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Actuarial AdvantEDGE

Exam P / CAS 1 Study Manual

George Barnidge, MS, FSA

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Contents

1 Introductory Probability Topics 11.1 Risk and Insurance Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Summation Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Measures of Central Tendency and Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4 Set Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.5 Probability Rules and Related Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.6 Counting Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.7 Conditional Probability and Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391.8 Bayes’ Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2 Discrete Probability 632.1 Introduction to Discrete Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . . 632.2 Expectation Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.3 The Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802.4 The Hypergeometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 832.5 The Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 872.6 The Geometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1002.7 The Negative Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1092.8 The Poisson Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1162.9 Discrete Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1242.10 Chapter Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

3 Continuous Probability 1333.1 Introduction to Continuous Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . 1333.2 Expectation Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1453.3 The Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1523.4 The Exponential Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1583.5 The Gamma Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1683.6 The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1743.7 Continuous Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1943.8 Order Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2003.9 Chapter Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

4 Multivariate Distributions 2074.1 Introduction to Joint Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . 2074.2 Expectation Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2514.3 Multivariate Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

5 Related Topics 2795.1 Linear Combinations of Indendepent Random Variables and the Central Limit Theorem . . . 2795.2 Normal Approximation of Discrete Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . 292

6 Practice Exam #1 2996.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2996.2 Solutions to Practice Exam #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305





Total Event Decomposition:

Suppose the sample space S is completely partitioned into N mutually exclusive (i.e. non-overlapping) events, witheach event labeled Ai (where i = 1, 2, 3, . . . , N). Further, suppose another event, B, intersects some (or all) of theseAi events. This setup is given in Figure 1.5.

B

A1 A2 A3

A4

A5. . .

AN−1

AN

S

Figure 1.5: Event B interweaving the partitioned sample space S

Figure 1.5 shows that event B can be equivalently expressed as the collection (i.e. union) of its overlap (i.e. intersec-tion) with each of the individual Ai events. Using set notation, this means:

B =(B

and︷︸︸︷∩ A1)

or︷︸︸︷∪ (B ∩A2) ∪ . . . ∪ (B ∩AN )

=

N⋃i=1

(B ∩Ai)(1.13)

In words, event B happens if. . .

B happens within A1 −→ intersectionOR −→ union

B happens within A2 −→ intersectionOR −→ union

B happens within A3 −→ intersectionOR. . . OR

B happens within AN −→ intersection

Expressing B in Formula 1.13 may seem long-winded at first, but it can be a convenient (and necessary) arrangementto solve many problems.

20

Example 1.21 (1 of 153; SOA May 2003 #1) A survey of a group’s viewing habits over the last year revealed thefollowing information:

i) 28% watched gymnasticsii) 29% watched baseballiii) 19% watched socceriv) 14% watched gymnastics and baseballv) 12% watched baseball and soccervi) 10% watched gymnastics and soccervii) 8% watched all three sports

Calculate the percentage of the group that watched none of the three sports during the last year.

Solution: Let us define the following events. Let. . .

G = watching gymnasticsB = watching baseballC = watching soccer

Note that I am using C rather than S for soccer since S is usually reserved to denote the sample space. If each eventcircle represents watching a sport and since we are asked to find the percentage who did not watch a sport, then weneed to find the percentage who are classified outside the event circles. For this problem, we recommend starting atthe most intersected region and working out to the least intersected region.

The Venn diagram region dealing with the intersection of all three event circles:Since 8% refers to those who watched all three sports, we label the innermost intersection with 0.08.

The Venn diagram region dealing with the intersection of two event circles:Since 10% watched gymnastics and soccer and since the 8% who watched all three sports is a subset of this 10%,then 10% − 8% = 2% watched just gymnastics and soccer. Similar logic shows that 12% − 8% = 4% watched justbaseball and soccer and that 14%− 8% = 6% watched just gymnastics and baseball.

The Venn diagram region dealing with the non-intersected event circles:Since 19% watched soccer and since the 8% who watched all three sports and since the 2% who watched justgymnastics and soccer and since the 4% who watched just baseball and soccer are subsets of this 19%, then19%− 8%− 2%− 4% = 5% watched just soccer. Similar logic shows that 29% − 8% − 4% − 6% = 11% watchedjust baseball and that 28%− 8%− 2%− 6% = 12% watched just gymnastics.

The seven percentages identified in the previous three paragraphs completely fill the event circles in the Venn diagram,summing to 0.08 + (0.02 + 0.04 + 0.06) + (0.05 + 0.11 + 0.12) = 0.48. Since 48% watched some type of sport,then Formula 1.14 shows that 1 − 48% = 52% did not watch a sport. The Venn diagram shows these percents.

G

B C

S

0.08

0.04

0.06 0.02

0.12

0.11 0.05

0.52

25

1.7 Conditional Probability and Independence

This section introduces some very simple yet powerful probability formulas that will be used throughout the rest ofthis manual. It is very important that you master these, for they are crucial to passing this exam.

Definition 1.23 An unconditional probability is a probability where no additional information or conditions areknown or expressed. The unconditional probability that event A happens is notated P (A).

Definition 1.24 A conditional probability is a probability where additional information or conditions are known orexpressed. The conditional probability that event A happens given that event B happens is notated P (A|B). Thevertical bar “|” is read as “given”.

Important note: With conditional probabilities, the event before the vertical bar “|” is the event whose probabilitywe are interested in finding. The event after the vertical bar is the additional information or condition that may (ormay not) affect the probability of interest. To determine if a probability is unconditional or conditional, look for thephrase “given” (if expressed verbally) or the vertical bar “|” (if expressed symbolically).

Suppose we have the following tree diagram where the first level expresses the occurrence of event A or its non-occurrence, A. The second level expresses the occurrence of event B or its non-occurrence, B.

Start

A

B → P (A ∩B) = P (A)P (B|A)

B → P (A ∩B) = P (A)P (B|A)

A

B → P (A ∩B) = P (A)P (B|A)

B → P (A ∩B) = P (A)P (B|A)

Looking at this picture, to get to the A ∩ B event endpoint, we have to first travel up to A (rather than down toA). Then, given that we are now standing at A, we have to find the probability that we further travel up to B(rather than down to B). Therefore, the probability that we observe event A ∩B is the product of an unconditionalprobability, P (A), and a conditional probability, P (B|A). The same patterns hold for the other three branches. Thegeneral theme of this movement is expressed here:

P (A ∩B) = P (A) · P (B|A) (1.21)

A rearranged version of Formula 1.21 is given here.

P (B|A) =P (A ∩B)

P (A)(1.22)

Example 1.26 A sample of 100 drivers was taken in which respondents were asked whether they passed or failedtheir drivers exam on their first attempt. The following table shows the results from this sample, broken down bygender.

Test outcomeGender Pass Fail Total

Male 25 35 60Female 30 10 40

Total 55 45 100

39

Chapter 3

Continuous Probability

3.1 Introduction to Continuous Probability Distributions

As we saw in the previous chapter, discrete random variables take on a countable number of possibilities, usually thewhole numbers. Each of these possibilities has an associated probability, determined by the pmf. In this chapter, westudy continuous random variables. Unlike discrete random variables, continuous random variables take on valuesover an interval and are therefore not countable.

Definition 3.1 A generic random variable X is continuous if it takes on values over an interval1.

Example 3.1 The following are examples of continuous random variables:

a) Let X measure the amount of rainfall (in inches) over the next year. Since X can be not only 1” or 2” or 3”, etc.of rainfall but also all the values in between, such as 0.71” or 3.14” of rainfall, X is a continuous random variable.

b) Let X measure the weight (in pounds) of a person during his next life insurance underwriting process. Since X canbe not only 185 pounds or 197 pounds but also all the values in between, such as 186.53 pounds, X is a continuousrandom variable.

Definition 3.2 A probability density function (pdf), f(x), is a function that assigns probabilities to continuousrandom variables2. A pdf can be presented as a formula or graph and is the analog of the discrete pmf3.

Example 3.2 The following are equivalent versions of the same pdf:

As a formula: f(x) = 3−x2

where 1 ≤ x ≤ 3.

As a graph:

1This is a “loose” definition of a continuous random variable but one that should suffice for our purposes.2This is a “loose” definition of a probability density function but one that should suffice for our purposes.3A pdf, unlike pmfs, is never presented in table form because continuous random variables do not take on a countable set of

possibilities (i.e. they can’t be listed).

133

f(x)

1

x1 2 3

In Example 3.2, notice that the area under f(x) above the x axis is 1. Also notice that f(x) ≥ 0. These twoproperties, given here, will be true for any pdf and is analogous to Formula 2.1:

1)

∫ ∞−∞

f(x) dx = 1

2) f(x) ≥ 0

(3.1)

To calculate probabilities associated with continuous random variables, the main theme is. . .

Probabilities are areas!!! Areas are probabilities!!!

