The variance of h(X) is the expected value of the squared difference between h(X) and its expected value:
(3.13) When , a linear function,
Substituting this into (3.13) gives a simple relationship between V[h(X)] and V(X):
h(x)2 E[h(X)]5ax1 b2 (am1 b)5 a(x2m) h(X)5aX1b
V[h(X)]5sh(X)2 5
g
D5h(x)2E[h(X)]62
#
p(x) 5E(X2)2 2m
#
m1m25 E(X2)2 m2 s25
g
D
x2
#
p(x)22m
# g
D
x
#
p(x)1m2
g
D
p(x) x22 2mx1m2 g
(x2 m)2 s2511.352(2.85)25 3.2275
E(X2)5
g
6x51
x2
#
p(x)5(12)(.30)1 (22)(.25)1c 1(62)(.15)5 11.35 m5 2.85 p(6)5 .15
p(5)5.10 p(4) 5.05
p(3)5 .15
p(2)5 .25 p(1)5 .30
In particular,
(3.14) saX5 uau
#
sX, sX1b5sX V(aX1 b)5 saX1b2 5a2
#
sX
2 and saX1b5 uau
#
sx
Example 3.26
The absolute value is necessary because a might be negative, yet a standard deviation cannot be. Usually multiplication by acorresponds to a change in the unit of measurement (e.g., kg to lb or dollars to euros). According to the first relation in (3.14), the sd in the new unit is the original sd multiplied by the conversion factor.
The second relation says that adding or subtracting a constant does not impact vari- ability; it just rigidly shifts the distribution to the right or left.
In the computer sales scenario of Example 3.23, and
so . The profit function then has variance
and standard deviation 800. ■ (800)2
#
V(X)5(640,000)(1)5 640,000
h(X)5800X2900 V(X)552(2)25 1
E(X2)5(0)2(.1)1(1)2(.2)1 (2)2(.3)1 (3)2(.4)5 5 E(X)5 2
EXERCISES Section 3.3 (29–45)
29. The pmf of the amount of memory X(GB) in a purchased flash drive was given in Example 3.13 as
a. Compute E(X), E(X2), and V(X).
b. If the price of a freezer having capacity Xcubic feet is , what is the expected price paid by the next customer to buy a freezer?
c. What is the variance of the price paid by the next customer?
d. Suppose that although the rated capacity of a freezer is X,the actual capacity is . What is the expected actual capacity of the freezer purchased by the next customer?
33. Let Xbe a Bernoulli rv with pmf as in Example 3.18.
a. Compute E(X2).
b. Show that .
c. Compute E(X79).
34. Suppose that the number of plants of a particular type found in a rectangular sampling region (called a quadrat by ecolo- gists) in a certain geographic area is an rv Xwith pmf
Is E(X) finite? Justify your answer (this is another distribu- tion that statisticians would call heavy-tailed).
35. A small market orders copies of a certain magazine for its magazine rack each week. Let for the maga- zine, with pmf
X5demand p(x)5 ec/x3 x51, 2, 3, . . .
0 otherwise V(X)5p(12p)
h(X)5X2.01X2 25X28.5 25X28.5
x 1 2 4 8 16
p(x) .05 .10 .35 .40 .10
y 0 1 2 3
p(y) .60 .25 .10 .05
x 1 2 3 4 5 6
p(x) 151 152 153 154 153 152
x 13.5 15.9 19.1
p(x) .2 .5 .3
Compute the following:
a. E(X)
b. V(X) directly from the definition c. The standard deviation of X d.V(X) using the shortcut formula
30. An individual who has automobile insurance from a certain company is randomly selected. Let Ybe the number of mov- ing violations for which the individual was cited during the last 3 years. The pmf of Yis
Suppose the store owner actually pays $2.00 for each copy of the magazine and the price to customers is $4.00. If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint:For both three and four copies ordered, express net revenue as a func- tion of demand X,and then compute the expected revenue.]
a. Compute E(Y).
b. Suppose an individual with Y violations incurs a sur- charge of $100Y2. Calculate the expected amount of the surcharge.
