1. A mail-order computer business has six telephone lines. Let X ...

1. A mail-order computer business has six telephone lines. Let X denote
the number of lines in use at a specified time. Suppose the pmf of X
is as given in the accompanying table.
x 0 1 2 3 4 5 6
p(x) .10 .15 .20 .25 .20 .06 .04
Calculate the probability of each of the following events.
a. {At most 3 lines are in use}
Ans P [X ≤ 3] = P [X = 0] + P [X = 1] + P [X = 2] + P [X = 3] =
0.10 + 0.15 + 0.20 + 0.25 = 0.70
b. {Fewer than 3 lines are in use}
Ans P [X < 3] = P [X = 0] + P [X = 1] + P [X = 2] = 0.10 +
0.15 + 0.20 = 0.45
c. {At least 3 lines are in use}
Ans P [X ≥ 3] = 1 − P [X < 3] = 1 − 0.45 = 0.55
d. {Between 2 and 5 lines, inclusive, are in use}
Ans P [2 ≤ X ≤ 5] = P [X = 2] + P [X = 3] + P [X = 4] + P [X =
5] = .20 + 0.25 + 0.20 + 0.06 = 0.71
e. {Between 2 and 4 lines, inclusive, are not in use}
Ans P [X = 2]+P [X = 3]+P [X = 4]) = .20+0.25+0.20 = 0.65
f. {At least 4 lines are not in use}
Ans P [X ≤ 2] = P [X = 0] + P [X = 1] + P [X = 2] = 0.45
2. An insurance company offers its policyholders a number of different
premium payment options. For a randomly selected policyholder, let
X = the number of months between successive payments. The cdf of
X is as follows:
⎧
⎪⎨
F (x) =
a. What is the pmf of X?
Ans
⎪⎩
⎪⎨
f(x) =
0 x < 1
.30 1 < x < 3
.40 3 < x < 4
.45 4 < x < 6
.60 6 < x < 12
1 12 ≤ x
1
⎧
⎪⎩
0 x = 0
.30 x = 1
.10 x = 3
.05 x = 4
.15 x = 6
0.4 x = 12

. Using just the cdf, compute P (3 < X < 6) and P (4 < X).
Ans
P (3 < X < 6) = P [X = 3, 4, 6] = 0.1 + 0.05 + 0.25 = 0.3
P (4 < X) = P [X = 6, 12] = 0.55.
3. 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 X = the amount of storage space purchased by the next customer
to buy a freezer. Suppose that X has probability mass function
x 13.5 15.9 19.1
p(x) .2 .5 .3
a. Compute E(X), E(X 2 ), and V (X).
Ans
E(X) = 16.38, E(X 2 ) = 272.298, and V (X) = 3.9936
b. If the price of a freezer having capacity X cubic feet is 25X −8.5,
what is the expected price paid by the next customer to buy a
freezer?
Ans
E[25X − 8.5] = (25)E[X] − 8.5 = 401,
c. What is the variance of the price 25X − 8.5 paid by the next
customer?
Ans
V [25X − 8.5] = [(25) 2 ]V [X] = 2496
d. Suppose that while the rated capacity of a freezer is X, the actual
capacity is h(X) = X − .01X 2 . What is the expected actual
capacity of the freezer purchased by the next customer?
Ans
E[h(X)] = E[X − .01X 2 ] = E[X] − (.01)E[X 2 ] = 13.66.
4. A chemical supply company currently has in stock 100 Kg of a certain
chemical, which it sells to customers 5 Kg lots. Let X = the number
of lots ordered by a randomly chosen customer, and suppose that X
has pmf
x 1 2 3 4
p(x) .2 .4 .3 .1
Compute E(X) and V (X). Then compute the expected number of
Kg’s left after the next customer’s order is shipped, and the variance
of the number of Kg’s left.
Ans
2

E[X] = (1 × 0.2) + (2 × 0.4) + (3 × 0.3) + (4 × 0.1)
= 0.2 + 0.8 + 0.9 + 0.4 = 2.3
V [X] = E[X 2 ]−(E[X]) 2 = (1×0.2)+(4×0.4)+(9×0.3)+(16×0.1)−2.3 2
= (0.2 + 1.6 + 2.7 + 1.6) − 2.3 2 = 0.81
The required r.v. is Y = 100 − 5X, therefore,
E[Y ] = 100 − 5 × E[X] = 100 − 5 × 2.3 = 100 − 11.5 = 88.5
V [Y ] = V [100 − 5 × V [X]] = (5 2 ) × V [X] = 20.25
5. When circuit boards used in the manufacture of compact disc players
are tested, the long-run percentage of defectives is 5%. Let X = the
number of defective boards in a random sample of size n = 25, so
X ∼ BIN(25, 0.05).
(a) Determine P (X ≤ 2)
(b) Determine P (X ≥ 5)
(c) Determine P (1 ≤ X ≤ 4)
(d) What is the probability that none of the 25 boards are defective?
(e) Calculate the expected value and standard deviation of X.
(a) Determine P [X ≤ 2].
=
10
0
P [X ≤ 2] =
(0.5) 0 0.5 10 +
=
(b) Determine P [X ≥ 5].
2
x=0
10
x
10
1
(0.5) x [1 − 0.5] 10−x
(0.5) 1 0.5 9
10
+
2
10!
0!(10 − 0)! [0.000976562]
10!
+
1!(10 − 1)! [0.000976562]
10!
+
2!(10 − 2)! [0.000976252]
= (1 + 10 + 45) × 0.000976562
= 56 × 0.000976562 = 0.0546875
3
(0.5) 2 0.5 8

