In practice we often construct new random variables by performing arithmetic operations on other random variables. For example, we might add a constant to a random variable, multiply a random variable by a constant, or add two or more random variables together.
In this section, we describe how to compute means and variances of random variables
constructed in these ways, and we present some practical examples. The presentation in this section is intuitive. A more rigorous presentation is provided in Section 2.6. For those desiring such a presentation, Section 2.6 may be covered in addition to, or in place of, this section.
Adding a Constant
When a constant is added to a random variable, the mean is increased by the value of the constant, but the variance and standard deviation are unchanged. For example, assume that steel rods produced by a certain machine have a mean length of 5.0 in. and a variance ofσ2=0.003 in2. Each rod is attached to a base that is exactly 1.0 in. long. The mean length of the assembly will be 5.0+1.0 = 6.0 in. Since each length is increased by the same amount, the spread in the lengths does not change, so the variance remains the same. To put this in statistical terms, letX be the length of a randomly chosen rod, and letY = X +1 be the length of the assembly. ThenμY = μX+1 = μX +1, and σY2 =σX+2 1 =σX2. In general, when a constant is added to a random variable, the mean is shifted by that constant, and the variance is unchanged.
Summary
IfXis a random variable andbis a constant, then
μX+b=μX+b (2.39)
σX2+b=σX2 (2.40)
Multiplying by a Constant
Often we need to multiply a random variable by a constant. This might be done, for example, to convert to a more convenient set of units. We continue the example of steel rod production to show how multiplication by a constant affects the mean, variance, and standard deviation of a random variable.
If we measure the lengths of the rods described earlier in centimeters rather than inches, the mean length will be (2.54 cm/in.)(5.0 in.)=12.7 cm. In statistical terms, let the random variableXbe the length in inches of a randomly chosen rod, and letY =2.54X be the length in centimeters. ThenμY =2.54μX. In general, when a random variable is multiplied by a constant, its mean is multiplied by the same constant.
Summary
IfXis a random variable andais a constant, then
μa X =aμX (2.41)
When the lengthXof a rod is measured in inches, the varianceσX2must have units of in2. IfY =2.54Xis the length in centimeters, thenσY2must have units of cm2. Therefore we obtainσY2by multiplyingσX2by 2.542, which is the conversion factor from in2to cm2. In general, when a random variable is multiplied by a constant, its variance is multiplied by thesquareof the constant.
Summary
IfX is a random variable andais a constant, then
σa X2 =a2σX2 (2.42)
σa X = |a|σX (2.43)
If a random variable is multiplied by a constant and then added to another constant, the effect on the mean and variance can be determined by combining Equations (2.39) and (2.41) and Equations (2.40) and (2.42). The results are presented in the following summary.
Summary
IfX is a random variable, andaandbare constants, then
μa X+b=aμX +b (2.44)
σa X+b2 =a2σX2 (2.45)
σa X+b= |a|σX (2.46)
Note that Equations (2.44) through (2.46) are analogous to results for the sample mean and standard deviation presented in Section 1.2.
E xample
2.48 The molarity of a solute in solution is defined to be the number of moles of solute per liter of solution (1 mole=6.02×1023molecules). If the molarity of a stock solution of concentrated sulfuric acid (H2SO4) is X, and if one part of the solution is mixed withNparts water, the molarityY of the dilute solution is given byY =X/(N+1).
Assume that the stock solution is manufactured by a process that produces a molarity with mean 18 and standard deviation 0.1. If 100 mL of stock solution is added to 300 mL of water, find the mean and standard deviation of the molarity of the dilute solution.
Solution
The molarity of the dilute solution isY =0.25X. The mean and standard deviation of XareμX =18 andσX =0.1, respectively. Therefore
μY =μ0.25X
=0.25μX (using Equation 2.41)
=0.25(18.0)
=4.5 Also,
σY =σ0.25X
=0.25σX (using Equation 2.43)
=0.25(0.1)
=0.025
Means of Linear Combinations of Random Variables
Consider the case of adding two random variables. For example, assume that there are two machines that fabricate a certain metal part. The mean daily production of machine A is 100 parts, and the mean daily production of machine B is 150 parts. Clearly the mean daily production from the two machines together is 250 parts. Putting this in mathematical notation, letXbe the number of parts produced on a given day by machine A, and letY be the number of parts produced on the same day by machine B. The total number of parts isX+Y, and we have thatμX+Y =μX +μY.
