Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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Electrical and Thermal Stress,” IEEE Trans. Electr. Insul ., 1985: 70–78).

A Weibull probability plot necessitates first computing the 5th, 15th, ..., and

95th percentiles of the standard extreme value distribution. The (100p)th percentile

( p ) satisfies

p ¼ F½ðpÞ ¼ 1  e

e

ðpÞ

from which ( p ) ¼ ln[ln(1  p)].

Percentile 2.97 1.82 1.25 .84 .51

x 282 501 741 851 1,072

ln(x) 5.64 6.22 6.61 6.75 6.98

Percentile .23 .05 .33 .64 1.10

x 1,122 1,202 1,585 1,905 2,138

ln(x) 7.02 7.09 7.37 7.55 7.67

The pairs (2.97, 5.64), (1.82, 6.22), ..., (1.10, 7.67) are plotted as points in

Figure 4.36. The straightness of the plot argues strongly for using the Weibull

distribution as a model for insulation life, a conclusion also reached by the author of

the cited article.

The gamma distribution is an example of a family involving a shape parame-

ter for which there is no transformation h(x) such that h(X) has a distribution that

depends only on location and scale parameters. Construction of a probability plot

necessitates first estimating the shape parameter from sample data (some methods

for doing this are described in Chapter 7).

Sometimes an investigator wishes to know whether the transformed

variable X

has a normal distribution for some value of y (by convention, y ¼ 0

is identified with the logarithmic transformation, in which case X has a lognormal

distribution). Th e book Graphical Methods for Data Analysis, listed in the

Chapter 1 bibliography, discusse s this type of problem as well as other refinements

of probability plotting.

−3 −2 −10 1

Percentile

ln(x)

Figure 4.36 A Weibull probability plot of the insulation lifetime data ■

218

CHAPTER 4 Continuous Random Variables and Probability Distributions

Exercises Section 4.6 (97–107)

97. The accompanying normal probability plot was

constructed from a sample of 30 readings on

tension for mesh screens behind the surface of

video display tubes. Does it appear plausible that

the tension distribution is normal?

−2 −10 1 2

200

250

300

350

z percentile

Tension

98. A sample of 15 female collegiate golfers was

selected and the clubhead velocity (km/h) while

swinging a driver was determined for each one,

resulting in the following data (“Hip Rotational

Velocities during the Full Golf Swing,” J. of

Sports Science and Medicine, 2009: 296–299):

69.0 69.7 72.7 80.3 81.0

85.0 86.0 86.3 86.7 87.7

89.3 90.7 91.0 92.5 93.0

The corresponding z percentiles are

1.83 1.28 0.97 0.73 0.52

0.34 0.17 0.0 0.17 0.34

0.52 0.73 0.97 1.28 1.83

Construct a normal probability plot and a dotplot.

Is it plausible that the population distribution is

normal?

99. Construct a normal probability plot for the fol-

lowing sample of observations on coating thick-

ness for low-viscosity paint (“Achieving a Target

Value for a Manufacturing Process: A Case

Study,” J. Qual. Tech., 1992: 22–26). Would

you feel comfortable estimating population

mean thickness using a method that assumed a

normal population distribution?

.83 .88 .88 1.04 1.09 1.12 1.29 1.31

1.48 1.49 1.59 1.62 1.65 1.71 1.76 1.83

100. The article “A Probabilistic Model of Fracture in

Concrete and Size Effects on Fracture Tough-

ness” (Mag. Concrete Res., 1996: 311–320)

gives arguments for why fracture toughness

in concrete specimens should have a Weibull

distribution and presents several histograms

of data that appear well fit by superimposed

Weibull curves. Consider the following sample

of size n ¼ 18 observations on toughness for

high-strength concrete (consistent with one of

the histograms); values of p

¼ (i  .5)/18 are

also given.

Observation .47 .58 .65 .69 .72 .74

.0278 .0833 .1389 .1944 .2500 .3056

Observation .77 .79 .80 .81 .82 .84

.3611 .4167 .4722 .5278 .5833 .6389

Observation .86 .89 .91 .95 1.01 1.04

.6944 .7500 .8056 .8611 .9167 .9722

Construct a Weibull probability plot and comment.