Recall from your calculus class that integration allows us to calculate areas under curves. Probabilities associatedwith continuous random variables are calculated by integrating the pdf over the given interval, shown here:

a b x

f(x)

P (a ≤ X ≤ b) =∫ baf(x) dx

The main theme from the previous picture is summarized here:

P (a ≤ X ≤ b) =

∫ b

a

f(x) dx (3.2)

In the previous chapter on discrete random variables, it was common to calculate probabilities associated with asingle possibility of the discrete random variable. For example, if X is a discrete random variable, it was commonto calculate positive probabilities for expressions such as P (X = 7) or P (X = 0) (just 7 or just 0). This does nottranslate over to continuous random variables. Recalling, from the continuous perspective, that “Probabilities areAreas, and Areas are Probabilities”, this next picture shows why probabilities associated with a single constant value,call it a, are 0.

a x

f(x)

P (X = a) =∫ aaf(x) dx = 0

(an infinitely-thin line has noarea and therefore no probability)

134

The main theme from the previous picture is summarized here:

P (X = a) = 0 (3.3)

Formula 3.3 may seem somewhat counterintuitive at first since X = a is clearly within the range of possibilities. Butalthough X = a is a possibility, it has no probability since it has no area (!!!). The answer therefore is 0. Continuousrandom variables have positive probability only over intervals, not single values. Therefore, with continuous randomvariables, the probability P (X > a) and P (X ≥ a) would have the same answer (since they differ only in the inclusionor exclusion of the single value x = a). The same can be said about P (X < a) and P (X ≤ a).

This next picture shows how P (a ≤ X ≤ b) can be expressed as the difference of two probabilities.

a b a b a bx

f(x)

= −

x

f(x)

P (a ≤ X ≤ b) =∫ baf(x) dx P (X ≤ b) =

∫ b−∞ f(x) dx P (X ≤ a) =

∫ a−∞ f(x) dx

x

f(x)

The previous picture can be expressed from a slightly different perspective, given here:

a b a b a bx

f(x)

= −

x

f(x)

P (a ≤ X ≤ b) =∫ baf(x) dx P (X ≥ a) =

∫∞a

f(x) dx P (X ≥ b) =∫∞b

f(x) dx

x

f(x)

The main theme from the previous two pictures is summarized here:

P (a ≤ X ≤ b) = P (X ≤ b)− P (X ≤ a)

= P (X ≥ a)− P (X ≥ b)(3.4)

For continuous distributions, Formula 3.4 is true regardless of the inequality signs that appear in P (a ≤ X ≤ b).For example, one could apply Formula 3.4 even if the original problem was P (a < X ≤ b) or P (a ≤ X < b) orP (a < X < b). These all give the same answer due to Formula 3.3.

Although the logic of Formula 3.4 can also apply to discrete random variables, the inequality sign can impact theanswer. Therefore, more care must be taken when applying Formula 3.4 to a discrete random variable.

Definition 3.3 The cumulative distribution function (CDF) of a random variable X, denoted FX(x), is the proba-bility that X takes on a values less than or equal to x. Symbolically, this means that FX(x) = P (X ≤ x).

The next picture shows the graphical relationship between the pdf and the cdf.

135

t x

f(x)

t

the density function, f(x) the cumulativedistribution function, FX(x)

1

x

FX(x)

P (X ≤ t) =∫ t−∞ f(x) dx = FX(t)

In the previous picture, notice that the graph of FX(x) (the right graph) begins with a height of 0. This happensbecause the graph of f(x) (the left graph) has not yet accumulated any area when t is to the left of the leftmostpossible x value. Also notice that FX(x) approaches a height of 1 as t moves to the right. This happens because thetotal accumulated area under f(x) is 1 (see Formula 3.1) once t moves to the far right. Finally, notice that FX(x)never decreases, which happens because f(x) ≥ 0 (see Formula 3.1). In summary, cdfs from continuous distributionsalways begin at 0 and end at 1.

These ideas are summarized here and are the analog of Formula 2.2:

FX(x) = P (X ≤ x) =

∫ x

−∞f(t) dt

FX(−∞) = 0

FX(∞) = 1

(3.5)

Example 3.3 Refer to Example 3.2.

a) Find FX(x) using Formula 3.5.b) Check that FX(x) satisfies the two conditions at the bottom of Formula 3.5.c) Graph FX(x)d) Find P (X ≤ 2.5)e) Find P (X ≥ 1.75)f) Find P (1.75 ≤ X ≤ 2.5)g) Find P (X = 2)

a) FX(x) = P (X ≤ x) =

∫ x

−∞

3− t2

dt =

∫ x

1

3− t2

dt =3

2t− t2

4

]x1

=6t− t2

4

]x1

=6x− x2 − 5

4

In the previous algebra, notice how the lower bound of integration was changed from −∞ (as appears in Formula 3.5)to 1. This is because the lowest value that X can assume in this example is x = 1.

b) When x = 1 (the leftmost bound for x), FX(1) should be 0.

FX(1) = 6(1)−(1)2−54

= 0

When x = 3 (the rightmost bound for x), FX(3) should be 1.

FX(1) = 6(3)−(3)2−54

= 18−9−54

= 44

= 1

c) The graph of FX(x) is given here:

136

x

F (x)

1

1 2 3 4 5

d) P (X ≤ 2.5) = FX(2.5) = 6(2.5)−2.52−54

= 0.9375

e) P (X ≥ 1.75) = 1− P (X < 1.75) = 1− FX(1.75) = 1−

0.6094︷︸︸︷6(1.75)− 1.752 − 5

4= 0.3906

f) P (1.75 ≤ X ≤ 2.5) =

F3.4︷︸︸︷P (X ≤ 2.5)− P (X ≤ 1.75) = 0.9375− 0.6094 = 0.3281

g) P (X = 2) = 0 (see Formula 3.3)

Formula 3.5 shows that FX(x) is calculated by integrating f(x). Since integration and differentiation are inverseprocesses, then the pdf is calculated by differentiating FX(x), given here:

f(x) =d

dxFX(x) (3.6)

Just as the previous chapter dedicated separate sections to specific “brand name” discrete random variables (i.e.binomial, geometric), so too will we devote separate sections in this chapter to specific “brand name” continuousrandom variables.

End-of-Section Practice Problems

Problem 3.1.1 Let f(x) = 1/10, where −7 ≤ x ≤ 3. Find. . .

a) the median.b) the 85th percentile.

Problem 3.1.2 The lifetime X (in years) of a machine has pdf f(x) = 2xθ2

exp[−(x/θ)2], where x > 0. If P (X >5) = 0.01, find θ.

Problem 3.1.3 The annual demand X for hand soap (in thousands of gallons) for a school district has the pdff(x) = 2x exp(−x2), where x > 0. If 3,000 gallons are in stock at the beginning of each year (with no additionalamount provided during the year), what is the probability that the soap will run out before the year is over?

Problem 3.1.4 Suppose that X assumes the density function f(x) = cx, where 0 ≤ x ≤ 4 and c is a constant.

a) Find the value of c that makes f(x) a pdf.b) Find FX(x).c) Find P (1 ≤ X ≤ 2) using FX(x).

Problem 3.1.5 A grain supplier for farms has a 250-gallon tank that is filled at the beginning of each month. Themonthly demand for grain shows a relative frequency behavior that linearly increases up to 100 gallons and then islevel between 100 and 250 gallons. If X denotes that monthly demand in hundreds of gallons, the relative frequency

137

3.5 The Gamma Distribution

A generic gamma random variable, X, is often (but not always) used to measure the waiting time for α (“alpha”)

observations of some event, where θ is a measure of the average waiting time between event observations7. Sincewaiting time can be any positive number, the range of possible values is X > 0. The symbols α and θ are consideredthe parameters of the gamma distribution since they are the only values needed to fully define the probability space.

Shorthand notation: For convenience, rather than tediously writing “X is distributed as a gamma random variablewith parameters α and θ”, we will condense this information into the equivalent expression “X ∼ Gamma(α, θ)”.The “∼” symbol should be interpreted as “is distributed as.”

Relationship to the negative binomial random variable: Recall from Section 2.7 that a negative binomial trials randomvariable counts the number of trials needed to observe r successes. Since the gamma random variable measures thewaiting time for α observations of some event, it can be seen as the continuous analog of the discrete negative binomialtrials random variable.

The pdf of a gamma random variable X is

f(x) =1

θαΓ(α)xα−1 exp(−x/θ)

where

x > 0

α > 0

θ > 0

Γ(α) = (α− 1)! when α is an integer

(3.33)

The graph of several gamma pdfs is given here:

θ = 1

α = 1

α = 2

α = 3

θ = 11θ

x

f(x)

Probability calculations involving the gamma random variable will require us to integrate the pdf from Formula 3.33.This requires the use of integration “by parts”. To save time, the following formulas show, for a given value of θ, the

7Compare this to the description of the exponential random variable in Section 3.4. The exponential random variablemeasures the waiting time for the first observation whereas the gamma random variable measures the waiting time for αobservations, where α is a positive whole number.