P [X ≤ 2] =
=
10
x=5
10
x
(0.5) x [1 − 0.5] 10−x
10!
5!(10 − 5)! [0.000976562]
10!
+
6!(10 − 6)! [0.000976562]
10!
+
7!(10 − 7)! [0.000976252]
=
10!
8!(10 − 8)! [0.000976562]
10!
+
9!(10 − 9)! [0.000976562]
10!
+
10!(10 − 10)! [0.000976252]
= (252 + 210 + 120 + 45 + 10 + 1) × 0.000976562
(c) Determine P [1 ≤ X ≤ 4].
Ans Required probability:
= 0.6230469
(1 + 10 + 45 + 120 + 210) × 0.000976562
= 386 × 0.000976562
= 0.3769531
Note that we could also have arrived at the same answer, without
computing binomial coefficients, by observing that,
P [1 ≤ X ≤ 4] = P [X = 0] + [1 − P [X ≥ 5]]
= 1 + 0.000976562 − 0.6230469 = 0.3769
(d) What is the probability that none of the 10 boards are defective?
Ans This is the probability
P [X = 0] = 0.0009765
(e) Calculate the expected value and standard deviation of X.
Ans We know that any Binomial r.v. can be defined as the
sum of independent Bernoulli trials, i.e., if X ∼ BIN(n, p), then
4

X = n
i=1 Xi, where each Xi ∼ BIN(1, p), (or Bernoulli with
parameter p - a r.v taking value 1 with probability p and 0 with
pr. 1 − p).
Therefore
=
E[X] = E[
=
n
Xi]
i=1
n
E[Xi] = np
i=1
= 10 × 0.5 = 5
V [X] = V [
=
n
Xi]
i=1
n
V [Xi]
i=1
n
p(1 − p) = np(1 − p)
i=1
= 10 × 0.5 × 0.5 = 2.5
Standard deviation is the square-root of variance, and hence equals
1.581139.
6. Show that E(X) = np when X is a binomial random variable. [HINT:
First express E(X) as a sum with lower limit x = 1. Then factor out
np, let y = x − 1 so that the sum is from y = 0 to y = n − 1, and show
that the sum equals 1.]
Ans
E[X] =
=
=
n
x=1
n
x=0
n
x
x
p x (1 − p) n−x
n n!
x
x!(n − x)! px (1 − p) n−x
x=0
n!
(x − 1)!(n − x)! px (1 − p) n−x
n−1 (n − 1!
= n
y!(n − y − 1)! py+1 (1 − p) n−y−1
y=0
= np(p + 1 − p) n−1 = np
5

7. Customers at a gas station select either regular (A), premium (B), or
diesel fuel (C). Assume that successive customers make independent
choices, with P (A) = 0.3, P (B) = 0.2, and P (C) = 0.5.
(a) Among the next 100 customers, what are the mean and variance
of the number who select regular fuel? Explain your reasoning.
Ans The no. who select regular do so with P (A) = 0.3, therefore
who do not select regular do so with 1 − P (A) = P (B) + P (C) =
0.7. Define Xi = 1 if i-th customer chooses regular, 0 otherwise.
This is therefore a Bernoulli r.v. with parameter, p = 0.3.
Therefore the total no. of customers among the 100 customers
who choose regular, is a Binomial r.v. = Xi. Therefore the
mean and variance are 100 × 0.3 = 30 and 100 × 0.3 × 0.7 = 21.
(b) Answer part (a) for the number among the 100 who select a
nondiesel fuel.
8. An airport limousine can accommodate up to four passengers on any
one trip. The company will accept a maximum of six reservations
for a trip, and a passenger must have a reservation. From previous
records, 20% of all those making reservations do not appear for the
trip. Answer the following questions, assuming independence wherever
appropriate.
(a) If six reservations are made, what is the probability that at least
one individual with a reservation cannot be accommodated on
the trip?
Ans At least one individual with a reservation cannot be accommodated
on the trip is equivalent to 5 or 6 individuals turning
up. This is a Binomial r.v., as can be seen by defining Xi = 1
if an individual turns up, and taking the total no. turning up as
Y =
i Xi. We now have to find, P [Y = 5] + P [Y = 6].
6
5
0.8 5 (1 − 0.8) 6−5 +
6
6
0.8 6 (1 − 0.8) 6−6
= 0.393216 + 0.262144 = 0.65536
(b) If six reservations are made, what is the expected number of
available places when the limousine departs?
Ans The available no. of places Z = max[4 − Y, 0]. Therefore,
P [Z = k] = P [Y = 4 − k], k = 1, . . . , 4, and P [Z = 0] = P [Y =
4] + P [Y = 5] + P [Y = 6]. The expected value is therefore,
1.P [Y = 3] + 2.P [Y = 2] + 3.P [Y = 1] + 4.P [Y = 0]
= 1.P [Y = 3] + 2.P [Y = 2] + 3.P [Y = 1] + 4.P [Y = 0]
6