101. Construct a normal probability plot for the

escape time data given in Exercise 33 of Chapter 1.

Does it appear plausible that escape time has a

normal distribution? Explain.

102. The article “The Load-Life Relationship for M50

Bearings with Silicon Nitride Ceramic Balls”

(Lubricat. Engrg., 1984: 153–159) reports the

accompanying data on bearing load life (million

revs.) for bearings tested at a 6.45-kN load.

47.1 68.1 68.1 90.8 103.6 106.0 115.0

126.0 146.6 229.0 240.0 240.0 278.0 278.0

289.0 289.0 367.0 385.9 392.0 505.0

a. Construct a normal probability plot. Is nor-

mality plausible?

b. Construct a Weibull probability plot. Is the

Weibull distribution family plausible?

103. Construct a probability plot that will allow you to

assess the plausibility of the lognormal distribu-

tion as a model for the rainfall data of Exercise 80

in Chapter 1.

104. The accompanying observations are precipita-

tion values during March over a 30-year period

in Minneapolis–St. Paul.

.77 1.20 3.00 1.62 2.81 2.48

1.74 .47 3.09 1.31 1.87 .96

.81 1.43 1.51 .32 1.18 1.89

1.20 3.37 2.10 .59 1.35 .90

1.95 2.20 .52 .81 4.75 2.05

a. Construct and interpret a normal probability

plot for this data set.

4.6 Probability Plots 219

b. Calculate the square root of each value and

then construct a normal probability plot based

on this transformed data. Does it seem plau-

sible that the square root of precipitation is

normally distributed?

c. Repeat part (b) after transforming by cube

roots.

105. Use a statistical software package to construct

a normal probability plot of the shower-flow

rate data given in Exercise 13 of Chapter 1, and

comment.

106. Let the ordered sample observations be denoted

by y

, y

, ..., y

being the smallest and y

the

largest). Our suggested check for normality is to

plot the (F

1

[(i  .5)/n], y

) pairs. Suppose we

believe that the observations come from a distri-

bution with mean 0, and let w

, ..., w

the ordered absolute values of the x

’s.Ahalf-

normal plot is a probability plot of the w

’s.

More specifically, since P(|Z|  w) ¼ P(w 

Z  w) ¼ 2F(w) 1, a half-normal plot is a plot

of the (F

1

[(p

+ 1)/2], w

) pairs, where p

¼ (i

 .5)/n. The virtue of this plot is that small or

large outliers in the original sample will now

appear only at the upper end of the plot rather

than at both ends. Construct a half-normal plot

for the following sample of measurement errors,

and comment: 3.78, 1.27, 1.44, .39, 12.38,

43.40, 1.15, 3.96, 2.34, 30.84.

107. The following failure time observations (1,000’s

of hours) resulted from accelerated life testing of

16 integrated circuit chips of a certain type:

82.8 11.6 359.5 502.5 307.8 179.7

242.0 26.5 244.8 304.3 379.1 212.6

229.9 558.9 366.7 204.6

Use the corresponding percentiles of the exponen-

tial distribution with l ¼ 1 to construct a proba-

bility plot. Then explain why the plot assesses the

plausibility of the sample having been generated

from any exponential distribution.

4.7

Transformations of a Random Variable

Often we need to deal with a transformation Y ¼ g(X) of the random variable X.

Here g(X) could be a simple change of time scale. If X is in hours and Y is in

minutes, then Y ¼ 60X. What happens to the pdf when we do this? Can we get the

pdf of Y from the pdf of X? Consider first a simple exam ple.

Example 4.38 The interval X in minutes between calls to a 911 center is exponentially distributed

with mean 2 min, so has pdf f

ðxÞ¼

x=2

for x > 0. Can we find the pdf of

Y ¼ 60X,soY is the number of seconds? In order to get the pdf, we first find the

cdf. The cdf of Y is

ðyÞ¼PðY  yÞ¼Pð60X  yÞ¼PðX  y=60Þ¼F

ðy=60Þ

y=60

u=2

du ¼ 1  e

y=120

Differentiating this with respect to y gives f

(y) ¼ (1/120)e

y/120

for y > 0.