This idea extends to any number of random variables.
IfX1,X2, . . . ,Xnare random variables, then the mean of the sumX1+X2+
ã ã ã +Xnis given by
μX1+X2+ ããã +Xn =μX1+μX2+ ã ã ã +μXn (2.47)
The sumX1+X2+ ã ã ã +Xnis a special case of alinear combination:
IfX1, . . . ,Xnare random variables andc1, . . . ,cnare constants, then the ran- dom variable
c1X1+ ã ã ã +cnXn is called alinear combinationofX1, . . . ,Xn.
To find the mean of a linear combination of random variables, we can combine Equations (2.41) and (2.47):
IfXandY are random variables, andaandbare constants, then
μa X+bY =μa X+μbY =aμX+bμY (2.48) More generally, ifX1,X2, . . . ,Xnare random variables andc1,c2, . . . ,cnare constants, then the mean of the linear combinationc1X1+c2X2+ ã ã ã +cnXn is given by
μc1X1+c2X2+ ããã +cnXn=c1μX1+c2μX2+ ã ã ã +cnμXn (2.49)
Independent Random Variables
The notion of independence for random variables is very much like the notion of inde- pendence for events. Two random variables are independent if knowledge concerning one of them does not affect the probabilities of the other. When two events are indepen- dent, the probability that both occur is found by multiplying the probabilities for each event (see Equations 2.19 and 2.20 in Section 2.3). There are analogous formulas for independent random variables. The notation for these formulas is as follows. LetX be a random variable and letSbe a set of numbers. The notation “X ∈ S” means that the value of the random variableXis in the setS.
Definition
IfXandY areindependentrandom variables, andSandT are sets of numbers, then
P(X∈ SandY ∈T)=P(X ∈S)P(Y ∈T) (2.50) More generally, ifX1, . . . ,Xnare independent random variables, andS1, . . . ,Sn are sets, then
P(X1∈S1andX2∈S2andã ã ã andXn∈ Sn)=
P(X1∈S1)P(X2∈S2)ã ã ãP(Xn ∈Sn) (2.51)
E xample
2.49 Rectangular plastic covers for a compact disc (CD) tray have specifications regarding length and width. Let X be the length and Y be the width, each measured to the nearest millimeter, of a randomly sampled cover. The probability mass function ofX is given byP(X =129)=0.2, P(X =130)=0.7, andP(X =131)=0.1. The probability mass function ofY is given byP(Y =15)=0.6 andP(Y =16)=0.4.
The area of a cover is given byA=XY. AssumeXandY are independent. Find the probability that the area is 1935 mm2.
Solution
The area will be equal to 1935 ifX =129 andY =15. Therefore P(A=1935)=P(X =129 andY =15)
=P(X =129)P(Y =15) sinceX andY are independent
=(0.2)(0.6)
=0.12
Equations (2.50) and (2.51) tell how to compute probabilities for independent random variables, but they are not usually much help in determining whether random variables actually are independent. In general, the best way to determine whether random variables are independent is through an understanding of the process that generated them.
Variances of Linear Combinations of Independent Random Variables
We have seen that the mean of a sum of random variables is always equal to the sum of the means (Equation 2.47). In general, the formula for the variance of a sum of random variables is a little more complicated than this. But when random variables are independent, the result is simple: the variance of the sum is the sum of the variances.
IfX1,X2, . . . ,Xnareindependentrandom variables, then the variance of the sumX1+X2+ ã ã ã +Xnis given by
σX21+X2+ ããã +Xn =σX21+σX22+ ã ã ã +σX2n (2.52) To find the variance of a linear combination of random variables, we can combine Equations (2.52) and (2.42):
If X1,X2, . . . ,Xn are independent random variables andc1,c2, . . . ,cn are constants, then the variance of the linear combinationc1X1+c2X2+ã ã ã+cnXn is given by
σc21X1+c2X2+ ããã +cnXn =c12σX21+c22σX22+ ã ã ã +c2nσX2n (2.53) Two frequently encountered linear combinations are the sum and the difference of two random variables. Interestingly enough, when the random variables are independent, the variance of the sum is the same as the variance of the difference.
