1.292 Probability and Random Processes
Exercise 1.6
1. Suppose that the life time of a certain kind of radioactive Materig
which follow Weibull distribution with parameters. (in ey,
a=6.25x10-? fB=05
(a) Determine the mean life of a radio active material.
(b) Calculate its variance.
Ans: (a) 512 years, () 1319759
Yen
2. If the random ‘variable follows exponential distribution with
prove that ¥ = X? follows a weibull distribution with parameters » oy
= 1 4-1 ewt i
[um 2(SJeho,
1.12 Normal Distribution
The Normal Distribution was first described by De Moive in 1933 as the int
form of Binorzial Distribution as the number of trials becomes infinite. This
covery came into limelight after its discovery by both Laplace and Cau ie
century later.
So this distribution is also called Gaussian distribution.
1.12.1 Definition
A continuous random variable X with probability density function.
-aesey? —0o<2r<00
f(z) = ae —00
0
is said to follow normal distribution.
Here the parameters are o and 4, where mean = pz and S.D =o.
Notes: The normal distribution is denoted by X ~ N(y,0) or X ~ N(u,0")
1.12.2 Characteristics of Normal Distribution and Normal pro>
ability curve
(@) The Curve is bell shaped and symmetrical about the line x = }1.
i). Mean, Median and Mode of the distribution coincide. Random Variables 1.233
(iii) As # increases numerically, F(x) decreases rapidly, the maximum proba-
1. given by [p(t)]ying =
bility occurring at the point x
[ a n o
(iv) Since f() is non - negative, the curve will not g0 below the
(v) z-axis is an asymptote to the curve.
(vi) wre = 0 (7 =0,1,2,3,..,)
and far = 1.3.5... (2r ~ 1) 92, (r =0,1,2,.
1.12.3 Moment Generating Function of Normal Distribution
By definition
Mx(t) = Efe'*]
oo
Put z
dz = odz
1 @
> | etter). 2
co
8
1
=e / en Ble-09? 027]
"00
°0
2
sett oP 1 feteovtas
00
Von
Put 2-ot=y
dz=dy
seme or Pe
‘Vin fF au
~00 1.234 Probability and Random Processes . ’
- 2
ote. z= edu
=
a
2
Put 57?
u atv? = du=dtv2
woe 2 fee
et vel dtV2
4
sh
2
noth feta one 2
Oo
[Jew
3
1 Mx(t) = ete
1.12.4 Moments of Normal distribution
Prove that all odd order moments of a Normal distribution N(y1, 07) about is
mean are zero and its even order moments about the mean are given by the recu-
rence relation.
) Han = 0?(2n — 1)pan-2-
Proof:
Odd order moments about mean are given by
pana = f (@- eP'se Ms
2
°°
fe = pen). en de
co
—
- [wore Foo dz is an odd function of z.
a2nt le
e integrand = i.
i h 7
a order moments about mean are given by
.
pon = | (e— WPLodr
oS
7
1 ag
ale f (o2)"" -6°F ode
avin :
a
2n va ae
7 | 2" .eF de
an J,
.
2n
fi 2". e"Fdz |." the integrand is an even function of z}.
ao
a Probability and Random Processes —
on gn
=F hos
‘Changing n to (n — 1), we get
en : gn-1¢g2n-2 T
Han-2 = FR 3,
In +
Loi
n—-4
= fon = Han—2° 0? (2n — 1)
1.12.5 Mean and Variance of Normal Distribution using M.G.F
Mx(é) = onthe
awxi=[$ ax(o)|
- te n+ o*} .
= [Pu +0] =H
[ween =]
E[X?] = [= Ix (€ -
cE d
a [e" oF ut oto] a
= ao (wt ot)? + euttee ©]
t=0
ee Oe
ELX?] a we +02
Variance (X) = E[X?] — [E[X]]?
=p +0? = [uP = 0?
Var (X) = 0?