The distribution of Y is exponential with mean 120 s (2 min).

Sometimes it isn’t possible to evaluate the cdf in closed form. Could we still

find the pdf of Y without evaluating the integral? Yes, and it involves differentiating

the integral with respect to the upper limit of integration. The rule, which is

sometimes presented as part of the Fundamental Theorem of Calculus, is

hðuÞdu ¼ hðxÞ:

220 CHAPTER 4 Continuous Random Variables and Probability Distributions

Now, setting x ¼ y/60 and using the chain rule, we get the pdf using the rule for

differentiating integrals:

ðyÞ¼

ðxÞ



x¼y=60

ðxÞ



x¼y=60

u=2

du ¼



x=2

120

y=120

y > 0:

Although it is useful to have the integral expression of the cdf here for clarity, it is

not necessary. A more abstract approach is just to use differentiation of the cdf to

get the pdf. That is, with x ¼ y/60 and again using the chain rule,

ðyÞ¼

ðxÞ



x¼y=60

ðxÞ¼

ðxÞ



x=2

120

y=120

y > 0:

Is it plausible that, if X ~ exponential with mean 2, then 60X ~ exponential

with mean 120? In terms of time between calls, if it is exponential with mean 2 min,

then this should be the same as exponential with mean 120 s. Generalizing, there is

nothing special here about 2 and 60, so it should be clear that if we multiply an

exponential random variable with mean m by a positive constant c we get another

exponential random variable with mean cm. This is also easily verified using a

moment generating function argument. ■

The method illustrated above can be applied to other transformations.

THEOREM

Let X have pdf f

(x) and let Y ¼ g(X), where g is monotonic (either strictly

increasing or strictly decreasing) so it has an inverse function X ¼ h(Y).

Assume that h has a derivative h

(y). Then f

(y) ¼ f

(h(y)) |h

(y)|

Proof Here is the proof assuming that g is monotonically increasing. The proof

for g monotonically decreasing is similar. We follow the last method in Example

4.38. First find the cdf.

ðyÞ¼PðY  yÞ¼PgðXÞy½¼PX hðyÞ½¼F

½hðyÞ:

Now differentiate the cdf, letting x ¼ h(y ).

ðyÞ¼

½hðyÞ ¼

ðxÞ¼h

ðyÞf

ðxÞ¼h

ðyÞf

½hðyÞ

The absolute value is needed on the derivative only in the other case where g is

decreasing. The set of possible values for Y is obtained by applying g to the set of

possible values for X. ■

4.7 Transformations of a Random Variable 221

A heuristic view of the theorem (and a good way to remember it) is to say that

ðxÞdx ¼ f

ðyÞdy

ðyÞ¼f

ðxÞ

¼ f

ðhðyÞÞh

ðyÞ

Of course, because the pdf’s must be nonnegative, the absolute value is required on

the derivative if it is negative.

Sometimes it is easier to find the derivative of g than to find the derivative of

h. In this case, remember that

Example 4.39 Let’s apply the theorem to the situation introduced in Example 4.38. There

Y ¼ g(X) ¼ 60X and X ¼ h(Y) ¼ Y/60.

ðyÞ¼f

½hðyÞ h

ðyÞ

x=2

120

y=120

y > 0

■

Example 4.40 Here is an even simpler example. Suppose the arrival time of a delivery truck will

be somewhere between noon and 2:00. We model this with a random variable X that

is uniform on [0, 2], so f

ðxÞ¼

on that interval. Let Y be the time in minutes,

starting at noon, Y ¼ g(X) ¼ 60X so X ¼ h(Y) ¼ Y/60.

ðyÞ¼f

½hðyÞ h

ðyÞ



120

0 < y < 120

Is this intuitively reasonable? Beginning with a uniform distribution on [0, 2],

we multiply it by 60, and this spreads it out over the interval [0, 120]. Notice that

the pdf is divided by 60, not multiplied by 60. Because the distribution is spread

over a wider interval, the density curve must be lower if the total area under the

curve is to be 1.