168

integration of the gamma pdf for α = 1, 2, 3

For α = 1:

1

θ1Γ(1)

∫ b

a

x1−1 exp(−x/θ) dx = −e−x/θ∣∣∣ba

For α = 2:

1

θ2Γ(2)

∫ b

a

x2−1 exp(−x/θ) dx = −e−x/θ(

1 +x

θ

) ∣∣∣ba

For α = 3:

1

θ3Γ(3)

∫ b

a

x3−1 exp(−x/θ) dx = −e−x/θ(

1 +x

θ+

x2

2θ2

) ∣∣∣ba

(3.34)

Example 3.11 Cars arrive at a border crossing at a mean rate of 180 cars every hour. Assume that it is thebeginning of the day, and no cars have yet arrived. Find the probability that the border station guard will have towait longer than 1 minute for the second car to arrive at the station.

Solution: Let X measure the waiting time for the second car to arrive. Since X measures the waiting time, we wantto express the given rate in terms of time, not cars. Since the given rate is 180 cars every 60 minutes, then this isthe same as 60 minutes every 180 cars, or 60/180 = 1/3. In other words, a car arrives, on average, every 1/3 minute;so θ = 1/3. Since we are waiting until the second car arrives, α = 2. Therefore, X ∼ Gamma(α = 2, θ = 1/3). Weare being asked to find P (X > 1), given here:

P (X > 1) =

F3.34 where α = 2 and θ = 1/3︷︸︸︷−e−3x(1 + 3x)

∣∣∣∞1

=

=0︷︸︸︷limx→∞

−e−3x(1 + 3x)−[− e−3(1)(1 + 3(1))

]= 0.1991

The mean and variance of a gamma random variable are:

µX =E(X) = α · θ

σ2X =V(X) = α · θ2

(3.35)

The moment generating function (mgf) of a gamma random variable is:

MX(t) = (1− θt)−α (3.36)

Example 3.12 If MX(t) = (1− 7t)−15, find. . .

a) the pdf f(x)b) µX = E(X)c) σ2

X = V (X)

Solution: Matching MX(t) with Formula 3.36, we see that α = 15 and θ = 7. Therefore X ∼ Gamma(α = 15, θ = 7).

a) Formula 3.33 gives f(x) = 1715Γ(15)

x15−1 exp(−x/7).

b) Formula 3.35 gives µX = E(X) = 15 · 7 = 105.

c) Formula 3.35 gives σ2X = V (X) = 15(7)2 = 735.

169

3.6 The Normal Distribution

A generic normal random variable is a continuous random variable that resembles the bell curve. It is the mostimportant distribution (for reasons we will see in Section 5.1 and since a lot of natural phenomenon is approximatelynormally distributed). The symbols µ (“mu”) and σ (“sigma”) represent the mean and standard deviation, respec-tively, of the distribution and are considered the parameters of the normal distribution since they are the only valuesneeded to fully define the probability space.

Shorthand notation: For convenience, rather than tediously writing “X is distributed as a normal random variablewith parameters µ and σ”, we will condense this information into the equivalent expression X ∼ N(µ, σ). The “∼”symbol should be interpreted as “is distributed as.”

The pdf of the normal random variable X is

f(x) =1√2πσ

exp

[−1

2

(x− µσ

)2]

where

−∞ < x <∞−∞ < µ <∞

0 < σ <∞

(3.39)

The graphs of several normal pdfs are given in Figure 3.1. Notice how the different values of µ (a location parameter)and σ (a dispersion parameter) affect the curve’s center and spread, respectively.

0 5 10 15 20

W

X Y

W ∼ N(µW = 5, σW = 0.8)X ∼ N(µX = 5, σX = 1.7)Y ∼ N(µY = 15, σY = 1.7)

Figure 3.1: pdf of several normal distributions

Definition 3.4 A standard normal random variable is a normal random variable with parameters µ = 0 and σ = 1.It is given the special symbol Z. Therefore, we write Z ∼ N(µZ = 0, σZ = 1).

There is nothing mathematically special about a standard normal random variable compared to any other normalrandom variable. However, the standard normal random variable comes up very often in practice, so it has beenassigned the special symbol Z. The graph of the standard normal random variable, Z, is given here. Notice how itis centered at its mean µZ = 0 with a spread of σZ = 1:

174

−3 −2 −1 µZ = 0 1 2 3 z

f(z)

σZ = 1 σZ = 1 σZ = 1

The parameters of the normal distribution are themselves the mean and variance:

E(X) =µ

V (X) =σ2(3.40)

The z-chart: Unlike the other continuous distributions, we will never be asked to integrate the normal pdf8. Instead,we will use a probability chart for a standard normal random variable (from here on simply called the “z-chart”) toequivalently determine areas under the normal pdf.

The z-chart that you will receive on exam day is constructed according to the following picture. You may have used az-chart that was constructed differently in your undergraduate statistics class. If so, you will want to become familiarwith the version given during the exam (given just before the End-of-Section Practice Problems in this section andat the beginning and end of this manual). The first column of the z-chart gives you the ones and tenths places ofa positive value along the horizontal z-axis. The first row of the z-chart give you the hundredths place of this samez value. The intersecting cell of the row (coming from the ones and tenths places) and column (coming from thehundredths place) give you the area (i.e. probability) under the z curve, from −∞ up to the positive z-value. Thiscan be seen in the following picture.

P(Z < 0.00) = 0.5

P(Z < +z) > 0.5

+z0 z

f(z)

When z = 0.00, for example, the chart gives you the area under the left half of the curve. Since the curve issymmetrical, then this left half is 0.5000 (verify this 0.5000 by looking up z = 0.00 in the top-left cell of the z-chart).Although the domain of the normal distribution is technically (−∞,+∞), the vast majority of the data falls between−3 ≤ z ≤ 3, which is why the z-chart does not give probabilities significantly past z = 3.00. Any area in the extremetails beyond values given in the z-chart is, for practical purposes, 0.

Example 3.14 Let Z ∼ N(µ = 0, σ = 1). Find P (Z < 1.52).

8The pdf of the normal distribution does not have an antiderivative.

175

µZ−

2·σ

Z=−

2µZ−

1·σ

Z=−

1µZ

=0

µZ

+1·σ

Z=

1µZ

+2·σ

Z=

2

σZ

=1

σZ

=1

σZ

=1

σZ

=1

X∼

N(µX,σX

)

Z∼

N(µZ

=0,σZ

=1)

X=µX

+Z·σ

XZ

=X

−µX

σX

∫ µ X+k2·σX

µX+k1·σX

f(x

)dx

=

∫ µ Z+k2·σZ=k2

µZ+k1·σZ=k1

f(z

)dz

wh

eref

(·)is

the

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rase

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X∼

N(µX,σX

)

Z∼

N(µZ

=0,σZ

=1)

µX−

2·σ

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XµX

µX

+1·σ

XµX

+2·σ

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σX

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Fig

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181

Problem 4.1.14 (77 of 153) A device runs until either of two components fails, at which point the device stopsrunning. The joint density function of the lifetimes of the two components, both measured in hours, is:

f(x, y) =x+ y

8for 0 < x < 2 and 0 < y < 2.

What is the probability that the device fails during its first hour of operation?

Problem 4.1.15 (88 of 153; SOA May 2001 #22) The waiting time for the first claim from a good driver and thewaiting time for the first claim from a bad driver are independent and follow exponential distributions with means 6years and 3 years, respectively.

What is the probability that the first claim from a good driver will be filed within 3 years and the first claim from abad driver will be filed within 2 years?

Problem 4.1.16 (89 of 153; SOA November 2000 #20) The future lifetimes (in months) of two components of amachine have the following joint density function:

f(x, y) =6

125, 000(50− x− y) for 0 < x < 50− y < 50.

What is the probability that both components are still functioning 20 months from now?

Problem 4.1.17 (90 of 153; SOA May 2000 #10) An insurance company sells two types of auto insurance policies:Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean two days.The time until the next Deluxe Policy claim is an independent exponential random variable with mean three days.

What is the probability that the next claim will be a Deluxe Policy claim?

Problem 4.1.18 (91 of 153; SOA November 2000 #36) An insurance company insures a large number of drivers.Let X be the random variable representing the company’s losses under collision insurance, and let Y represent thecompany’s losses under liability insurance. X and Y have joint density function:

f(x, y) =2x+ 2− y

4for 0 < x < 1 and 0 < y < 2.

What is the probability that the total loss is at least 1?

Problem 4.1.19 (92 of 153; SOA November 2001 #28) Two insurers provide bids on an insurance policy to a largecompany. The bids must be between 2,000 and 2,200. The company decides to accept the lower bid if the two bidsdiffer by 20 or more. Otherwise, the company will consider the two bids further. Assume that the two bids areindependent and are both uniformly distributed on the interval 2,000 to 2,200.

Determine the probability that the company considers the two bids further.

Problem 4.1.20 (93 of 153; SOA May 2003 #20) A family buys two policies from the same insurance company.Losses under the two policies are independent and have continuous uniform distributions on the interval from 0 to10. One policy has a deductible of 1 and the other has a deductible of 2. The family experiences exactly one lossunder each policy.