1.12.6 Median and Mode of Normal distribution
Median of normal distribution
M
Mis the median of X if J f(x)de = f f(x)de = 4
0° M Random Variables 1.237
oV2n
fp 1 Gag? Pq hha ;
i 7 dx + i pee
Bee da On wae 5 a)
i
Now, since the normal curve is symmetrical about z = 1, and
oo : 2 0° 1 ]
1 -S#dc=1, wehave / es
ovr oV2n 2
te i
Using this in (1), we get :
j=
02 dr =
[ avon
M =». i.e., Median =
| Mode of normal distribution
mm variable X is defined as the value of x for which
Mode of a continuous randot
He)is maximum, ie., f(z) = 0, £"(@) <0. For the normal distribution,
aoe 09 rap
f(z) = dite = p)f'(x) + F(z)
Peay = GW = Sp x ee <0 hability sind Random Processes
is maximum ate
. Mode = j1
12.7. Standard Normal Distribution
Standard normal distribution: If X is # random variable Following normal dig
X =H jg called the standard noma,
bution with parameter jr and 7 then Z = —G
is given by
variate and the p.d.f of the standard variate Zi
- Fee 0
0 Z,
‘Shaded area = P(Z Z,)
=0.5+P(0 o?.
Solved Problem 1.203
IFX is (3,4), find k so that P(X — 3] > &) = 0.05.
Solution
Here p=3,0=2
goxXt#e x28 -
o 2
0.05 = P(|X — 3| > k)
X-3/_k
2 (FRI 2 4) = 0.05
ie, P(izi > 5) = 0.05 1.240 Probability and Random Processes
bd 5
je.. 2P (7 > 5) = 0.05
k r
ie, P (2 > 5) = 0.025
P (0
Pe Bo) =P so 150 =,” 150 a
sto — 500
=P|zZ> a
[ T60 2 0.05
ey 500) a ate
plos= < 21500 0.5 — 0.05
= 0.45.
the value is approximately 1.645.
P ao — 500
fen | 700) =P [
X — 662 . 700 623
32 32
= P(Z > 1.187)
=0.5— P(0< Z< 1.187)
= 0.5 — 0.3830
= 0.1170
.. The number of plots expected to yield 700 kgs = 1000 x 0.1170
=117:
‘X — 662 _ 650 — 662
p(x < 650) = P| 32 < a
= P(Z.< —0.38) = P(Z > 0.38)
=0.5—P(0 < Z < 0.38)
= 0.5 — 0.148 = 0.352
di)
The number of plots expected to yield below 650 kgs.
= 1000 x 0.352 = 352 plots Random Variables 1,253
Let 21 be the lowest yield of the best 100 plots
‘Then P(X >) = yum = 0.1
Now P(X aaa [A = 662 . 3
ie, -P(Z>m)=01
-P(0 4 = 1.28 (From tables)
. 2-662,"
i.e., —»2 1.28
= 702.96
‘The lowest yield of best 100 plots = 702.96 kgs.
Solved Problem 1.221
1f-X isa random variable with normal distribution and p = 1, 0 = 2. Find
P|ix- a< s|x>0|
Solution
Given p = 1,0 =2
To.find
Piix-a)< 3[x>]
X-1_0-1
a |
_ Pik-3] \ 984 Pohabitity and Random Processes
po
=Po 80) = 0.05 Random Voriables 1.235
Jj so -
p(z< a) =01
su
Let SoH,
7
P(Z@<2Ay=01
. PWS Z<-2) = 04
f.. _ SUH so-
‘Also p(x > Sar) =0.05 + Let “ "zy
P(Z > Zz) = 0.05
=05-P(0< Z < 23) = 0.05
ie,
+ P(O |
1.256 Probability and Random Processes
Solved Problem 1.224
arks obtained by the students in Maths, pj.
istributed about mean 50, 52, 48 and S.p 1 et a
bility of securing a total mark of (i) 180 or walby
In an examination the m
Chemistry are normally di
spectively. Find the probill
90 or below.
8, (4
Solution
Let X.Y; Z be the marks of respective subjects.