■

Example 4.41 This being a special day (an A in statistics!), you plan to buy a steak (substitute five

Portobello mushrooms if you are a vegetarian) for dinner. The weight X of the steak

is normally distributed with mean m and variance s

. The steak costs a dollars per

pound, and your other purchases total b dollars. Let Y be the total bill at the cash

X  Nðm; s

Þ and Y ¼ aX + b, where a 6¼ 0. In our example a is positive, but we

will do a more general calculation that allows negative a. Then the inverse function

is x ¼ h(y) ¼ (y  b)/a.

ðyÞ¼f

½hðyÞjh

ðyÞj ¼

ﬃﬃﬃﬃﬃﬃ

 f½ðybÞ=amg=sðÞ

jaj

ﬃﬃﬃﬃﬃﬃ

jajs

½ðybamÞ=ðajsjÞ

222 CHAPTER 4 Continuous Random Variables and Probability Distributions

Thus, Y is normally distributed with mean am + b and standard deviation |a |s.

The mean and standard deviation did not require the new theory of this section

because we could have just calculated E(Y) ¼ E(aX + b) ¼ am + b, V(Y) ¼

V(aX + b) ¼ a

, and therefore s

¼ |a|s.

As a special case, take Y ¼ (X  m)/s,sob ¼m/s and a ¼ 1/s. Then Y is

normal with mean value am + b ¼ m/s  m/s ¼ 0 and standard deviation |a|s ¼

|1/s| s ¼ 1. Thus the transformation Y ¼ (X  m)/s creates a new normal random

variable with mean 0 and standard deviation 1. That is, Y is standard normal. This is

the first proposition in Section 4.3.

On the other hand, suppose that X is already standard normal, X ~ N(0, 1).

If we let Y ¼ m + sX, then a ¼ s and b ¼ m,soY will have mean 0 ·s + m ¼ m,

and standard deviation |

a|·1¼ s. If we start with a standard normal, we can obtain

any other normal distribution by means of a linear transformation.

■

Example 4.42 Here we want to see what can be done with the simple uniform distribution. Let

X have uniform distribution on [0, 1], so f

(x) ¼ 1 for 0 < x < 1. We want to

transform X so that g(X) ¼ Y has a specified distribution. Let’s specify that f

(y)

¼ y/2 for 0 < y < 2. Integrating this, we get the cdf F

(y) ¼ y

/4, 0 < y < 2.

The trick is to set this equal to the inverse function h(y). That is, x ¼ h(y) ¼ y

/4.

Inverting this (solving for y, and using the positive root), we get

y ¼ gðxÞ¼F

1

ðxÞ¼

ﬃﬃﬃﬃﬃ

¼ 2

ﬃﬃﬃ

. Let’s apply the foregoing theorem to see if

Y ¼ gðXÞ¼2

ﬃﬃﬃﬃ

has the desired pdf:

ðyÞ¼f

½hðyÞjh

ðyÞj ¼ ð1Þ

0 < y < 2

A graphical representation may help in understanding why the transform

Y ¼ 2

ﬃﬃﬃﬃ

yields f

(y) ¼ y/2 if X is uniform on [0, 1]. Figure 4.37(a) shows the

uniform distribution with [0, 1] partitioned into ten subi ntervals. In Figure 4.37(b)

the endpoints of these intervals are shown after transforming according to y ¼ 2

ﬃﬃﬃ

The heights of the rectangles are arranged so each rectangle still has area .1, and

therefore the probability in each interval is preserved. Notice the close fit of the

dashed line, which has the equatio n f

(y) ¼ y/2.

1.0

pdf of X

1.0

pdf of Y

.5 1.0 1.5 2.0

Figure 4.37 The effect on the pdf if X is uniform on [0, 1] and Y ¼ 2

ﬃﬃﬃﬃ

4.7 Transformations of a Random Variable 223

Can the method be generalized to produce a random variable with any desired

pdf? Let the pdf f

(y) be specified along with the corresponding cdf F

(y). Define g

to be the inverse function of F

,soh(y) ¼ F

(y). If X is uniformly distributed on

[0, 1], then using the theorem, the pdf of Y ¼ gðXÞ¼F

1

ðX Þ is

½hðyÞ h

ðyÞ

¼ð1Þf

ðyÞ¼f

ðyÞ

This says that you can build any random variable you want from humble uniform

variates. Uniformly distributed random variables are available from almost any

calculator or computer language, so our method enables you to produce values of

any continuou s random variable, as long as you know its cdf.