Calculate the probability that the total benefit paid to the family does not exceed 5.

Problem 4.1.21 (108 of 153; SOA November 2001 #37) A device containing two key components fails when, andonly when, both components fail. The lifetimes, T1 and T2, of these components are independent with commondensity function f(t) = e−t, t > 0. The cost, X, of operating the device until failure is X = 2T1 + T2.

Write an expression for the density function of X for x > 0.

219

y

x

(0, 2)

(1, 0)

(1, 2)

(0, 0)

(0, 1) y = 1− xy ≥ 1− x

Looking at the shaded region, integrating over y first, we see that y goes as low as the line y = 1−x and is as high asy = 2. Once we integrate over y, we then integrate over x, which goes as far left as x = 0 and as far right as x = 1.

P (X + Y ≥ 1) = 14

∫ 1

0

[∫ 2

1−x(2x+ 2− y) dy]

dx = 14

∫ 1

0

[2xy + 2y − 1

2y2∣∣∣y=2

y=1−x

]dx

= 14

∫ 1

0

(2x[2− (1− x)] + 2[2− (1− x)]− 1

2[22 − (1− x)2]

)dx

= 14

∫ 1

0

(52x2 + 3x+ 1

2

)dx = 1

4

(56x3 + 3

2x2 + 1

2x) ∣∣∣x=1

x=0= 1

4

[56(13 − 03) + 3

2(12 − 02) + 1

2(1− 0)

]= 17

24

Solution 4.1.19 Let X and Y be the insurer’s bids. The company considers the two bids further if the differencebetween the two bids is less than 20. Mathematically, this happens when

|x− y| < 20→ −20 < x− y < 20→ y > x− 20 and y < x+ 20.

The shaded region over which we need to integrate to find the answer is given in the following picture.

y

x

(2200, 2200)

(2200, 2000)

(2000, 2200)(0, 2200)

(0, 2000)

(2200, 0)(2000, 0)(0, 0)

y = x+ 20

y = x− 20

(2180, 2200)

(2200, 2180)(2000, 2020)

(2020, 2000)

Since X and Y are each uniformly distributed over the interval (2000, 2200) and since x and y are independent, thenFormula 4.7 says that the joint pdf is the product of the marginal pdfs:

f(x, y) =

(1

2, 200− 2, 000

)︸︷︷︸

F3.20

(1

2, 200− 2, 000

)︸︷︷︸

F3.20

= 12002

A quick, geometric approach for the area of the shaded region uses the complement property. For this, recall thatthe area of a triangle is “ 1

2(base × height)”.

P (|x− y| < 20) = 1− P (volume above two unshaded triangles)

= 1−

symmetry of triangles and f(x, y)︷︸︸︷2P (volume above one unshaded triangles)

236

= 1− 2( [ 1

2(2, 200− 2, 020)(2, 180− 2, 000)︸︷︷︸

area of triangle

][ 1

2002︸︷︷︸f(x,y)

]︸︷︷︸

volume above one unshaded triangle

)= 0.19

Solution 4.1.20 Let X and Y be the loss amounts from the policies with deductibles of 1 and 2, respectively. SinceX and Y are each uniformly distributed over the interval (0, 10) and since x and y are independent, then Formula4.7 says that the joint pdf is the product of the marginal pdfs:

f(x, y) =

(1

10− 0

)︸︷︷︸

F3.20

(1

10− 0

)︸︷︷︸

F3.20

= 1100

The following graph will be helpful to determine the region where the total post-deductible benefit is at most 5.

y

x

(0, 10) (10, 10)

(10, 0)

(0, 0)

C(1, 0)

(6, 0)

(0, 2)

(0, 7)D

(y − 2) + (x− 1) = x+ y − 3 ≤ 5→ y ≤ 8− x

A

B

The letters indicate the four regions where the total post-deductible benefit is at most 5:

Region A: The payment is y − 2 because x < 1 (no payment) and y > 2.Region B: The payment is 0 because x < 1 (no payment) and y < 2 (no payment).Region C: The payment is x− 1 because x > 1 and y < 2 (no payment).Region D: The payment is (y − 2) + (x− 1) = x+ y − 3 because x > 1 and y > 2.

All the other unshaded regions represent situations where the total post-deductible benefit exceeds 5. Let us usegeometric formulas for the area of a rectangle and triangle to find the answer:

P (post-deductible benefit ≤ 5) = P (volume above shaded region)

=[(

A︷︸︸︷5× 1) +

B︷︸︸︷(2× 1) +

C︷︸︸︷(5× 2) +

D︷︸︸︷1

2(5× 5)

][ f(x,y)︷︸︸︷1

100

]= 0.295

Solution 4.1.21 We are told that T1 ∼ Exp(θ = 1) and T2 ∼ Exp(θ = 1). Further, since T1 and T2 are independent,then f(t1, t2) = [exp(−t1)︸︷︷︸

F3.24

][exp(−t2)︸︷︷︸F3.24

] for t1 > 0 and t2 > 0. We are being asked to find g(x), the pdf of X = 2T1 +T2.

Let G(x) be the cdf of X.

g(x) =

F3.6︷︸︸︷d

dxG(x) = d

dxP (X ≤ x) = d

dxP (

X︷︸︸︷2T1 + T2 ≤ x) = d

dxP (T2 ≤ x− 2T1).

The following picture shows the integration region needed to calculate P (T2 ≤ x− 2T1):

237

T2

T1

(0, x)

( 12x, 0)(0, 0)

T2 = x− 2T1

Looking at the shaded region, integrating over T2 first, we see that T2 goes as low as T2 = 0 and is as high as the lineT2 = x− 2T1. Once we integrate over T2, we then integrate over T1, which goes as far left as T1 = 0 and as far rightas T1 = 1

2x. Therefore. . .

g(x) = ddxP (T2 ≤ x− 2T1) = d

dx

[ ∫ 12x

0exp(−t1)

[ ∫ x−2t10

exp(−t2) dt2]

dt1]

= ddx

[ ∫ 12x

0exp(−t1)

[ F3.27︷︸︸︷1− exp[−(x− 2t1)]

]dt1]

= ddx

[ ∫ 12x

0

[exp(−t1)− exp(t1 − x)

]dt1]

= ddx

[− exp(−t1)− exp(t1 − x)

∣∣∣t1= 12x

t1=0

]= d

dx

[−[exp(− 1

2x)− exp(−0)

]−[exp( 1

2x− x)− exp(0− x)

] ]= d

dx

[1− 2 exp(− 1

2x) + exp(−x)

]= exp(− 1

2x)− exp(−x)

Solution 4.1.22 Let P and C be the premiums and claims, respectively. We are told that P ∼ Exp(θ = 2) andC ∼ Exp(θ = 1). Further, since P and C are independent, then f(p, c) = [(1/2) exp(−p/2)︸︷︷︸

F3.24

][exp(−c)︸︷︷︸F3.24

] for p > 0 and

c > 0. We are being asked to find g(x), the pdf of X = CP

. Let G(x) be the cdf of X.

g(x) =

F3.6︷︸︸︷d

dxG(x) = d

dxP (X ≤ x) = d

dxP( X︷︸︸︷C

P≤ x

)= d

dxP (C ≤ Px).

The following picture shows the integration region needed to calculate P (C ≤ Px):

C

P

C = Px

C ≤ Px

(0, 0)

Looking at the shaded region, integrating over C first, we see that C goes as low as C = 0 and is high as the lineC = Px. Once we integrate over C, we then integrate over P , which goes as far left as P = 0 and as far right asP =∞. Therefore. . .

g(x) = ddxP (C ≤ Px) = d

dx

[ ∫∞0

12

exp(−p/2)[ ∫ px

0exp(−c) dc

]dp]

= ddx

[ ∫∞0

12

exp(−p/2)[ F3.27︷︸︸︷

1− exp(−px)]

dp]

= ddx

[ ∫∞0

12

exp(−p/2)− 12

exp(−p(1+2x)

2

)dp]

238

For constants c and d and random variables X and Y , we have:

Cov(c,X) = 0 (4.22)

andCov(cX, dY ) = cdCov(X,Y ) (4.23)

Whereas the covariance of two random variables X and Y can be negative, zero, or positive of any magnitude, thecorrelation coefficient (from now on, simply called “correlation”) of two random variables X and Y , defined here andoften denoted ρX,Y , is a standardized number between -1 and 1:

ρX,Y =Cov(X,Y )

σXσY(4.24)

The following picture shows the values of ρX,Y for various (x, y) scatter-plots:

y

x

ρX,Y = 1

y

x

0 < ρX,Y < 1

y

x

ρX,Y = −1

y

x

−1 < ρX,Y < 0

y

x

ρX,Y ≈ 0

y

x

ρX,Y ≈ 0

Correlation, much like covariance, measures the strength of the linear relationship between X and Y . Of these twomeasures, correlation is a much more interpretable metric because it is constrained between -1 and 1. On one extreme,the correlation is exactly 1 when an increase of x is associated with an exact linear increase in y. This is seen in thefirst graph on the top row. Correlation is positive (but not exactly 1) when an increase of x is generally associatedwith a linear increase in y. This is seen in the second graph on the top row. Similar arguments can be made fornegative correlations, as seen in the last two graphs in the top row.

