  “There is about a 13% chance a battery will only last 10 to 15 hours”
  
  Exponential Distribution
  • Example 8.7
  • The service rate at a supermarket checkout is 6 customers per hour.
  • If the service time is exponential, find the following probabilities:
  • A service is completed in 5 minutes,
  • A customer leaves the counter more than 10 minutes after arriving
  • A service is completed between 5 and 8 minutes.
  Exponential Distribution
  • Solution
  • A service rate of 6 per hour = A service rate of .1 per minute (l = .1/minute).
  • P(X < 5) = 1-e-lx = 1 – e-.1(5) = .3935
  • P(X >10) = e-lx = e-.1(10) = .3679
  • P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572
  Exponential Distribution
  • Cars arrive randomly and independently to a tollbooth at an average of 360 cars per hour.
  • Use the exponential distribution to find the probability that the next car will not arrive within half a minute.
  • Solution
  • Let X denote the time (in minutes) that elapses before the next car arrives. X is exponentially distributed with l = 360/60 = 6 cars per minute.P(X>.5) = e-6(.5) = .0498.
  Exponential Distribution vs. Poisson Distribution
  • What is the probability that no car will arrive within the next half minute?
  • Solution
  • If Y counts the number of cars that will arrive in the next half minute, then Y is a Poisson variable with m = (.5)(6) = 3 cars per half a minute.P(Y = 0) = e-3(30)/0! = .0498.
  
  Comment:
  
  If the first car will not arrive within the next half a minute then no car will arrives within the next half minute.
  
  Therefore, not surprisingly, the probability found here is the exact same probability found in the previous question.
  
  8.4 Other Continuous Distribution
  • Three other important continuous distributions which will be used extensively in later sections are introduced here:
  • Student t Distribution
  • Chi-Squared Distribution
  • F Distribution
  Student t Distribution（t分配）
  • Here the letter t is used to represent the random variable, hence the name. The density function for the Student t distribution is as follows:
  • (nu) is called the degrees of freedom（自由度）
  • (Gamma function) is (k)=(k-1)(k-2)…(2)(1)
  Student t Distribution
  • Much like the standard normal distribution, the Student t distribution is “mound” shaped and symmetrical about its mean of zero:
  • The mean and variance of a Student t random variable are
  
  E(t) = 0 and
  
  Student t Distribution
  • In much the same way that  and  define the normal distribution,  , the degrees of freedom, defines the Student t Distribution:
  • As the number of degrees of freedom increases, the t distribution approaches the standard normal distribution.
  Determining Student t Values
  • The student t distribution is used extensively in statistical inference.
  • Thus, it is important to determine values of tA associated with a given number of degrees of freedom.
  • We can do this using
  • t tables
  • Excel
  • Minitab
  Determining Student t Values
  • The student t distribution is used extensively in statistical inference. Table 4 in Appendix B lists values of .
  • That is, values of a Student t random variable with degrees of freedom  such that:
  • The values for A are pre-determined “critical” values, typically in the 10%, 5%, 2.5%, 1% and 0.5% range.
  
  A
  
  A
  
  = .05
  
  = .05
  
  -tA
  
  Using the t Table
  • The table provides the t values (tA) for which P(tn > tA) = A
  
  t
  
  t
  
  t
  
  t
  
  The t distribution is
  
  symmetrical around 0
  
  tA
  
  =-1.812
  
  =1.812
  
  t.100
  
  t.05
  
  t.025
  
  t.01
  
  t.005
  
  Student t Probabilities and Values
  • Excel can calculate Student distribution probabilities and values.
  Chi – Squared Distribution（卡方分配）
  • The chi-squared density function is given by:
  • As before, the parameter  is the number of degrees of freedom.
  The Chi – Squared Distribution
  • The chi-squared distribution is not symmetrical
  • The square, i.e. , forces non-negative values (e.g. finding is illogical).
  • Chi squared values can be found from the chi squared table, from Excel, or from Minitab.
  The Chi – Squared Distribution
  • Table 5 in Appendix B makes it easy to look-up the right hand tail probability (A), for which
  The Chi – Squared Distribution
  • For probabilities of this sort:
  • We use 1–A,
  
  i.e. we are determining
  
  A
  
  c2A
  
  Using the Chi-Squared Table
  
  To find c2 for which P(c2n<c2)=.01, lookup the column labeledc21-.01 or c2.99
  
  =.05
  
  A =.99
  
  c2.05
  
  c2.995 c2.990 c2.05 c2.010 c2.005
  
  For Example
  • To find the point in a chi-squared distribution with 8 degrees of freedom, such that the area to the right is .05,
  • Look up the intersection of the 8 d.f. row with the 0.05 column, yielding a value of 15.5073
  For Example
  • To find the point in a chi-squared distribution with 8 degrees of freedom, such that the area to the left is .05,
  • Look up the intersection of the 8 d.f. row with the 0.95 column, yielding a value of 2.73264
  Chi-Squared Probabilities and Values
  • Excel can calculate chi-squared distribution probabilities and values.
  F Distribution（F分配）
  • The F density function is given by:
  • Two parameters define this distribution, and like we’ve already seen these are again degrees of freedom.
  • is the “numerator” degrees of freedom（分子的自由度）
  • is the “denominator” degrees of freedom （分母的自由度）
  F Distribution
  • The mean and variance of an F random variable are given by:
  • and
  • The F distribution is similar to the distribution in that its starts at zero (is non-negative) and is not symmetrical.
  Determining Values of F
  • The values of the F variable can be found in the F table, Excel, or from Minitab.
  • The entries in the table are the values of the F variable of the right hand tail probability (A), for which
  Determining Values of F
  • For example, what is the value of F for 5% of the area under the right hand “tail” of the curve, with a numerator degree of freedom of 3 and a denominator degree of freedom of 7?
  • Solution: use the F look-up (Table 6)
  
  There are different tables
  
  for different values of A.
  
  the correct table!!
  
  F.05,3,7
  
  Denominator Degrees of Freedom : ROW
  
  Numerator Degrees of Freedom : COLUMN
  
  Determining Values of F
  • For areas under the curve on the left hand side of the curve, we can leverage the following relationship:
  
  Pay close attention to the order of the terms!
  
  F Probabilities and Values
  • Excel can calculate F distribution probabilities and values.
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