31. Refer to Exercise 12 and calculate V(Y) and sY. Then deter- mine the probability that Yis within 1 standard deviation of its mean value.
32. An appliance dealer sells three different models of upright freezers having 13.5, 15.9, and 19.1 cubic feet of storage space, respectively. Let of storage space purchased by the next customer to buy a freezer. Suppose that Xhas pmf
X5the amount
36. Let Xbe the damage incurred (in $) in a certain type of acci- dent during a given year. Possible X values are 0, 1000, 5000, and 10000, with probabilities .8, .1, .08, and .02, respectively. A particular company offers a $500 deductible policy. If the company wishes its expected profit to be $100, what premium amount should it charge?
37. The ncandidates for a job have been ranked 1, 2, 3, . . . , n.
Let of a randomly selected candidate, so that Xhas pmf
(this is called the discrete uniform distribution). Compute E(X) and V(X) using the shortcut formula. [Hint:The sum of the first npositive integers is , whereas the
sum of their squares is .]
38. Let when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or dollars, would you accept the guaranteed amount or would you gamble? [Note:It is not generally true that .]
39. A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let of batches ordered by a ran- domly chosen customer, and suppose that Xhas pmf
X5the number 1/E(X)5E(1/X) h(X)51/X
X5the outcome
n(n11)(2n11)/6 n(n11)/2 p(x)5 e1/n x51, 2, 3, . . . , n
0 otherwise X5the rank
40. a.Draw a line graph of the pmf of Xin Exercise 35. Then determine the pmf of and draw its line graph. From these two pictures, what can you say about V(X) and
?
b. Use the proposition involving to establish a general relationship between V(X) and .
41. Use the definition in Expression (3.13) to prove that
. [Hint: With ,
where .]
42. Suppose and . What is
a. E(X2)? [Hint:
]?
b. V(X)?
c. The general relationship among the quantities E(X), , and V(X)?
43. Write a general rule for where c is a constant.
What happens when you let , the expected value of X?
44. A result called Chebyshev’s inequalitystates that for any probability distribution of an rv Xand any number kthat is
at least 1, . In words, the proba-
bility that the value of Xlies at least kstandard deviations from its mean is at most 1/k2.
a. What is the value of the upper bound for ? ?
? ? ?
b. Compute m and s for the distribution of Exercise 13.
Then evaluate for the values of k
given in part (a). What does this suggest about the upper bound relative to the corresponding probability?
c. Let Xhave possible values , 0, and 1, with probabilities , , and , respectively. What is , and how does it compare to the corresponding bound?
d.Give a distribution for which .
45. If , a#X#b show that a#E(X)#b.
P(uX2mu $5s)5.04 P(uX2mu $3s) 1
18 8 9 1 18
21 P(uX2mu $ks) k510
k55 k54
k53 k52 P(uX2mu $ks)#1/k2
c5m E(X2c) E[X(X21)]
E(X2)2E(X)
E[X(X21)]5E[X22X]5 E[X(X21)]527.5 E(X)55
m5E(X) E[h(X)]5am1b
h(X)5aX1b V(aX1b)5a2
#
sX 2
V(2X) V(aX1b) V(2X)
2X
x 1 2 3 4
p(x) .2 .4 .3 .1
Compute E(X) and V(X). Then compute the expected num- ber of pounds left after the next customer’s order is shipped and the variance of the number of pounds left. [Hint:The number of pounds left is a linear function of X.]
3.4 The Binomial Probability Distribution
There are many experiments that conform either exactly or approximately to the fol- lowing list of requirements:
1. The experiment consists of a sequence of nsmaller experiments called trials, where nis fixed in advance of the experiment.
2. Each trial can result in one of the same two possible outcomes (dichotomous trials), which we generically denote by success (S) and failure (F).