IfX andY areindependentrandom variables with variancesσX2 andσY2, then the variance of the sumX+Y is
σX2+Y =σX2+σY2 (2.54) The variance of the differenceX−Y is
σX2−Y =σX2+σY2 (2.55)
The fact that the variance of the difference is thesumof the variances may seem coun- terintuitive. However, it follows from Equation (2.53) by settingc1=1 andc2 = −1.
E xample 2.50 A piston is placed inside a cylinder. The clearance is the distance between the edge of the piston and the wall of the cylinder and is equal to one-half the difference between the cylinder diameter and the piston diameter. Assume the piston diameter has a mean of 80.85 cm with a standard deviation of 0.02 cm. Assume the cylinder diameter has a mean of 80.95 cm with a standard deviation of 0.03 cm. Find the mean clearance.
Assuming that the piston and cylinder are chosen independently, find the standard deviation of the clearance.
Solution
LetX1 represent the diameter of the cylinder and letX2the diameter of the piston.
The clearance is given byC = 0.5X1 −0.5X2. Using Equation (2.49), the mean clearance is
μC =μ0.5X1−0.5X2
=0.5μX1−0.5μX2
=0.5(80.95)−0.5(80.85)
=0.050
Since X1 andX2 are independent, we can use Equation (2.53) to find the standard deviationσC:
σC =
σ0.5X2 1−0.5X2
=
(0.5)2σX21+(−0.5)2σX22
=
0.25(0.02)2+0.25(0.03)2
=0.018
Independence and Simple Random Samples
When a simple random sample of numerical values is drawn from a population, each item in the sample can be thought of as a random variable. The items in a simple random sample may be treated as independent, except when the sample is a large proportion (more than 5%) of a finite population (see the discussion of independence in Section 1.1). From here on, unless explicitly stated to the contrary, we will assume this exception has not occurred, so that the values in a simple random sample may be treated as independent random variables.
Summary
If X1,X2, . . . ,Xn is a simple random sample, then X1,X2, . . . ,Xn may be treated as independent random variables, all with the same distribution.
WhenX1, . . . ,Xnare independent random variables, all with the same distribution, it is sometimes said thatX1, . . . ,Xnareindependent and identically distributed (i.i.d.).
The Mean and Variance of a Sample Mean
The most frequently encountered linear combination is the sample mean. Specifically, if X1, . . . ,Xnis a simple random sample from a population with meanμand variance σ2, then the sample meanX is the linear combination
X = 1
nX1+ ã ã ã + 1 nXn From this fact we can compute the mean and variance ofX.
μX =μ1nX1+ ããã +1nXn
= 1
nμX1+ ã ã ã +1
nμXn (using Equation 2.49)
= 1
nμ+ ã ã ã + 1 nμ
=(n) 1
n
μ
=μ
As discussed previously, the items in a simple random sample may be treated as inde- pendent random variables. Therefore
σX2 =σ21n X1+ ããã +1 n Xn
= 1
n2σX21+ ã ã ã + 1
n2σX2n (using Equation 2.53)
= 1
n2σ2+ ã ã ã + 1 n2σ2
=(n) 1
n2
σ2
= σ2 n
Summary
If X1, . . . ,Xn is a simple random sample from a population with meanμand varianceσ2, then the sample meanX is a random variable with
μX =μ (2.56)
σX2 = σ2
n (2.57)
The standard deviation ofXis
σX = σ
√n (2.58)
E xample 2.51 A process that fills plastic bottles with a beverage has a mean fill volume of 2.013 L and a standard deviation of 0.005 L. A case contains 24 bottles. Assuming that the bottles in a case are a simple random sample of bottles filled by this method, find the mean and standard deviation of the average volume per bottle in a case.
Solution
LetV1, . . . ,V24represent the volumes in 24 bottles in a case. This is a simple random sample from a population with meanμ=2.013 and standard deviationσ =0.005.
The average volume isV =(V1+ ã ã ã +V24)/24. Using Equation (2.56), μV =μ=2.013
Using Equation (2.58),
σV = σ
√24 =0.001
Exercises for Section 2.5
1. If X and Y are independent random variables with meansμX = 9.5 andμY = 6.8, and standard devi- ationsσX = 0.4 andσY = 0.1, find the means and standard deviations of the following:
a. 3X b. Y−X c. X+4Y
2. The bottom of a cylindrical container has an area of 10 cm2. The container is filled to a height whose mean is 5 cm, and whose standard deviation is 0.1 cm. Let Vdenote the volume of fluid in the container.
a. FindμV. b. FindσV.