To get a sequence of random values with the pdf f

(y) ¼ y/2, 0 < y < 2, start

with a sequence of random values from the uniform distribution on [0, 1]: .529, .043,

.294, .... Then take Y ¼ gðXÞ¼F

1

ðX Þ¼2

ﬃﬃﬃﬃ

to get 1.455, .415, 1.084, .... ■

Can the process be reversed, so we start with any continuous random variable

and transform to a uniform variable? Let X have pdf f

(x) and cdf F

(x). Transform

X to Y ¼ g(X) ¼ F

(X), so g is F

. The inverse functi on of g ¼ F

is h. Again

apply the theorem to show that Y is uniform:

ðyÞ¼f

hðyÞ½h

ðyÞ

¼ f

ðxÞ=f

ðxÞ¼10 x  1

This works because h and F are inverse functions, so their derivatives are reciprocals.

Example 4.43 To illustrate the transformation to uniformity, assume that X has pdf f

(x) ¼

x/2, 0 < x < 2. Integrating this, we get the cdf F

(x) ¼ x

/4, 0 < x < 2. Let

Y ¼ g(X) ¼ F

(X) ¼ X

/4. Then the inverse function is hðyÞ¼

ﬃﬃﬃﬃﬃ

¼ 2

ﬃﬃﬃ

and

ðyÞ¼f

ðxÞ



ðxÞ



x=2

¼ 10< y < 1

■

The foregoing theorem requires a monotonic transformation, but there are

important applications in which the transformation is not monotonic. Nevertheless,

it may be possible to use the theorem anyway with a little trickery.

Example 4.44 In this example, we start with a standard normal random variable X, and we transform

to Y ¼ X

. Of course, this is not monotonic over the interval for X,(1, 1).

However, consider the transformation U ¼ |X|. Can we obtain the pdf of this intui-

tively, without recourse to any theory? Because X has a symmetric distribution, the

pdf of U is f

(u) ¼ f

(u)+f

(u) ¼ 2 f

(u). Don’t despair if this is not intuitively

clear, because we’ll verify it shortly. For the time being, assume it to be true. Then

Y ¼ X

¼ |X|

¼ U

, and the transformation in terms of U is monotonic because

its set of possible values is [0, 1). Thus we can use the theorem with h(y) ¼ y

ðyÞ¼f

½hðyÞh

ðyÞ¼2f

½hðyÞh

ðyÞ

ﬃﬃﬃﬃﬃﬃ

:5ðy

ð:5y

:5

Þ¼

ﬃﬃﬃﬃﬃﬃﬃﬃ

2py

y=2

y>0

This is the chi-squared distribution (with 1 degree of freedom) introduced in Sec-

tion 4.4. The squares of normal random variables are important because the sample

variance is built from squares, and we will need the distribution of the variance. The

variance for normal data is proportional to a chi-squared rv.

224 CHAPTER 4 Continuous Random Variables and Probability Distributions

You were asked to believe intuitively that f

(u) ¼ 2 f

(u) on an intuitive

basis. Here is a little derivation that works as long as f

(x) is an even function, [i.e.

(x) ¼ f

(x)]. If u > 0,

ðuÞ¼PðU  uÞ¼Pð XjjuÞ¼Pðu  X  uÞ¼2P ð0  X  uÞ

¼ 2 F

ðuÞF

ð0Þ½:

Differentiating this with respect to u gives f

(u) ¼ 2 f

(u). ■

Example 4.45 Sometimes the theorem cannot be used at all, and you need to use the cdf. Let

(x) ¼ (x + 1)/8, 1 < x < 3, and Y ¼ X

. The transformation is not monotonic

and f

(x) is not an even function. Possible values of Y are {y:0 y  9}.