The graphs in the second row show correlations around 0. In the first of these two graphs, x and y appear tobe randomly distributed. Although the second graph shows a strong parabolic relationship between x and y, itis important to recall that covariance and correlation measure the strength of the linear relationship between twovariables. Since none of these bottom two graphs exhibit a strong linear relationship, their correlations are around 0.

The sum of a fixed number of random variables

For random variables Xi and Yj and constants ai, bj , m, and n, let us define the following linear combinations:

∑mi=1 aiXi = a1X1 + a2X2 + · · ·+ amXm∑nj=1 bjYj = b1Y1 + b2Y2 + · · ·+ bnYn

253

Problem 4.2.19 (104 of 153; SOA May 2001 #7) A joint density function is given by:

f(x, y) = kx for 0 < x < 1, 0 < y < 1,

where k is a constant. What is Cov(X,Y )?

Problem 4.2.20 (105 of 153; SOA May 2003 #6) Let X and Y be continuous random variables with joint densityfunction:

f(x, y) =8

3xy for 0 ≤ x < 1, x ≤ y ≤ 2x.

What is Cov(X,Y )?

Problem 4.2.21 (107 of 153; SOA November 2001 #7) Let X denote the size of a surgical claim and let Y denotethe size of the associated hospital claim. An actuary is using a model in which E(X) = 5, E(X2) = 27.4, E(Y ) = 7,E(Y 2) = 51.4, and V (X + Y ) = 8.

Let C1 = X + Y denote the size of the combined claims before the application of a 20% surcharge on the hospitalportion of the claim, and let C2 denote the size of the combined claims after the application of that surcharge.

Calculate Cov(C1, C2).

Problem 4.2.22 (121 of 153; SOA May 2003 #24) Let X represent the age of an insured automobile involved in anaccident. Let Y represent the length of time the owner has insured the automobile at the time of the accident. Xand Y have joint probability density function:

f(x, y) =1

64(10− xy2) for 2 ≤ x < 10 and 0 ≤ y ≤ 1.

Calculate the expected age of an insured automobile involved in an accident.

Problem 4.2.23 (122 of 153; SOA November 2000 #40) A device contains two circuits. The second circuit is abackup for the first, so the second is used only when the first has failed. The device fails when and only when thesecond circuit fails. Let X and Y be the times at which the first and second circuits fail, respectively. X and Y havejoint probability density function:

f(x, y) = 6e−xe−2y for 0 < x < y <∞.

What is the expected time at which the device fails?

Problem 4.2.24 (137 of 153) Let X and Y be identically distributed independent random variables such that themoment generating function of X + Y is:

M(t) = 0.09e−2t + 0.24e−t + 0.34 + 0.24et + 0.09e2t for −∞ < t <∞.

Calculate P (X ≤ 0).

Problem 4.2.25 (138 of 153) A machine consists of two components, whose lifetimes have the joint density function:

f(x, y) =1

50for x > 0, y > 0, x+ y < 10.

The machine operates until both components fail. Calculate the expected operational time of the machine.

Problem 4.2.26 (148 of 153) The number of hurricanes that will hit a certain house in the next ten years is Poissondistributed with mean 4. Each hurricane results in a loss that is exponentially distributed with mean 1,000. Lossesare mutually independent and independent of the number of hurricanes. Calculate the variance of the total loss dueto hurricanes hitting this house in the next ten years.

Problem 4.2.27 (106 of 153; SOA May 2000 #20) Let X and Y denote the values of two stocks at the end of afive-year period. X is uniformly distributed on the interval (0, 12). Given X = x, Y is uniformly distributed on theinterval (0, x). Determine Cov(X,Y ) according to this model.

262

Solution 4.2.19 Since f(x, y) =

kx︷︸︸︷g(x)

1︷︸︸︷h(y) can be reexpressed as the product of a function of just x and a function of

just y and since the (x, y) space is a product space, then X and Y are independent random variables (see Formula 4.7).Therefore, Formula 4.21 gives Cov(X,Y ) = 0. You can verify this by calculating Cov(X,Y ) = E(XY )−E(X)E(Y ).

Solution 4.2.20 The (x, y) space is given here:

y

x

(1, 2)

(1, 1)

(0, 0) (1, 0)

y = 2x

x ≤ y ≤ 2x

y = x

Looking at the shaded region, integrating over y first, we see that y goes as low as the line y = x and is as highas the line y = 2x. Once we integrate over y, we then integrate over x, which goes as far left as x = 0 and as far

right as x = 1. We are being asked to find Cov(X,Y ) = E(XY )− E(X)E(Y ). Although f(x, y) =

83x︷︸︸︷

g(x)

y︷︸︸︷h(y) can be

expressed as the product of a function of just x and a function of just y, the (x, y) integration space is not rectangular.Therefore, unlike Problem 4.2.19, we cannot assume that X and Y are independent nor that Cov(X,Y ) = 0. Wemust therefore calculate the various moments in the covariance formula:

E(XY ) =∫ 1

0

∫ 2x

xxy

83xy︷︸︸︷

f(x, y) dy dx = 83

∫ 1

0x2[ ∫ 2x

xy2 dy

]dx = 8

3

∫ 1

0x2[

13y3∣∣∣y=2x

y=x

]dx

= 89

∫ 1

0x2[(2x)3 − x3

]dx = 56

9

∫ 1

0x5 dx = 56

9

[16x6∣∣∣x=1

x=0

]= 56

54

[16 − 06

]= 56

54

E(X) =∫ 1

0

∫ 2x

xx

83xy︷︸︸︷

f(x, y) dy dx = 83

∫ 1

0x2[ ∫ 2x

xy dy

]dx = 8

3

∫ 1

0x2[

12y2∣∣∣y=2x

y=x

]dx

= 86

∫ 1

0x2[(2x)2 − x2

]dx = 24

6

∫ 1

0x4 dx = 24

6

[15x5∣∣∣x=1

x=0

]= 24

30

[15 − 05

]= 24

30

E(Y ) =∫ 1

0

∫ 2x

xy

83xy︷︸︸︷

f(x, y) dy dx = 83

∫ 1

0x[ ∫ 2x

xy2 dy

]dx = 8

3

∫ 1

0x[

13y3∣∣∣y=2x

y=x

]dx

= 89

∫ 1

0x[(2x)3 − x3

]dx = 56

9

∫ 1

0x4 dx = 56

9

[15x5∣∣∣x=1

x=0

]= 56

45

[15 − 05

]= 56

45

Therefore, Cov(X,Y ) = 56/54− (24/30)(56/45) = 0.0415.

Solution 4.2.21 Since the 20% surcharge only applies to Y , then C2 = X+1.2Y . We can use the first two momentsof X and Y to find V (X) and V (Y ):

V (X) = E(X2)− [E(X)]2 = 27.4− [5]2 = 2.4V (Y ) = E(Y 2)− [E(Y )]2 = 51.4− [7]2 = 2.4

Formula 4.27 gives:

8︷︸︸︷V (X + Y ) =

2.4︷︸︸︷V (X) +

2.4︷︸︸︷V (Y ) +2

?︷︸︸︷Cov(X,Y )→ Cov(X,Y ) = 1.6

Therefore. . .

Cov(C1, C2) = Cov(

C1︷︸︸︷X + Y ,

C2︷︸︸︷X + 1.2Y ) =

F4.29︷︸︸︷Cov(X,X) + Cov(X, 1.2Y ) + Cov(Y,X) + Cov(Y, 1.2Y )

273

2.8 The Poisson Distribution

A generic poisson random variable, X, counts the number of events occurring in a fixed interval of time (or space,distance, volume, or some other measure), where the average observation rate for the given interval is denoted λ.The λ symbol is considered the parameter of the poisson distribution since it is the only value needed to fully definethe probability space. The range of possible values is X = 0, 1, 2, . . .

Shorthand notation: For convenience, rather than tediously writing “X is distributed as a poisson random variablewith parameter λ”, we will condense this information into the equivalent expression “X ∼ Pois(λ)”. The “∼” symbolshould be interpreted as “is distributed as”.

Example 2.24 Assume that copper wire is known to have irregularities, on average, once every 12 feet. Let X countthe number of defects in 12 feet of copper wire. Assume that X is distributed as a poisson random variable.

a) Express the poisson random variable using shorthand notation.b) Adjust your answer to a) if X counts the number of defects in 30 feet of copper wire.c) Adjust your answer to a) if X counts the number of defects in 1 foot of copper wire.

