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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(a)… • 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 daily sales will fall between 2,500 and 3,000 gallons. • Algebraically: what is P(2,500 ≤ X ≤ 3,000) ? f(x) 2,000 5,000 x

  7. Example 8.1(a)… • P(2,500 ≤ X ≤ 3,000) = (3,000 – 2,500) x = .1667 • “there is about a 17% chance that between 2,500 and 3,000 gallons of gas will be sold on a given day” f(x) 2,000 5,000 x

  8. 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) ? f(x) 2,000 5,000 x

  9. Example 8.1(b)… • P(X ≥ 4,000) = (5,000 – 4,000) x = .3333 • “There is a one-in-three chance the gas station will sell more than 4,000 gallons on any given day” f(x) 2,000 5,000 x

  10. Example 8.1(c)… • 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 station will sell exactly 2,500 gallons? • Algebraically: what is P(X = 2,500) ? f(x) 2,000 5,000 x

  11. Example 8.1(c)… • P(X = 2,500) = (2,500 – 2,500) x = 0 • “The probability that the gas station will sell exactly 2,500 gallons is zero” f(x) 2,000 5,000 x

  12. 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 …

  13. 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

  14. 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.

  15. 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…

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

  17. 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!

  18. Calculating Normal Probabilities… • We can use the following function to convert any normal random variable to a standard normal random variable… This shifts the mean of X to zero… 0

  19. Calculating Normal Probabilities… • We can use the following function to convert any normal random variable to a standard normal random variable… 0 This changes the shape of the curve…

  20. 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

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

  22. 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?!

  23. 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

  24. Calculating Normal Probabilities… • We can use Table 3 in • Appendix B to look-up • probabilities P(0 < Z < z) • We can break up P(–.5 < Z < 1) into: • P(–.5 < Z < 0) + P(0 < Z < 1) • The distribution is symmetric around zero, so (multiplying through by • minus one and re-arranging the terms) we have: • P(–.5 < Z < 0) = P(.5 > Z > 0) = P(0 < Z < .5) • Hence: P(–.5 < Z < 1) = P(0 < Z < .5) + P(0 < Z < 1)

  25. Calculating Normal Probabilities… • How to use Table 3… 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

  26. 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”

  27. 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

  28. 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

  29. 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)

  30. 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

  31. 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

  32. Finding Normal Probabilities 0% 10% • Example 8.2 • The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of 5%, • What is the probability of losing money? 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

  33. Find Normal Probabilities Finding Normal Probabilities 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 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

  34. Finding Values of Z… • 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

  35. Finding the values of z (zA) zA represents the value of z such that the area to its right under the standard normal curve is A. P(Z > zA) = A To find zA we read the standard normal table backwards. (That is, we start in the middle of the table and read outwards.)

  36. 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

  37. Normal Distribution – Backwards Problems These exercises give you a probability (often as a %), and ask you to find the value of “a” which makes a probability statement true. (ie: P(X > a) = 0.90 for what value of “a”?) Example: Suppose the amount of hours worked per week for full time students follows a normal distribution with m = 20 hours and s = 4.5 hours. 90% of all students work less than how many hours per week? Here, P(X < a) = 0.90 (from 90%). We want to find “a”, but we must begin with the standard normal distribution.

  38. Normal Distribution – Backwards Problems P(Z < b) = 0.90 when P(0 < Z < b) = 0.90 – 0.5 = 0.40. Look for 0.40 in the body of your normal probability table. You should find b = 1.28 (yes, it is positive) Now recall Z = but b = Z and X = a for our exercise. Then 1.28 = , so a = 1.28(4.5) + 20 = 25.76 hours. So, 90% of all full time students work less than 25.76 hours per week.

  39. Finding Values of Z 0.05 • 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

  40. Example 8.3 • Determine z exceeded by 5% of the population • If .05 is the area in the tail, then the area between 0 and z.05 must be .45. To find this, we look inside the table for a probability of .4500. It isn’t there, but the two closest are .4495 and .4505. These correspond to z values of 1.64 and 1.65. The average of these two is 1.645 • So P(Z>1.645) = .05

  41. Example 8.5 (page 252) • A store that carries electric fans has a policy to order a new shipment of 250 fans when the inventory level falls to the reorder point, which is 100 fans. However, the policy has resulted in frequent shortages. The manager would like to reduce the incidence of shortages so that only 5% of orders will arrive after the inventory drops to 0. This policy is expressed as a 95% service level. From the previous periods, the company has determined that demand is normally distributed with a mean of 200 and a standard deviation of 50. Determine the reorder point.

  42. 0.05 Example 8.5 • The reorder point is set so that demand during lead time exceeds this quantity by 5%. 0.45 0.05 1.645 Z0.05 -Z0.05 0

  43. Example 8.5 • We know from the previous problem that Z.05 = 1.645. • But we don’t want Z, we want the X value of the reorder point where Z = 1.645. • Z = • 1.645 = (ROP – 200) / 50 • Solving the algebra gives us ROP = 282.25 which we round up to 283. The new policy will be to order a new batch of fans when there are 283 fans left in the inventory.

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

  45. 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

  46. Summary • From this chapter, the most important things to remember are: • -the area under the normal distribution curve gives the probability that a certain event will occur. • -z is the number of standard deviations from the mean that a measurement has. • -z = • To find the probability from the z value, you use a table. • -The equation and the table can be used to calculate the probability, or working backwards to find a value if we are given the probability or percentage. • -You should be able to work any problems like those on pages 253-255.

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