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Chapter 8

Chapter 8. Continuous Probability Distributions. Probability Density Functions…. Unlike a discrete random variable which we studied in Chapter 7, a continuous random variable is one that can assume an uncountable number of values.

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Chapter 8

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  1. Chapter 8 Continuous Probability Distributions

  2. Probability Density Functions… • Unlike a discrete random variable which we studied in Chapter 7, a continuous random variable is one that can assume an uncountable number of values. •  We cannot list the possible values because there is an infinite number of them. •  Because there is an infinite number of values, the probability of each individual value is virtually 0.

  3. 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).

  4. Probability Density Function… • A function f(x) is called a probability density function (over the range a ≤ x ≤ b if it meets the following requirements: • f(x) ≥ 0 for all x between a and b, and • The total area under the curve between a and b is 1.0 f(x) area=1 a b x

  5. Uniform Distribution… • Consider the uniform probability distribution (sometimes called the rectangular probability distribution). • It is described by the function: f(x) a b x area = width x height = (b – a) x = 1

  6. Example 8.1(b)… • 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. • What is the probability that the service station will sell at least 4,000 gallons? • Algebraically: what is P(X ≥ 4,000) ? • P(X ≥ 4,000) = (5,000 – 4,000) x (1/3000) = .3333 f(x) 2,000 5,000 x

  7. The Normal Distribution… • The normal distribution is the most important of all probability distributions. The probability density function of a normal random variable is given by: • It looks like this: • Bell shaped, • Symmetrical around the mean …

  8. 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 Unlike the range of the uniform distribution (a ≤ x ≤ b) Normal distributions range from minus infinity to plus infinity

  9. 0 1 1 Standard Normal Distribution… • A normal distribution whose mean is zero and standard deviation is one is called the standard normal distribution. • As we shall see shortly, any normal distribution can be converted to a standard normal distribution with simple algebra. This makes calculations much easier.

  10. Normal Distribution… • The normal distribution is described by two parameters: • its mean and its standard deviation . Increasing the mean shifts the curve to the right…

  11. Normal Distribution… • The normal distribution is described by two parameters: • its mean and its standard deviation . Increasing the standard deviation “flattens” the curve…

  12. Calculating Normal Probabilities… • Example: The time required to build 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? • Algebraically speaking, what is P(45 < X < 60) ? 0

  13. Calculating Normal Probabilities… • P(45 < X < 60) ? …mean of 50 minutes and a standard deviation of 10 minutes… 0

  14. Calculating Normal Probabilities… • P(–.5 < Z < 1) looks like this: • The probability is the area • under the curve… • We will add up the • two sections: • P(–.5 < Z < 0) and • P(0 < Z < 1) 0 –.5 … 1

  15. Calculating Normal Probabilities… • How to use Table 3… [other forms of normal tables exist] This table gives probabilities P(0 < Z < z) First column = integer + first decimal Top row = second decimal place P(0 < Z < 0.5) P(0 < Z < 1) P(–.5 < Z < 1) = .1915 + .3414 = .5328 The probability time is between 45 and 60 minutes = .5328

  16. 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

  17. 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

  18. 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

  19. 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

  20. Example 8.2 • The return on investment is normally distributed with a mean of 10% and a standard deviation of 5%. What is the probability of losing money? • We want to determine P(X < 0). Thus,

  21. Finding Values of ZA… • Often we’re asked to find some value of Z for a given probability, i.e. given an area (A) under the curve, what is the corresponding value of z (zA) on the horizontal axis that gives us this area? That is: • P(Z > zA) = A

  22. Finding Values of Z… • What value of z corresponds to an area under the curve of 2.5%? That is, what is z.025 ? Area = .50 Area = .025 Area = .50–.025 = .4750 If you do a “reverse look-up” on Table 3 for .4750,you will get the corresponding zA = 1.96 Since P(z > 1.96) = .025, we say: z.025 = 1.96

  23. Finding Values of Z… • Other Z values are • Z.05 = 1.645 • Z.01 = 2.33 • Will show you shortly how to use the “t-tables” with infinite degrees of freedom to find a bunch of these standard values for Zα

  24. 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 • The old Empirical Rule stated about 95% within + 2σ • P(-2 < Z < 2) = .95 • From now on we will use the 1.96 number for this statement unless you are just talking in general terms about how much of a population in with + 2σ • Similarly • P(-1.645 < Z < 1.645) = .90

  25. Exponential Distribution…[Not on test] • Another important continuous distribution is the exponential distribution which has this probability density function: • 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 )

  26. Exponential Distribution…[Not on test] • The exponential distribution depends upon the value of • Smaller values of “flatten” the curve: • (E.g. exponential • distributions for • = .5, 1, 2)

  27. Other Continuous Distributions… • Three other important continuous distributions which will be used extensively in later sections are introduced here: • Student t Distribution, Looks like the standard normal distribution (Z) after someone sat on it • Chi-Squared Distribution, • F Distribution.

  28. Student t Distribution…[don’t really need to know formula] • 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, and • (Gamma function) is (k)=(k-1)(k-2)…(2)(1)

  29. Student t Distribution…[1 parameter] • In much the same way that and define the normal distribution [2 parameters], , the degrees of freedom, defines the Student [will use df] • t Distribution: • As the number of degrees of freedom increases, the t distribution approaches the standard normal distribution. Figure 8.24

  30. Using the t table (Table 4) for values… • For example, if we want the value of t with 10 degrees of freedom such that the area under the Student t curve is .05: Area under the curve value (t) : COLUMN t.05,10 t.05,10=1.812 Degrees of Freedom : ROW

  31. Student t Probabilities and Values • Excel can calculate Student distribution probabilities and values. Warning: Excel will give you the value for “t” where  is the area in “BOTH” tails

  32. Chi-Squared Distribution…[Not on test] • The chi-squared density function is given by: • As before, the parameter is the number of degrees of freedom. Figure 8.27

  33. F Distribution…[Not on test] • The F density function is given by: • F > 0. Two parameters define this distribution, and like we’ve already seen these are again degrees of freedom. • is the “numerator” degrees of freedom and • is the “denominator” degrees of freedom.

  34. Problems: Standard Normal “Z” • If the random variable Z has a standard normal distribution, calculate the following probabilities. • P(Z > 1.7) = • P(Z < 1.7) = • P(Z > -1.7) = • P(Z < -1.7) = • P(-1.7 < Z < 1.7)

  35. Problems: Normal Distribution • If the random variable X has a normal distribution with • mean 40 and std. dev. 5, calculate the following • probabilities. • P(X > 43) = • P(X < 38) = • P(X = 40) = • P(X > 23) =

  36. Problem: Normal • The time (Y) it takes your professor to drive home each • night is normally distributed with mean 15 minutes and • standard deviation 2 minutes. Find the following • probabilities. Draw a picture of the normal distribution and • show (shade) the area that represents the probability you are • calculating. • P(Y > 25) = • P( 11 < Y < 19) = • P (Y < 18) =

  37. Problem: Target the Mean • The manufacturing process used to make “heart pills” is • known to have a standard deviation of 0.1 mg. of active ingredient. • Doctors tell us that a patient who takes a pill with over 6 mg. of • active ingredient may experience kidney problems. Since you want to • protect against this (and most likely lawyers), you are asked to • determine the “target” for the mean amount of active ingredient in each • pill such that the probability of a pill containing over 6 mg. is 0.0035 ( • 0.35% ). You may assume that the amount of active ingredient in a pill • is normally distributed. • *Solve for the target value for the mean. • *Draw a picture of the normal distribution you came up with and show the 3 sigma limits.

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