Solution:

a) We are told that, on average, 1 irregularity shows up every 12 feet. Since X counts the number of defects in 12feet of wire, then X ∼ Pois(λ = 1).

b) If there is, on average, 1 copper irregularity every 12 feet, then there must be, on average, 2.5 copper irregularitiesevery 30 feet. Therefore, X ∼ Pois(λ = 2.5). This 2.5 is calculated using proportions, given here:

1 irregularity

12 feet=x irregularities

30 feet→ 12x = 30(1)→ x = 2.5

c) If there is, on average, 1 copper irregularity every 12 feet, then there must be, on average, 1/12 copper irregularityevery 1 foot. Therefore, X ∼ Pois(λ = 1/12). This 1/12 is calculated using proportions, given here:

1 irregularity

12 feet=x irregularities

1 feet→ 12x = 1(1)→ x = 1/12

Note: In a), b), and c), the possible values that X can assume are X = 0, 1, 2, . . . This is true for any poisson randomvariable, regardless of how λ is adjusted using proportions.

Each of the possible X values has an associated probability, given by the following pmf:

P (X = x) =λx exp(−λ)

x!

where

x = 0, 1, 2, . . .

λ > 0

(2.37)

Note: The “exp” phrase in Formula 2.37 simply refers to the irrational number e = 2.718 . . .

The mean and variance of a poisson random variable are given here:

µX = E(X) =λ

σ2X = V(X) =λ

(2.38)

Formula 2.38 shows that the parameter of the poisson distribution, λ, is also the mean and variance of the distribution.

116

Example 2.25 Customers arrive at a grocery store at a mean rate of 3 per hour. Assuming that the number ofarrivals per time period has a Poisson distribution, find the probability that at least 1 customer arrives in the next. . .

a) hour.b) 10 minutes.

Solution: In a) and b), we are being asked to find P (X ≥ 1). Formula 1.14 gives P (X ≥ 1) = 1− P (X = 0).

a) Let X count the number of customers arriving per hour. Since customers arrive at a rate of 3 per hour,X ∼ Pois(λ = 3). Formula 2.37 gives:

P (X ≥ 1) = 1− P (X = 0) = 1− 30 exp(−3)0!

= 1− 0.0498 = 0.9502.

b) Let X count the number of customers arriving every ten minutes. Because λ is presented to us as a “per hour” rateand because b) phrases this problem over a 10 minute interval, we need to use proportions to adjust λ accordingly:

3 customers

60 minutes=x customers

10 minutes→ 60x = 10(3)→ x = 0.5

This implies that λ = 0.5 customer arrives, on average, every 10 minutes. Therefore, X ∼ Pois(λ = 0.5).

P (X ≥ 1) = 1− P (X = 0) = 1− 0.50 exp(−0.5)0!

= 1− 0.6065 = 0.3935.

The following picture shows the graphical probability distribution of X ∼ Pois(λ = 3). Since there are infinitely-manyvalues that X can assumed (X = 0, 1, 2, . . .), only the first five are shown. The height of each rectangle representsthe probability that X assumes that particular value. Also the mean is labeled.

Probability

0.05

0.10

0.15

0.20

0.25

1 2 3 4

µX = E(X) = λ = 3

The moment generating function (mgf) of a poisson random variable is:

MX(t) = exp[λ(et − 1)

](2.39)

Example 2.26 In the context of Example 2.25 part a), identify the mgf of X.

Solution: Substituting λ = 3 into Formula 2.39 gives MX(t) = exp[3(et − 1)

].

Sums of poisson random variables

Let Xi ∼ Pois(λi) for i = 1, 2, . . . ,m represent m independent poisson random variables. Because λ is subscripted,this implies that it can be different from one Xi to the next. Let X =

∑mi=1 Xi = X1 +X2 + · · ·+Xm be the sum of

117

these m random variables. Then X ∼ Pois(λ = λ1 + λ2 + · · ·+ λm). In words, the sum of m poisson random variablesis a poisson random variable whose λ parameter is the sum of the individual λ parameters. This is summarized here:

Let Xi ∼ Pois(λi) where i = 1, 2, . . . ,m be m independent poisson random variables.

Let X = X1 +X2 + · · ·+Xm. Then. . .

X ∼ Pois(λ = λ1 + λ2 + · · ·+ λm)

(2.40)

End-of-Section Practice Problems

Problem 2.8.1 Let X have a Poisson distribution with a mean of 3. Find. . .

a) P (2 ≤ X < 5)b) P (X < 2)c) P (X ≥ 2)

Problem 2.8.2 Let X have a Poisson distribution with a standard deviation of 4. Find P (X = 18).

Problem 2.8.3 The random variable X has a Poisson distribution such that P (X = 3) = 0.60 · P (X = 2). FindP (X = 5).

Problem 2.8.4 Flaws in linen produced at a certain textile factory appear at a rate of 3 in every 204 square feet.If we assume the Poisson distribution, find the probability of at most one flaw in 80 square feet.

Problem 2.8.5 The mean of a Poisson random variable X is E(X) = µX = 3. Find P (µX − σX ≤ X < µX + 2σX).

Problem 2.8.6 The opening day of a golf season at a country club gets postponed if it rains on that day. Theclub loses $15,000 for each consecutive day that it rains. Say X, the number of consecutive days that it rains at thebeginning of the golf season, has a Poisson distribution with a mean of 0.3. What is the expected loss before theopening golf day?

Problem 2.8.7 Let X denote a random variable that has a Poisson distribution with mean 3. Find. . .

a) P (X ≥ 3)b) P (X ≥ 1)c) P (X ≥ 3|X ≥ 1)

Problem 2.8.8 The number of errors that an actuarial intern makes when designing spreadsheets has a Poissondistribution with an average of five errors per spreadsheet. If more than two errors appear in a given spreadsheet,the intern must do 100 pushups. What is the probability that the intern does not have to do 100 pushups for herdesign of a certain spreadsheet?

Problem 2.8.9 The number of defects per square foot of sheet metal, X, is assumed to have a Poisson distributionwith a standard deviation of 1.35. The profit, P , per square foot when the sheet metal is sold is P = 100− 4X −X2.Find the expected profit per square foot.

Problem 2.8.10 Let X ∼ Pois(λ). Use Formula 2.39 to. . .

a) Show that the first moment is E(X) = λ.b) Show that the second moment is E(X2) = λ2 + λ.c) Use your answers to a) and b) to show that V(X) = λ.

Note: This is not an exam-type problem but is important because it shows the link between Formulas 2.38 and 2.39.

118

Problem 2.8.11 The moment generating function of a random variable is MX(t) = exp[12(et − 1)

].

a) Identify the distribution of X, including its parameters.b) Calculate P (X = 15).

Problem 2.8.12 (30 of 153; SOA May 2000 #24) An actuary has discovered that policyholders are three times aslikely to file two claims as to file four claims. If the number of claims filed has a Poisson distribution, what is thevariance of the number of claims filed?

Problem 2.8.13 (50 of 153; SOA November 2000 #23) A company buys a policy to insure its revenue in the eventof a major snowstorm that shuts down business. The policy pays nothing for the first such snowstorm of the yearand 10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shutdown business is assumed to have a Poisson distribution with mean 1.5.

What is the expected amount paid to the company under this policy during a one-year period?

Problem 2.8.14 (67 of 153; SOA November 2001 #19) A baseball team has scheduled its opening game for April1. If it rains on April 1, the game is postponed and will be played on the next day that it does not rain. The teampurchases insurance against rain. The policy will pay 1,000 for each day, up to 2 days, that the opening game ispostponed. The insurance company determines that the number of consecutive days of rain beginning on April 1 isa Poisson random variable with mean 0.6.

What is the standard deviation of the amount the insurance company will have to pay?

Problem 2.8.15 (153 of 153) Let X represent the number of customers arriving during the morning hours and letY represent the number of customers arriving during the afternoon hours at a diner. You are given:

i) X and Y are Poisson distributed.ii) The first moment of X is less than the first moment of Y by 8.iii) The second moment of X is 60% of the second moment of Y .

Calculate the variance of Y .

End-of-Section Practice Solutions

Solution 2.8.1 Formula 2.38 shows us that the mean of the Poisson random variable, λ, is also the parameter ofthe Poisson distribution. Since the mean is 3, then X ∼ Pois(λ = 3).

a) We are being asked to find P (X = 2) + P (X = 3) + P (X = 4). Formula 2.37 gives:

P (X = 2) = 32 exp(−3)2!

= 0.2240

P (X = 3) = 33 exp(−3)3!

= 0.2240

P (X = 4) = 34 exp(−3)4!

= 0.1680

Therefore P (2 ≤ X < 5) = 0.2240 + 0.2240 + 0.1680 = 0.6160.

b) We are being asked to find P (X < 2) = P (X = 0) + P (X = 1). Formula 2.37 gives:

P (X = 0) = 30 exp(−3)0!

= 0.0498

P (X = 1) = 31 exp(−3)1!

= 0.1494

119

Therefore P (X < 2) = 0.0498 + 0.1494 = 0.1992.

c) P (X ≥ 2) = 1− P (X < 2) = 1− 0.1992 = 0.8008

Solution 2.8.2 If the standard deviation is 4, then the variance is 42 = 16. Formula 2.38 shows us that the varianceof the Poisson random variable, λ, is also the parameter of the Poisson distribution. Therefore, X ∼ Pois(λ = 16).Formula 2.37 gives:

P (X = 18) = 1618 exp(−16)18!

