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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. • 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).

  3. 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

  4. 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 …

  5. 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 Normal distributions range from minus infinity to plus infinity

  6. 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.

  7. 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…

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

  9. Calculating Normal Probabilities… • We can use the following function to convert any normal random variable to a standard normal random variable… 0 Some advice: always draw a picture!

  10. 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

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

  12. Calculating Normal Probabilities… • OK, we’ve converted P(45 < X < 60) for a normal distribution with mean = 50 and standard deviation = 10 • to • P(–.5 < Z < 1) [i.e. the standard normal distribution with mean = 0 and standard deviation = 1] • so • Where do we go from here?!

  13. 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) • We can use Table 3 in Appendix B to look-up probabilities P(Z < z) 0 –.5 … 1

  14. Calculating Normal Probabilities… • Recap: 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? P(45 < X < 60) = P(–.5 < Z < 1) = .5328 “Just over half the time, 53% or so, a computer will have an assembly time between 45 minutes and 1 hour”

  15. Using the Normal Table (Table 3)… • What is P(Z > 1.6) ? P(Z < 1.6) = .9452 z 0 1.6 P(Z > 1.6) = 1.0 – P(Z < 1.6) = 1.0 – .9452 = .0548

  16. Using the Normal Table (Table 3)… • What is P(0.9 < Z < 1.9) ? P(Z < 0.9) P(Z < 1.9) z 0 0.9 1.9 P(0.9 < Z < 1.9) = P(Z < 1.9) – P(Z < 0.9) =.9713 – .8159 = .1554

  17. 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 .9750,you will get the corresponding zA = 1.96 Since P(z > 1.96) = .025, we say: z.025 = 1.96

  18. Finding Values of Z… • Other Z values are • Z.05 = 1.645 • Z.01 = 2.33

  19. 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 • Recall that the empirical rule stated that approximately 95% would be within + 2 standard deviations. From now on we use 1.96 instead of 2. • Similarly • P(-1.645 < Z < 1.645) = .90

  20. Other Continuous Distributions… • Three other important continuous distributions which will be used extensively in later sections are introduced here: • Student t Distribution, [will use in this class] • Chi-Squared Distribution, • F Distribution, [will use in this class] • Exponential.

  21. Student t Distribution… • Much like the standard normal distribution, the Student t distribution is “mound” shaped and symmetrical about its mean of zero: Looks like the normal distribution after an elephant sat on it [flattened out/spread out more than a normal]

  22. 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. Figure 8.24

  23. 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 1/2% range.

  24. 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 tail area under the Student t curve is .05: Area under the curve value (tA) : COLUMN t.05,10 t.05,10=1.812 Degrees of Freedom : ROW

  25. F Distribution… • 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.

  26. 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. Make sure you start with the correct table!! F.05,3,7=4.35 F.05,3,7 Denominator Degrees of Freedom : ROW Numerator Degrees of Freedom : COLUMN

  27. Problem: Normal Distribution • 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)

  28. Problem: Normal • 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) =

  29. 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) =

  30. Problem: Normal – Targeting 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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