Problem set hso201

1. 1
Indian Institute of Technology Kanpur
Department of Humanities and Social Sciences
HSO201A Applied Probability and Statistics (2015-16-II)
Instructor: Professor Praveen Kulshreshtha
SUGGESTED SOLUTIONS TO
PRACTICE PROBLEM SET (PPS) # 3
1. Let X be a random variable with finite variance 2
. Show that for any real numbers
(constants) a and b, Var(aX + b) = a2
2
.
Let Y = aX + b. By definition,
Var(Y) = E[Y  E(Y)]2
= E(Y2
)  [E(Y)]2
= E[(aX + b)2
]  [E(aX + b)]2
= E(a2
X2
+ 2abX + b2
)  [aE(X) + b]2
= a2
E(X2
) + 2abE(X) + b2
 [a2
{E(X)}2
+ 2abE(X) + b2
]
= a2
[E(X2
)  {E(X)}2
]
= a2
Var(X) = a2
2
2. In 80% of all solar-heat installations, the utility bill is reduced by at least one-third.
Find the probability that the utility bill will be reduced by at least one-third in:
(a) five of six installations.
Let E = The event that the utility bill is reduced by at least one-third, in a solar-
heat installation. Then, clearly, P(E) = p (say) = 80% = 0.8. Let us call the
event E a “success”. Assume that the events E for successive solar-heat
installations are independent.
If the number of installations is n (positive integer), then clearly, the random
variable X, where X = No. of “successes” in n installations, follows the
Binomial distribution B(n, p). Here, n = 6 and p = 0.8 Using the p.m.f of the
Binomial(6, 0.8) distribution, we get:
P(X = 5) = (0.2) = 6(0.8)5
(0.2) = 0.393216
(b) at least five of six installations.
P(X  5) = P(X = 5) + P(X = 6) = (0.2) +
= 6(0.8)5
(0.2) + (0.8)6
= 0.393216 + 0.262144
= 0.65536
3. A shipment of 15 digital voice recorders contains 4 that are defective. If 5 of them are
randomly chosen for inspection, what is the probability that 2 of the 5 will be
defective?
Clearly, if 5 digital voice recorders, in a shipment containing only 15 digital voice
recorders, are randomly chosen for inspection, they are chosen without
replacement. A fixed number (4) of the voice recorders are ‘defective’ (of one
type), while the remaining number (11) of voice recorders are ‘non-defective’ (of
another type).

2. 2
Let X be the number of ‘defective’ voice recorders, in a randomly chosen sample
of size n, where the total number of voice recorders in N (> n), and a fixed
number K of the voice recorders are ‘defective’. Since the sampling is done
without replacement, clearly X follows the hypergeometric distribution, with
parameters (N, n, K). Here, N = 15, n = 5 and K = 4. Using the p.m.f. of the
hypergeometric distribution, we obtain:
P(X = 2) = = ] ] / ]
= ](120) = 900/2730 = 0.32967
4. Show that the Moment Generating Function (m.g.f.) for a random variable:
(a) X  B(n, p); n > 1 (integer) and 0 < p < 1 (Binomial Distribution), is given by:
MX(t) = [p + (1-p)]n
(b) X  Po();  > 0 (Poisson Distribution), is given by: MX(t) = 
(c) Y  NB(r, p); r  1 (integer) and 0 < p < 1, where Y = No. of failures before the rth
success occurs (Negative Binomial Distribution), is given by: MY(t) =
(d) Y  g(p); 0 < p < 1, where Y = No. of failures before the 1st
success occurs
(Geometric Distribution), is given by: MY(t) =
(e) X  Discrete Uniform(1, N); N > 1 (integer) (Discrete Uniform Distribution), is
given by: MX(t) =
(f) X  Uniform(a, b); a < b (Continuous Uniform Distribution), is given by:
MX(t) = , for t  0; and MX(0) = 1
(g) X  Gamma(, );  > 0 and  > 0 (Gamma Distribution), is given by:
MX(t) =

, for t < 
(h) X  2
(p); p > 0 (integer) (Chi Square Distribution), is given by:
MX(t) = , for t <
(i) X  exp();  > 0 (Exponential Distribution), is given by: MX(t) =