Considering first 0  y  1,

ðyÞ¼PðY  yÞ¼PðX

 yÞ¼Pð

ﬃﬃﬃ

 X 

ﬃﬃﬃ

Þ¼

ﬃﬃ



ﬃﬃ

u þ 1

du ¼

ﬃﬃﬃ

Then, on the other subinterval, 1 < y  9,

ðyÞ¼PðY  yÞ¼PðX

 yÞ¼P 

ﬃﬃﬃ

 X 

ﬃﬃﬃ



¼ P 1  X 

ﬃﬃﬃ



ﬃﬃ

1

u þ 1

du ¼ð1 þ y þ 2

ﬃﬃﬃ

Þ=16

Differentiating, we get

ðyÞ¼

ﬃﬃﬃ

0 < y < 1

y þ

ﬃﬃﬃ

16y

1 < y < 9

0 otherwise

■

If X is discrete, what happens to the pmf when we do a monotonic trans-

formation?

Example 4.46 Let X have the geometric distribution, with pmf p

(x) ¼ (1  p)

p, x ¼ 0, 1, 2, ...,

and define Y ¼ X/3. Then the pmf of Y is

ðyÞ¼PðY ¼ yÞ¼P

¼ y



¼ PðX ¼ 3yÞ¼p

ð3yÞ¼ð1  pÞ

y ¼ 0; 1=3; 2=3; :::

Notice that there is no need for a derivative in finding the pmf for transfor-

mations of discrete random variables.

To put this on a more general basis in the discrete case, if Y ¼ g(X) with

inverse X ¼ h(Y), then

ðyÞ¼PðY ¼ yÞ¼PgðXÞ¼hðy Þ½¼PX¼ hðyÞ½¼p

hðyÞ½;

and the set of possible values of Y is obtained by applying g to the set of possible

values of X.

■

4.7 Transformations of a Random Variable 225

Exercises Section 4.7 (108–126)

108. Relative to the winning time, the time X of

another runner in a 10 km race has pdf f

(x) ¼

2/x

, x > 1. The reciprocal Y ¼ 1/X represents

the ratio of the time for the winner divided by the

time of the other runner. Find the pdf of Y.

Explain why Y also represents the speed of the

other runner relative to the winner.

109. If X has the pdf f

(x) ¼ 2x,0< x < 1, find the

pdf of Y ¼ 1/X. The distribution of Y is a special

case of the Pareto distribution (see Exercise 10).

110. Let X have the pdf f

(x) ¼ 2/x

, x > 1. Find the

pdf of Y ¼

ﬃﬃﬃﬃ

111. Let X have the chi-squared distribution with

2 degree of freedom, so f

ðxÞ¼

x=2

; x>0.

Find the pdf of Y ¼

ﬃﬃﬃﬃ

. Suppose you choose

a point in two dimensions randomly, with the

horizontal and vertical coordinates chosen inde-

pendently from the standard normal distribution.

Then X has the distribution of the squared dis-

tance from the origin and Y has the distribution

of the distance from the origin. Because Y is

the length of a vector with normal components,

there are lots of applications in physics, and its

distribution has the name Rayleigh.

112. If X is distributed as N(m, s

), find the pdf of

Y ¼ e

. The distribution of Y is lognormal, as

discussed in Section 4.5.

113. If the side of a square X is random with the pdf

(x) ¼ x/8, 0 < x < 4, and Y is the area of the

square, find the pdf of Y.

114. Let X have the uniform distribution on [0, 1].

Find the pdf of Y ¼ln(X).

115. Let X be uniformly distributed on [0, 1]. Find the

pdf of Y ¼ tan[p(X  .5)]. The random variable

Y has the Cauchy distribution after the famous

mathematician.

116. If X is uniformly distributed on [0, 1], find a

linear transformation Y ¼ cX + d such that Y is

uniformly distributed on [a, b], where a and b are

any two numbers such that a < b. Is there

another solution? Explain.

117. If X has the pdf f

(x) ¼ x/8, 0 < x < 4, find a

transformation Y ¼ g(X) such that Y is uniformly

distributed on [0, 1].

118. If X is uniformly distributed on [1, 1], find the

pdf of Y ¼ |X|.