= 1 × 0.081920 + 2 × 0.015360 + 3 × 0.001536 + 4 × 0.000064
= 0.117504
(c) Suppose the probability distribution of the number of reservations
made is given in the accompanying table.
Number of reservations 3 4 5 6
Probability .1 .2 .3 .4
Let X denote the number of passengers on a randomly selected
trip. Obtain the probability mass function of X.
Ans
Define the r.v. R to be the no. of reservations. First consider the
case, X = 0, this can happen even if R = 3, 4, 5, 6. Therefore
P [X = 0] = P [X = 0, R = 3]+P [X = 0, R = 4]+P [X = 0, R = 5]+P [X = 0, R = 6]
3
= (0.1) 0.8
0
0 (1 − 0.8) 3−0
4
+ (0.2) 0.8
0
0 (1 − 0.8) 4−0
5
+(0.3) 0.8
0
0 (1 − 0.8) 5−0
6
+ (0.4) 0.8
0
0 (1 − 0.8) 6−0
= 0.0012416
P [X = 1] = P [X = 1, R = 3]+P [X = 1, R = 4]+P [X = 1, R = 5]+P [X = 1, R = 6]
3
= (0.1) 0.8
1
1 (1 − 0.8) 3−1
4
+ (0.2) 0.8
1
1 (1 − 0.8) 4−1
5
+(0.3) 0.8
1
1 (1 − 0.8) 5−1
6
+ (0.4) 0.8
1
1 (1 − 0.8) 6−1
0.0172544
P [X = 2] = P [X = 2, R = 3]+P [X = 2, R = 4]+P [X = 2, R = 5]+P [X = 2, R = 6]
= 0.090624
P [X = 3] = P [X = 3, R = 3]+P [X = 3, R = 4]+P [X = 3, R = 5]+P [X = 3, R = 6]
= 0.227328
P [X = 4] = 1−P [X = 0]−P [X = 1]−P [X = 2]−P [X = 3] = 0.663552
9. The article ”Modeling Sediment and Water Column Interactions for
Hydrophobic Pollutants” (Water Research, 1984, pp. 1169-1174) suggests
the uniform distribution on the interval (7.5, 20) as a model for
depth (cm) of the bioturbation layer in sediment in a certain region.
a. What are the mean and variance of depth?
b. What is the cdf of depth?
7

c. What is the probability that observed depth is at most 10? Between
10 and 15?
d. What is the probability that the observed depth is within 1 SD
of the mean value? Within 2 SD’s?
10. Let X have the Pareto pdf given by
kθk f(x; k, θ) = xk+1 x ≥ θ
0 x < θ
a. If k > 1 , compute E(X).
b. What can you say about E(X) if k = 1 ?
c. If k > 2, show that V (X) = kθ 2 (k − 1) −2 (k − 2) −1
d. If k = 2, what can you say about V (X)?
e. What conditions on k are necessary to ensure that E(X n ) is finite?
11. If bearing diameter is normally distributed, what is the probability
that the diameter of a randomly selected bearing is
a. Within 1.5 SD’s of its mean value? (Φ(1.5)−Φ(−1.5) = 0.8663856)
b. Farther than 2.5 SD’d from its mean value? (1 − [Φ(2.5) −
Φ(−2.5)]= 1 − 0.9875807= 0.0124193)
c. Between 1 and 2 SD’s from its mean value? ([Φ(2) − Φ(1)] +
[Φ(−1) − Φ(−2)] = 0.2718102)
12. The inside diameter of a randomly selected piston ring is a r.v. with
mean value 12 cm and standard deviation 0.04 cm.
a. If ¯ X is the sample of n = 16 rings, where is the sampling distribution
of ¯ X centred, and what is the standard deviation of the
¯X distribution?
Ans ¯ X is centred at 12, V [ ¯ X] = σ 2 /n = (0.04) 2 /16 or, s.d. is
0.04/4 = 0.01.
b. Answer the questions posed in part (a) for a sample of size n = 64
rings.
Ans ¯ X is centred at 12, V [ ¯ X] = σ 2 /n = (0.04) 2 /64 or, s.d. is
0.04/8 = 0.005.
c. For which of the two random samples, the one of part (a), or the
one of part (b), is ¯ X more likely to be within 0.01 cm of 12 cm?
Explain your reasoning.
Ans part (b), because, (b)has smaller variance/s.d.
8