= 0.0830.

Solution 2.8.3 We first need to use the given equality to solve for λ. Using Formula 2.37 gives:

P (X = 2) = λ2 exp(−λ)2!

P (X = 3) = λ3 exp(−λ)3!

Substituting these expressions into the given equality yields:P (X=3)︷︸︸︷

λ3��

�exp(−λ)

3!= 0.60 ·

P (X=2)︷︸︸︷λ2��

�exp(−λ)

2!→��λ

2·λ6

= 0.60��λ2

2→ λ = 1.8→ X ∼ Pois(λ = 1.8)

P (X = 5) = 1.85 exp(−1.8)5!

= 0.0260

Solution 2.8.4 Let X count the number of flaws that appear in 80 square feet. We are told the rate for every 204square feet. We need to find λ, the number of flaws every 80 square feet. Using proportions gives:

3 flaws

204 square feet=

λ flaws

80 square feet=→ λ =

3

204(80) = 1.1765

Therefore, there are 1.1765 flaws every 80 square feet. Therefore X ∼ Pois(λ = 1.1765). We are being asked to findP (X ≤ 1) = P (X = 0) + P (X = 1). Formula 2.37 gives:

P (X = 0) = 1.17650 exp(−1.1765)0!

= 0.3084

P (X = 1) = 1.17651 exp(−1.1765)1!

= 0.3628

Therefore, P (X ≤ 1) = 0.3084 + 0.3628 = 0.6712.

Solution 2.8.5 Formula 2.38 shows that λ is both the mean and variance of a Poisson random variable. Thereforeλ = E(X) = V (X) = 3. Therefore, X ∼ Pois(λ = 3). Since V (X) = σ2

X = 3, then SD(X) = σX =√

3 = 1.7321.Substituting µX = 3 and σX = 1.7321 into the expression gives:

P (µX − σX ≤ X < µX + 2σX) = P (3− 1.7321 ≤ X < 3 + 2(1.7321)) = P (1.2679 ≤ X < 6.4642).

Because X can only assume integer values, the previous expression simplifies to P (2 ≤ X ≤ 6). Formula 2.37 gives:

P (X = 2) = 32 exp(−3)2!

= 0.2240

P (X = 3) = 33 exp(−3)3!

= 0.2240

P (X = 4) = 34 exp(−3)4!

= 0.1680

P (X = 5) = 35 exp(−3)5!

= 0.1008

P (X = 6) = 36 exp(−3)6!

= 0.0504

Therefore P (2 ≤ X ≤ 6) = 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 = 0.7672

120

Solution 2.8.6 Let X count the number of consecutive days of rain at the beginning of the golf season. Formula2.38 shows us that the mean of the Poisson random variable, λ, is also the parameter of the Poisson distribution.Since the mean is E(X) = 0.3, then X ∼ Pois(λ = 0.3). Let Y measure the total loss due to postponing the openingday due to rain, so Y = 15, 000X. We are being asked to find E(Y ):

E(Y ) = E(15, 000X) = 15, 000E(X) = 15, 000(0.3) = 4, 500.

Solution 2.8.7 Formula 2.38 shows us that the mean of the Poisson random variable, λ, is also the parameter ofthe Poisson distribution. Since the mean is E(X) = 3, then X ∼ Pois(λ = 3). We will use Formula 2.37 for theprobability calculations.

a) P (X ≥ 3) = 1− P (X < 3) = 1− P (X = 0)− P (X = 1)− P (X = 2).

P (X = 0) = 30 exp(−3)0!

= 0.0498

P (X = 1) = 31 exp(−3)1!

= 0.1494

P (X = 2) = 32 exp(−3)2!

= 0.2240

Therefore, P (X ≥ 3) = 1− 0.0498− 0.1494− 0.2240 = 0.5768.

b) P (X ≥ 1) = 1− P (X = 0) = 1− 0.0498 = 0.9502.

c) Formula 1.22 expands P (X ≥ 3|X ≥ 1) = P [(X≥3)∩(X≥1)]P (X≥1)

= P (X≥3)P (X≥1)

= 0.57680.9502

= 0.6070

Solution 2.8.8 Let X count the number of errors that the intern makes on a given spreadsheet. Formula 2.38 showsus that the mean of the Poisson random variable, λ, is also the parameter of the Poisson distribution. Since the meanis E(X) = 5, then X ∼ Pois(λ = 5). Since the intern does not have to do pushups if at most 2 errors are made, weare being asked to find P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2). Formula 2.37 gives:

P (X = 0) = 50 exp(−5)0!

= 0.0067

P (X = 1) = 51 exp(−5)1!

= 0.0337

P (X = 2) = 52 exp(−5)2!

= 0.0842

Therefore P (X ≤ 2) = 0.0067 + 0.0337 + 0.0842 = 0.1246.

Solution 2.8.9 Formula 2.38 shows us that the variance of the Poisson random variable, λ, is also the parameterof the Poisson distribution. Since the standard deviation of X is SD(X) = σX = 1.35, then the variance of X isV (X) = σ2

X = 1.352 = 1.8225. Therefore, X ∼ Pois(λ = 1.8225). We are being asked to find E(P ):

E(P ) = E(100− 4X −X2) = 100− 4E(X)− E(X2)

Let us now find the first two moments of X, using Formula 2.38:E(X) = µX = λ = 1.8225V (X) = σ2

X = E(X2)− [E(X)]2︸︷︷︸F2.10

= E(X2)− [1.8225]2 = 1.8225→ E(X2) = 5.1440

Substituting these values gives E(P ) = 100− 4(1.8225)− 5.1440 = 87.5660

Solution 2.8.10 Formula 2.15 gives us the patterns to find the first two moments.

a) ddtMX(t) = d

dt

F2.39︷︸︸︷eλ(et−1) =

u︷︸︸︷eλ(et−1)

v︷︸︸︷λet E(X) = d

dtMX(t)

∣∣∣t=0

=

1︷︸︸︷eλ(e0−1) λe0 = λ

b) To find the second moment, we have to differentiate MX(t) twice. Using the differentiation rule for products,

121

where u and v are identified above, we have (uv)′ = u′v + uv′:

d2

dt2MX(t) = d

dt

[ddtMX(t)

]= d

dt

[eλ(et−1)λet

]=

u′︷︸︸︷eλ(et−1)λet

v︷︸︸︷λet +

u︷︸︸︷eλ(et−1)

v′︷︸︸︷λet = λeteλ(et−1)(λet + 1)

E(X2) = d2

dt2MX(t)

∣∣∣t=0

= λ

1︷︸︸︷e0eλ(e0−1)

λ+1︷︸︸︷(λe0 + 1) = λ(λ+ 1) = λ2 + λ

c) Formula 3.14 gives V (X) = σ2X = E(X2)− [E(X)]2 = λ2 + λ− [λ]2 = λ

Solution 2.8.11

a) Matching MX(t) with Formula 2.39 shows that λ = 12. Therefore, X ∼ Pois(λ = 12).

b) P (X = 15) =

F2.37︷︸︸︷1215 exp(−12)

15!= 0.0724

Solution 2.8.12 Let X represent the number of claims filed. We are told that X is distributed as a Poisson randomvariable but are not given its parameter λ. Therefore, we have X ∼ Pois(λ). We are being asked to find the varianceof X, which Formula 2.38 tells us is V (X) = λ. We are told that P (X = 2) = 3P (X = 4). Substituting the pmf intothis expression gives:P (X=2)︷︸︸︷

λ2��

�exp(−λ)

2!= 3

P (X=4)︷︸︸︷(λ4��

�exp(−λ)

4!

)→ λ2

2= 3λ4

24→ λ = V (X) = 2

Solution 2.8.13 Let X count the number of major snowstorms that shut down business during a given year, whereX = 0, 1, 2, . . .. We are told that E(X) = 1.5 which, based on Formula 2.38, is also the value of λ. Therefore,X ∼ Pois(λ = 1.5). Let Y be the amount paid to the company under this policy. Based on the given information, wehave

Y =

{0 if x = 0, 1

10, 000(X − 1) if x ≥ 2

We are being asked to find E(Y ) =∑y yP (Y = y), calculated here:

E(Y ) =

=0︷︸︸︷0P (X = 0) + 0P (X = 1) +

∞∑x=2

10, 000(x− 1)P (X = x)

=

∞∑x=2

10, 000(x− 1)P (X = x)

=

=0︷︸︸︷10, 000(0− 1)P (X = 0)− 10, 000(0− 1)P (X = 0)

+

=0︷︸︸︷10, 000(1− 1)P (X = 1)− 10, 000(1− 1)P (X = 1)

+

∞∑x=2

10, 000(x− 1)P (X = x)

=− 10, 000(0− 1)P (X = 0)−

0︷︸︸︷10, 000(1− 1)P (X = 1) +

E[10,000(X−1)]︷︸︸︷∞∑x=0

[10, 000(x− 1)]P (X = x)

Note that the overbrace expression in lines 3 and 4 above were included so as to allow the summation to go over allpossible X values (from 0 to ∞) so as to create E[10, 000(X − 1)] in the last line. Using the pmf of X, we have:

P (X = 0) = 1.50 exp(−1.5)0!