3. The trials are independent, so that the outcome on any particular trial does not influence the outcome on any other trial.
4. The probability of success P(S) is constant from trial to trial; we denote this probability by p.
DEFINITION An experiment for which Conditions 1–4 are satisfied is called a binomial experiment.
Example 3.29 Example 3.28
Example 3.27 The same coin is tossed successively and independently ntimes. We arbitrarily use Sto denote the outcome H(heads) and Fto denote the outcome T(tails). Then this experiment satisfies Conditions 1–4. Tossing a thumbtack n times, with and , also results in a binomial experiment. ■ Many experiments involve a sequence of independent trials for which there are more than two possible outcomes on any one trial. A binomial experiment can then be created by dividing the possible outcomes into two groups.
The color of pea seeds is determined by a single genetic locus. If the two alleles at this locus are AA or Aa (the genotype), then the pea will be yellow (the pheno- type), and if the allele is aa, the pea will be green. Suppose we pair off 20 Aa seeds and cross the two seeds in each of the ten pairs to obtain ten new genotypes. Call each new genotype a success Sif it is aa and a failure otherwise. Then with this identification of S and F, the experiment is binomial with and . If each member of the pair is equally likely to contribute a or
A, then . ■
Suppose a certain city has 50 licensed restaurants, of which 15 currently have at least one serious health code violation and the other 35 have no serious violations. There are five inspectors, each of whom will inspect one restaurant during the coming week. The name of each restaurant is written on a different slip of paper, and after the slips are thoroughly mixed, each inspector in turn draws one of the slips without replacement. Label the ith trial as a success if the ith restaurant selected
has no serious violations. Then
and
Similarly, it can be shown that for . However,
whereas
The experiment is not binomial because the trials are not independent. In gen- eral, if sampling is without replacement, the experiment will not yield independent trials. If each slip had been replaced after being drawn, then trials would have been independent, but this might have resulted in the same restaurant being inspected by
more than one inspector. ■
P(S on fifth trialuFFFF)535 465 .76 P(S on fifth trialuSSSS)5 31
465.67 i 53, 4, 5 P(S on ith trial)5.70
534 49
#
35501 35 49
#
15505 35 50a34
491 15 49b 535
505 .70 1P(second S | first F)P(first F)
5P(second S | first S)P(first S) P(S on second trial)5P(SS)1P(FS)
P(S on first trial)535 505 .70 (i51, . . . , 5)
p5P(a)
#
P(a)5 Q1 2RQ1
2R5 14 p5 P(aa genotype)
n510 F5 point down
S5 point up
DEFINITION RULE
A certain state has 500,000 licensed drivers, of whom 400,000 are insured. A sam- ple of 10 drivers is chosen without replacement. The ith trial is labeled Sif the ith driver chosen is insured. Although this situation would seem identical to that of Example 3.29, the important difference is that the size of the population being sam- pled is very large relative to the sample size. In this case
and
These calculations suggest that although the trials are not exactly independent, the conditional probabilities differ so slightly from one another that for practical purposes the trials can be regarded as independent with constant . Thus, to a very good approximation, the experiment is binomial with and . ■
We will use the following rule of thumb in deciding whether a “without- replacement” experiment can be treated as a binomial experiment.
p5.8 n510
P(S)5 .8 P(S on 10 | S on first 9)5399,991
499,9915.799996<.80000 P(S on 2 | S on 1)5 399,999
499,9995 .80000
Consider sampling without replacement from a dichotomous population of size N.If the sample size (number of trials) nis at most 5% of the population size, the experiment can be analyzed as though it were exactly a binomial experiment.
By “analyzed,” we mean that probabilities based on the binomial experiment assump- tions will be quite close to the actual “without-replacement” probabilities, which are typically more difficult to calculate. In Example 3.29, , so the binomial experiment is not a good approximation, but in Example 3.30,
.