119. If X is uniformly distributed on [1, 1], find the

pdf of Y ¼ X

120. Ann is expected at 7:00 pm after an all-day drive.

She may be as much as 1 h early or as much as

3 h late. Assuming that her arrival time X is

uniformly distributed over that interval, find the

pdf of |X  7|, the unsigned difference between

her actual and predicted arrival times.

121. If X is uniformly distributed on [1, 3], find the

pdf of Y ¼ X

122. If X is distributed as N(0, 1), find the pdf of |X|.

123. A circular target has radius 1 ft. Assume that you

hit the target (we shall ignore misses) and that the

probability of hitting any region of the target is

proportional to the region’s area. If you hit the

target at a distance Y from the center, then let

X ¼ p Y

be the corresponding area. Show that

(a) X is uniformly distributed on [0, p]. [Hint:

Show that F

(x) ¼ P(X  x) ¼ x/p.]

(b) Y has pdf f

(y) ¼ 2y,0< y < 1.

124. In Exercise 123 suppose instead that Y is

uniformly distributed on [0,1]. Find the pdf of

X ¼ p Y

. Geometrically speaking, why should X

have a pdf that is unbounded near 0?

125. Let X have the geometric distribution with pmf

(x) ¼ (1  p)

p, x ¼ 0, 1, 2, .... Find the pmf

of Y ¼ X + 1. The resulting distribution is also

referred to as geometric (see Example 3.10).

126. Let X have binomial distribution with n ¼ 1, (a

Bernoulli rv). That is, X has pmf b(x;1,p).

If Y ¼ 2X  1, find the pmf of Y.

226

CHAPTER 4 Continuous Random Variables and Probability Distributions

Supplementary Exercises (127–155)

127. Let X ¼ the time it takes a read/write head to

locate a desired record on a computer disk mem-

ory device once the head has been positioned over

the correct track. If the disks rotate once every

25 ms, a reasonable assumption is that X is uni-

formly distributed on the interval [0, 25].

a. Compute P(10  X  20).

b. Compute P(X  10).

c. Obtain the cdf F(X).

d. Compute E(X) and s

128. A 12-in. bar clamped at both ends is subjected to

an increasing amount of stress until it snaps. Let

Y ¼ the distance from the left end at which the

break occurs. Suppose Y has pdf

f ðyÞ¼

1 



0  y  12

0 otherwise

Compute the following:

a. The cdf of Y, and graph it.

b. P(Y  4), P(Y > 6), and P(4  Y  6).

c. E(Y), E(Y

), and V(Y).

d. The probability that the break point occurs

more than 2 in. from the expected break

point.

e. The expected length of the shorter segment

when the break occurs.

129. Let X denote the time to failure (in years) of a

hydraulic component. Suppose the pdf of X is

f(x) ¼ 32/(x +4)

for x > 0.

a. Verify that f(x) is a legitimate pdf.

b. Determine the cdf.

c. Use the result of part (b) to calculate the

probability that time to failure is between

2 and 5 years.

d. What is the expected time to failure?

e. If the component has a salvage value equal to

100/(4 + x) when its time to failure is x, what

is the expected salvage value?

130. The completion time X for a task has cdf F(x)

given by

0 x < 0

0  x < 1

1 

 x







1  x 

1 x 

a. Obtain the pdf f(x) and sketch its graph.

b. Compute P(.5  X  2).

c. Compute E( X).

131. The breakdown voltage of a randomly chosen

diode of a certain type is known to be normally

distributed with mean value 40 V and standard

deviation 1.5 V.

a. What is the probability that the voltage of a

single diode is between 39 and 42?

b. What value is such that only 15% of all

diodes have voltages exceeding that value?

c. If four diodes are independently selected,

what is the probability that at least one has a

voltage exceeding 42?

132. The article “Computer Assisted Net Weight

Control” (Qual. Prog., 1983: 22–25) suggests a

normal distribution with mean 137.2 oz and stan-

dard deviation 1.6 oz, for the actual contents of

jars of a certain type. The stated contents was

135 oz.

a. What is the probability that a single jar con-

tains more than the stated contents?

b. Among ten randomly selected jars, what is

the probability that at least eight contain