= 0.2231E[10, 000(X − 1)] = 10, 000E(X)− 10, 000 = 10, 000(1.5)− 10, 0000 = 5, 000

122

Therefore, E(Y ) = −10, 000(−1)(0.2231) + 5, 000 = 7, 231.

Solution 2.8.14 Let X count the number of consecutive days of rain beginning on April 1, where X = 0, 1, 2, . . .We are told that E(X) = 0.6 which, based on Formula 2.38, is also the value of λ. Therefore, X ∼ Pois(λ = 0.6). LetY be the amount paid to the team under its insurance policy. Based on the given information, we have

Y =

{1, 000x if x ≤ 1

2, 000 if x ≥ 2

We are being asked to find SD(Y ) =√V (Y ). Since V (Y ) = E(Y 2) − [E(Y )]2, we will need to find the first two

moments of Y . Let us first find E(Y ) =∑y yP (Y = y):

E(Y ) =

0︷︸︸︷1, 000(0)

[0.60 exp(−0.6)

0!

]+1, 000(1)

[0.61 exp(−0.6)

1!

]+

∞∑x=2

2, 000P (X = x)

=600 exp(−0.6) + 2, 000P (X ≥ 2)

=600 exp(−0.6) + 2, 000[1− P (X ≤ 1)]

=600 exp(−0.6) + 2, 000[1−

P (X=0)︷︸︸︷0.60 exp(−0.6)

0!−

P (X=1)︷︸︸︷0.61 exp(−0.6)

1!

]︸︷︷︸

=0.12190138

=573

Let us now find E(Y 2) =∑y y

2P (Y = y):

E(Y 2) =

0︷︸︸︷[1, 000(0)]2

[0.60 exp(−0.6)

0!

]+[1, 000(1)]2

[0.61 exp(−0.6)

1!

]+

∞∑x=2

2, 0002P (X = x)

=600, 000 exp(−0.6) + 2, 0002P (X ≥ 2)

=600, 000 exp(−0.6) + 2, 0002(0.12190138)

=816, 893

Therefore, V (Y ) = 816, 893− [573]2 = 488, 564. Therefore, SD(Y ) =√V (Y ) =

√488, 564 = 699.

Solution 2.8.15 We are told that both X and Y are Poisson random variables but with different parameters.Therefore, we can write: X ∼ Pois(λx) and Y ∼ Pois(λy). Formula 2.38 gives the first moment for a Poisson randomvariable: E(X) = λx and E(Y ) = λy. Statement (ii) tells us that λx = λy − 8.

Formula 2.10 tells us that V (X) = E(X2) − [E(X)]2 → E(X2) = V (X) + [E(X)]2. Formula 2.38 tells us thatV (X) = λx and V (Y ) = λy. Therefore, the second moments are E(X2) = λx + λ2

x and E(Y 2) = λy + λ2y. Statement

(iii) tells us that E(X2) = 0.60E(Y 2)→ λx+λ2x = 0.60(λy+λ2

y). Substituting λx = λy−8 into the previous equationgives:

(λy − 8︸︷︷︸λx

) + (λy − 8︸︷︷︸λx

)2 = 0.60(λy + λ2y),

which is a quadratic equation. Rearranging this gives 0.4λ2y − 15.6λy + 56 = 0.

Substituting a = 0.4, b = −15.6, and c = 56 into Formula 1.4 gives λy = 4 and λy = 35. If λy = 4, then the statementfrom (ii) gives λx = 4− 8 = −4. Since the Poisson parameter must be positive, λy 6= 4. Therefore λy = 35. We areasked to find V (Y ). Since V (Y ) = λy, then V (Y ) = 35.

123

5.2 Normal Approximation of Discrete Probabilities

Sometimes we are asked to calculate a very tedious probability involving a discrete random variable. For example, ifX ∼ Binom(n = 750, p) and if we are asked to calculate the probablity that X is at least 500, we would need to find:

P (X ≥ 500) = P (X = 500) + P (X = 501) + P (X = 502) + · · ·+ P (X = 749) + P (X = 750),

where each of the 251 probabilities on the right side must be evaluated using Formula 2.21. This would take a verylong time to calculate exactly. Under certain conditions, the continuous normal distristribution provides a very close(and much faster!) approximation to the exact discrete calculation. To make these approxmations a little moreaccurate, we need to apply a “continuity correction” as given by the following table.

For a generic discrete random variable X and constant c, do the following:

If problem is originally Apply this adjustment forphrased like this. . . the “continuity correction”

P (X < c) P (X < c− 0.5)P (X ≤ c) P (X ≤ c+ 0.5)P (X > c) P (X > c+ 0.5)P (X ≥ c) P (X ≥ c− 0.5)P (X = c) P (c− 0.5 ≤ X ≤ c+ 0.5)

(5.8)

Note: Since X assumes only one value in the last row of the previous formula, in this case, one would probablycalculate the probability exactly rather than approximating it with the normal distribution.

Example 5.4 Assume X is a generic discrete random variable. Apply Formula 5.8 to the following inequalities as ifwe were asked to approximate a discrete probability calculation using the normal curve.

a) P (X > 5)b) P (X ≤ 71)c) P (14 ≤ X)d) P (X < 12)e) P (3 < X < 10)f) P (X = 20)

Solution:

a) P (X > 5.5)b) P (X ≤ 71.5)c) P (14 ≤ X) = P (X ≥ 14)→ P (X ≥ 13.5)d) P (X < 11.5)e) P (3 < X) = P (X > 3)→ P (X > 3.5)→ P (3.5 < X < 9.5)f) P (19.5 ≤ X ≤ 20.5)

You may be interested in how the normal curve (a continuous distribution) can provide a reasonable approxima-tion to a discrete probability calculation. Under certain mathematical conditions, the probability distribution ofthe discrete random variable is reasonably symmetric and bell-shaped, closely mimicking the normal distribution(albeit from a discrete perspective). The following picture shows two distributions. The discrete distributionis of X ∼ Binom(n = 10, p = 0.5), whose mean and standard deviation (given by Formula 2.22) are µX = 5 andσX = 1.5811. The continuous normal distribution is of Y ∼ N(µY = 5, σY = 1.5811).

292

Probability

0.05

0.10

0.15

0.20

0.25

0 1 2 3 4 5 6 7 8 9 10

Y ∼ N(µY = 5, σY = 1.5811)

X ∼ Binom(n = 10, p = 0.5)

Since the two curves in this picture closely mimic each other, we should have confidence that the normal curve can beused to calculate reasonable approximations to discrete calculations. Note that the discussion so far in this sectionhas referenced the binomial distribution. The normal curve, however, can be used to approximate any discreteprobability, not just the binomial distribution. Here are the steps. . .

Step #1: Apply the continuity correction (Formula 5.8) to the discrete probability expression.

Step #2: Calculate the mean and standard deviation of the discrete random variable.

Step #3: Apply the z-transformation formula (Formula 3.43) to all sides of the probability expression.

Step #4: Carry out the calculation from the Z perspective, calculating the answer exactly like

the normal probability calculations from Section 3.6.

(5.9)

Example 5.5 Assume that 15% of all college students would answer “yes” to the question “Would you drive homefrom a party if you had been drinking excessively?” Let X count the number of students from a random sample ofsize n = 250 who would say “yes” to this question. Suppose we are interested in P (X ≤ 45).

a) How would you calculate P (X ≤ 45) exactly?b) Approximate P (X ≤ 45) using the method from this section.

Solution: Assuming that responses are independent, X is a binomial random variable since it counts the numberof “successes” (driving while being intoxicated, with probability p = 0.15) from a fixed number of people (n = 250)and because any one person would eith drive home or not drive home. Here, q = 1− 0.15 = 0.85.

a) To calculate this probability exactly, we would tediously apply Formula 2.21 to each of the following probabilities:

P (X ≤ 45) = P (X = 0) + P (X = 1) + · · ·+ P (X = 44) + P (X = 45)

b) Let us first find the mean and standard deviation of the binomial random variable X using Formula 2.22:

µX = E(X) = np = 250(0.15) = 37.5σ2X = V (X) = npq = 250(0.15)(0.85) = 31.875→ σX = SD(X) =

√V (X) =

√31.875 = 5.65

Now let us calculate the probability of interest:

P (X ≤ 45) =

F5.8︷︸︸︷P (X ≤ 45.5) = P

( Z︷︸︸︷X − µXσX

≤ 45.5−37.55.65

)≈ P (Z ≤ 1.42) = 0.9222

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Transcript of Actuarial AdvantEDGE Exam P / CAS 1 Study Manual · Actuarial AdvantEDGE Exam P / CAS 1 Study...