13. The lifetime of a certain type of battery is normally distributed with
mean value 8 hours and standard deviation 1 hour. There are four
batteries in a package. What lifetime value is such that the total
lifetime of all batteries in a package exceeds that value for only 5% of
all packages?
Ans
Individual lifetimes be denoted by Xi ∼ N(µ, σ 2 ). Then the total
lifetime of 4 batteries is given by Y = X1 + X2 + X3 + X4. Therefore,
Y ∼ N(4µ, 4σ 2 ) or Y ∼ N(4 × 8, 4 × 1 2 ) or Y ∼ N(32, 4). Therefore,
[Y − 32]/2 ∼ N(0, 1). Let z0 be the required critical point, then
z0 − 32
Φ
= 1 − 0.05 = 0.95
2
We know that Φ(1.644854) = 0.95. Therefore,
z0 − 32
2
= 1.644854 ⇒ z0 = 32 + 2 × 1.644854 = 35.28971
14. Each of 150 newly manufactured items is examined and the number
of scratches per item is recorded (the items are supposed to be free of
scratches), yielding the following data:
Scratches/item 0 1 2 3 4 5 6 7
Observed freq 18 37 42 30 13 7 2 1
Let X = the number of scratches on a randomly selected item and
assume that X has a Poisson distribution with parameter λ (P (X =
k) = exp(−λ)λ k /k!).
Find an unbiased estimator of λ and compute the estimate for the
above data. [Hint: E(X) = λ)
Ans If E[X] = λ, implies
E[ ¯ X = E[ 1
n
n
i=1
Xi] = 1
n
= [nλ]/n = λ
n
E[Xi]
Therefore ¯ X is an unbiased estimator of λ. The estimate is simply,
the mean of the above data,
= (0 × 18 + 1 × 37 + 2 × 42 + 3 × 30 + 4 × 13 + 5 × 7 + 6 × 2 + 7 × 1)
(18 + 37 + 42 + 30 + 13 + 7 + 2 + 1)
= 2.113333
9
i=1

15. A random sample of n bike helmets manufactured by a certain company
is selected. Let X = the number among the n that are flawed
and let p = P (flawed). Assume that only X is observed, rather than
the sequence of Successes and Failures. Derive the Maximum Likelihood
Estimator of p. If n = 20 and x = 3, what is the estimate? Is
this estimator unbiased?
Ans This is a Binomial model. Therefore, p.d.f. is
P [X = x] =
n
x
p x (1 − p) n−x
Because we have just a single observation, the jt. p.d.f itself is this,
and probability in this case is not 0, we can take log,
n
ln P [X = x] = ln[ ]x ln p + (n − x) ln(1 − p)
x
Maximizing w.r.t. p,
∂ ln P [X = x]
∂p
= x
p
n − x
+ (−1) = 0
1 − p
(n − x)p = x(1 − p) ⇒ np = x ⇒ ˆp = x
n
To show unbiasedness, use the fact that X is binomial,
E[X/n] = (1/n)E[X] = np/n = p
The following page contains a table of cumulative standard normal probabilities
x
Φ(x) = P [Z < x] =
For x < 0, Φ(−x) = 1 − Φ(x).
10
−∞
1 x2
− √ e 2 dx
2π

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0 0.5 0.504 0.508 0.512 0.516 0.5199 0.5239 0.5279 0.5319 0.5359
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753
0.2 0.5793 0.5832 0.5871 0.591 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.648 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.67 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.695 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.719 0.7224
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
0.7 0.758 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852
0.8 0.7881 0.791 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.834 0.8365 0.8389
1 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.877 0.879 0.881 0.883
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.898 0.8997 0.9015
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319
1.5 0.9332 0.9345 0.9357 0.937 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.975 0.9756 0.9761 0.9767
2 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817
2.1 0.9821 0.9826 0.983 0.9834 0.9838 0.9842 0.9846 0.985 0.9854 0.9857
2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.989
2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916
2.4 0.9918 0.992 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936
2.5 0.9938 0.994 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952
2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.996 0.9961 0.9962 0.9963 0.9964
2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.997 0.9971 0.9972 0.9973 0.9974
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.998 0.9981
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
3 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.999 0.999
3.1 0.999 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993
3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995
3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997
3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998
11