3. The lifetime of a certain lightbulb in a certain ap- plication has mean 700 hours and standard deviation 20 hours. As each bulb burns out, it is replaced with a new bulb. Find the mean and standard deviation of the length of time that five bulbs will last.
4. Two resistors, with resistances R1 and R2, are con- nected in series. The combined resistance Ris given byR=R1+R2. Assume thatR1has mean 50and standard deviation 5, and that R2has mean 100 and standard deviation 10.
a. FindμR.
b. AssumingR1andR2to be independent, findσR.
5. A piece of plywood is composed of five layers. The layers are a simple random sample from a population whose thickness has mean 0.125 in. and standard de- viation 0.005 in.
a. Find the mean thickness of a piece of plywood.
b. Find the standard deviation of the thickness of a piece of plywood.
6. Two independent measurements are made of the life- time of a charmed strange meson. Each measurement has a standard deviation of 7×10−15 seconds. The lifetime of the meson is estimated by averaging the two measurements. What is the standard deviation of this estimate?
7. The molarity of a solute in solution is defined to be the number of moles of solute per liter of solution (1 mole = 6.02 ×1023 molecules). If X is the molarity of a solution of magnesium chloride (MgCl2), andY is the molarity of a solution of ferric chloride (FeCl3), the molarity of chloride ion (Cl−) in a solu- tion made of equal parts of the solutions of MgCl2and FeCl3is given byM= X+1.5Y. Assume thatXhas mean 0.125 and standard deviation 0.05, and thatY has mean 0.350 and standard deviation 0.10.
a. FindμM.
b. AssumingXandY to be independent, findσM.
8. A machine that fills cardboard boxes with cereal has a fill weight whose mean is 12.02 oz, with a stan- dard deviation of 0.03 oz. A case consists of 12 boxes randomly sampled from the output of the machine.
a. Find the mean of the total weight of the cereal in the case.
b. Find the standard deviation of the total weight of the cereal in the case.
c. Find the mean of the average weight per box of the cereal in the case.
d. Find the standard deviation of the average weight per box of the cereal in the case.
e. How many boxes must be included in a case for the standard deviation of the average weight per box to be 0.005 oz?
9. The four sides of a picture frame consist of two pieces selected from a population whose mean length is 30 cm with standard deviation 0.1 cm, and two pieces selected from a population whose mean length is 45 cm with standard deviation 0.3 cm.
a. Find the mean perimeter.
b. Assuming the four pieces are chosen indepen- dently, find the standard deviation of the perimeter.
10. A gas station earns $2.60 in revenue for each gallon of regular gas it sells, $2.75 for each gallon of midgrade gas, and $2.90 for each gallon of premium gas. Let X1,X2, andX3denote the numbers of gallons of reg- ular, midgrade, and premium gasoline sold in a day.
Assume thatX1,X2, andX3have meansμ1=1500, μ2 = 500, andμ3 = 300, and standard deviations σ1=180,σ2=90, andσ3=40, respectively.
a. Find the mean daily revenue.
b. AssumingX1,X2, andX3to be independent, find the standard deviation of the daily revenue.
11. The number of miles traveled per gallon of gasoline for a certain car has a mean of 25 and a standard de- viation of 2. The tank holds 20 gallons.
a. Find the mean number of miles traveled per tank.
b. Assume the distances traveled are independent for each gallon of gas. Find the standard deviation of the number of miles traveled per tank.
c. The car owner travels X miles on 20 gallons of gas, and estimates her gas mileage asX/20. Find the mean of the estimated gas mileage.
d. Assuming the distances traveled are independent for each gallon of gas, find the standard deviation of the estimated gas mileage.
12. The Needleman-Wunsch method for aligning DNA sequences assigns 1 point whenever a mis- match occurs, and 3 points whenever a gap (insertion or deletion) appears in a sequence. Assume that un- der certain conditions, the number of mismatches has mean 5 and standard deviation 2, and the number of gaps has mean 2 and standard deviation 1.
a. Find the mean of the Needleman-Wunsch score.
b. Assume the number of gaps is independent of the number of mismatches. Find the variance of the Needleman-Wunsch score.