, for t < 
Can be shown by using the definition of MX(t) = E(etX
). You may also refer to
Casella and Berger, Chapters 2 and 3 for many of the above derivations.
5. Derive the Expectation (mean ) and Variance (2
) for a random variable X, where:
(a) X  B(n, p); n > 1 (integer) and 0 < p < 1 (Binomial Distribution)
E(X) =  = np ; Var(X) = 2
= np(1-p)
(b) X  Po();  > 0 (Poisson Distribution)
E(X) =  =  ; Var(X) = 2
= 
(c) X  NB(r, p); r  1 (integer) and 0 < p < 1 (Negative Binomial Distribution)
E(X) =  = r(1-p)/p ; Var(X) = 2
= r(1-p)/p2

3. 3
(see Casella and Berger, Chap. 3, pp. 96)
(d) X  g(p); 0 < p < 1 (Geometric Distribution)
E(X) =  = (1-p)/p ; Var(X) = 2
= (1-p)/p2
(since g(p)  NB(1, p))
(e) X  Discrete Uniform(1, N); N > 1 (integer) (Discrete Uniform Distribution)
E(X) =  = (N+1)/2 ; Var(X) = 2
= (N+1)(N-1)/12
(see Casella and Berger, Chap. 3, pp. 86)
(f) X  Hypergeometric (N, n, K); N > 1 (integer), n (integer) < N; K (integer) < N
(Hypergeometric Distribution)
E(X) =  = (KM)/N ; Var(X) = 2
= [(KM)/N][(N-M)(N-K)/N(N-1)]
(see Casella and Berger, Chap. 3, pp. 87-88)
(g) X  Uniform(a, b); a < b (Continuous Uniform Distribution)
E(X) =  = (b + a)/2 ; Var(X) = 2
= (b - a)2
/12
(see Casella and Berger, Chap. 3, pp. 99)
(h) X  Gamma(, );  > 0 and  > 0 (Gamma Distribution)
E(X) =  =  ; Var(X) = 2
= 2
(see Casella and Berger, Chap. 3, pp. 100)
(i) X  2
(p); p > 0 (integer) (Chi Square Distribution)
E(X) =  = p ; Var(X) = 2
= 2p (since 2
(p) Gamma(p/2, 2))
(j) X  exp();  > 0 (Exponential Distribution)
E(X) =  =  ; Var(X) = 2
= 2
(since exp() Gamma(1, ))
- Can be derived by using the definitions of expectation and variance. You may
also refer to Casella and Berger, Chapters 2 and 3 for many of the above
derivations.
6. A counter records an average of 1.3 gamma particles per millisecond coming from a
radioactive substance. Let X denote the random variable, which equals the count of
gamma particles during the next millisecond.
(a) Find a suitable probability distribution for X. What are the parameters of this
distribution?
Since X denotes the count of gamma particles per millisecond, X is a discrete
random variable. Moreover, the average number of gamma particles is fixed
(1.3). Therefore, we can model X as following a Poisson distribution, with
mean 1.3. Clearly, the parameter of this distribution is:  = 1.3 (E(X) = Var(X)
= )
(b) What are the mean and variance of the above distribution?
Since X  Po( =1.3), E(X) = Var(X) =  = 1.3
(c) Determine the probability that one or more gamma particles will be emitted during
the next millisecond.
P(X  1) = e-1.3
(1.3) = 0.35429

4. 4
7. In the city of Kanpur, the proportion of road sections requiring repairs in any given
year is a random variable having the beta distribution, with shape parameters  = 3
and  = 2.
(a) On average, what percentage of the road sections in Kanpur city require repairs in
any given year?
If X  Beta(, ), then E(X) = /(+). Here,  = 3 and  = 2.
Therefore, E(X) = 3/5 = 0.6
(b) Find the probability that at least half of the road sections in Kanpur city will
require repairs in any given year.
P(X  (1/2) = 0.5) = =
= = 12[{1-(0.5)3
}/3] - 12[{1-(0.5)4
}/4] = 3.5 – 2.8125 = 0.6875
8. In the city of Kanpur, the daily consumption of electricity (in thousands of megawatt-
hours) can be treated as a random variable having the gamma distribution, with shape
parameter  = 2 and scale parameter  = 5. If the power plant of the city has a daily
capacity of 100 megawatt-hours, what is the probability that this power supply will be
inadequate on any given day?
P(X  100) = = = (0.04)[500e-20
+ 25e-20
]
using integration by parts. Hence,
P(X  100) = (0.04)(525)e-20
= 4.3284 x (10)-8