13. In the article “An Investigation of the Ca–CO3–CaF2– K2SiO3–SiO2–Fe Flux System Using the Submerged Arc Welding Process on HSLA-100 and AISI-1018 Steels” (G. Fredrickson, M.S. thesis, Colorado School of Mines, 1992), the carbon equivalentP of a weld metal is defined to be a linear combination of the weight percentages of carbon (C), manganese (Mn), copper (Cu), chromium (Cr), silicon (Si), nickel (Ni), molybdenum (Mo), vanadium (V), and boron (B). The carbon equivalent is given by
P =C+Mn+Cu+Cr
20 +Si
30+Ni 60+Mo
15 +V 10+5B Means and standard deviations of the weight per- cents of these chemicals were estimated from mea- surements on 45 weld metals produced on HSLA-100 steel base metal. Assume the means and standard de- viations (SD) are as given in the following table.
Mean SD
C 0.0695 0.0018
Mn 1.0477 0.0269
Cu 0.8649 0.0225
Cr 0.7356 0.0113
Si 0.2171 0.0185
Ni 2.8146 0.0284
Mo 0.5913 0.0031
V 0.0079 0.0006
B 0.0006 0.0002
a. Find the mean carbon equivalent of weld metals produced from HSLA-1000 steel base metal.
b. Assuming the weight percents to be independent, find the standard deviation of the carbon equiv- alent of weld metals produced from HSLA-1000 steel base metal.
14. The oxygen equivalence number of a weld is a number that can be used to predict properties such as hard- ness, strength, and ductility. The article “Advances in Oxygen Equivalence Equations for Predicting the Properties of Titanium Welds” (D. Harwig, W.
Ittiwattana, and H. Castner, The Welding Journal, 2001:126s–136s) presents several equations for com- puting the oxygen equivalence number of a weld. One equation, designed to predict the hardness of a weld, is X = O+2N+(2/3)C, where X is the oxygen equivalence, and O, N, and C are the amounts of oxygen, nitrogen, and carbon, respectively, in weight percent, in the weld. Suppose that for welds of a cer- tain type,μO=0.1668,μN =0.0255,μC =0.0247, σO=0.0340,σN =0.0194, andσC =0.0131.
a. FindμX.
b. Suppose the weight percents ofO, N, andCare independent. FindσX.
15. Measurements are made on the length and width (in cm) of a rectangular component. Because of measure- ment error, the measurements are random variables.
LetXdenote the length measurement and letYdenote the width measurement. Assume that the probability density function ofXis
f(x)=
10 9.95<x<10.05 0 otherwise
and that the probability density function ofY is g(y)=
5 4.9<y<5.1 0 otherwise
Assume that the measurements X and Y are independent.
a. FindP(X<9.98). b. FindP(Y >5.01).
c. FindP(X<9.98 andY >5.01).
d. FindμX. e. FindμY.
16. The thicknessXof a wooden shim (in mm) has prob- ability density function
f(x)= 3
4−3(x−5)2
4 4≤x≤6
0 otherwise
a. FindμX. b. FindσX2.
c. Let Y denote the thickness of a shim in inches (1 mm = 0.0394 inches). FindμYandσY2. d. If three shims are selected independently and
stacked one atop another, find the mean and vari- ance of the total thickness.
17. The article “Abyssal Peridotites > 3800 Ma from Southern West Greenland: Field Relationships, Pe- trography, Geochronology, Whole-Rock and Mineral Chemistry of Dunite and Harzburgite Inclusions in the Itsaq Gneiss Complex” (C. Friend, V. Bennett, and A. Nutman,Contrib Mineral Petrol, 2002:71–92) describes the chemical compositions of certain min- erals in the early Archaean mantle. For a certain type of olivine assembly, the silicon dioxide (SiO2) con- tent (in weight percent) in a randomly chosen rock has mean 40.25 and standard deviation 0.36.
a. Find the mean and standard deviation of the sam- ple mean SiO2content in a random sample of 10 rocks.
b. How many rocks must be sampled so that the stan- dard deviation of the sample mean SiO2content is 0.05?
18. The number of bytes downloaded per second on an information channel has mean 105and standard de- viation 104. Among the factors influencing the rate is congestion, which produces alternating periods of faster and slower transmission. Let X represent the number of bytes downloaded in a randomly chosen five-second period.
a. Is it reasonable to assume thatμX = 5×105? Explain.
b. Is it reasonable to assume thatσX = √ 5×104? Explain.