The total marks T= X+Y+Z
p= E(T) = 2(X+¥ +2)
= E{X]+ ElY] + £[Z]
= 50 +52 +48 = 150
= Var (X + ¥ + Z)
= Var (X) + Var (Y) + Var (Z)
= 225 + 144 + 256 = 625
Var (T)
«By additive property, T is a normal variate with pp = 150; 0? = 625, ie,¢=2
To find
a >
P(T > 180) =P [ B27
30
- [222]
= P(Z > 1.2) =0.5— PIKE Z $14)
=0.5— 0.3849 = 0.1151 *
Gi) oe T —150 _ 90-150
PIT < 90) =P |95 = 95- :
= P[Z < -2.4)=0.5- P[0< Z< 24]
= 0.5 — 0.4918 = 0.0082.
@ T —150 _ 180— 122)
Solved Problem 1.225 :
th inde
The skulls are classified as A,B and C according to the length bread
as under 75, between 75 and 80, and above 80 respectively. Find the nee
standard deviation, assuming the distribution is normal in which #1 is 58%
38% and C is 4%. Also given that
is Random Variables 1.257
t
1 f(-#)
j= /e dx then (0.20) = 0,
LO Qn [ and ii 5 a
*y pe the random variable denoting the length and breadth index.
Xe =
yet n= pand S.D =o
Let mem
oz, Z2
oe P(X < 75) = 0.58
[> < Bee) = 0.58
o o
75 —
ie,s P(B<%)=0.58 — where Z) = —> H
=> P(0< Z< Z) = 0.08 (4)
Also.’ P(Z > 80) = 0.04
P [= > s=4] =0.04
o o
80=
P(Z > Zz) =0.04 | Z,y=—— B
> P(02
Gi). X < 20
Gi) O< X S12
(iv) Find @ when P(X > a) = 0.24
(v) Find b and.c when P(b< X c) = 0.25
Solution
Given p = 12,0 =4
@ Pox > 20) = [=5™ ad
4 2 4
=P(Z>2)
=05-P(0a) =0.24
=12a-12) 9,
7 Z| = 0.24
P [2 — 2) =0.24
=> P(Z>2Z)=0.24
Also P(Z > 2) =0.5- P(0 2)
= 0.5 — 0.24 = 0.26
rom the tables Zi = 0-71 (approx)
o-¥ son > a=1484
>
(y Given P(< X <0) = 0.50 and P(X > ce) = 0.25
p-12_ X-12 £512) Z) = 0.25
Now P(Z> 2) =05-P(0 Z, = 0.67 from the tables
: c—¥ = 0.07 > c= 14.68
From (1) and (2) P(Z < Z <0) = 0.25
“s = -0.67 > b=9.32
c= 14.68, b = 9.32
“Wed Problem 1.227
hay, 64
‘Ormal distributi
tle a pala 31% Of the items are under 45 and 8% are over 64. Find
= ‘ard deviation of the distribution.
a 1.260 Probability and Random Processes
Solution
Let mean be jt and standard deviation
X-1
Z=
Here 31% of the items are under 45.
i.e, oe < 45] = 0.31
p(z<5 < S=1) = 0.31
ofthe ordinate at Z = “854 is 0.31 and i
ordinate upto mean is 0.5-0.31=0.19,.
ing to this area is 0.5. (from Normal tables)
The area lying to the left of
area lying to the right of the
The value of Z, correspondit
<—S =<
Su Oty
7 s
45 -
Hence Z = —— #05 q
8% of the items are above 64.
ie, P[X < 64] = 0.08
p(2< e aon) = 0.08
‘Area to the left of the ordinate at Z = S4=# upto the mean is 0.5-0.0
the value of Z corresponding to this area is 1.4.
Hence Z = S47 # = 14
o
From equations (1) and (2) . *
=p + 0.50 = —45
—p- 140 = —64
Solving for jz and c, we get
: =50 and o=10 Random Variables 1.261
erciso 17 ation with mean 15.
tion with mean 15.00 and $.D 3.5, iti
normal popula a .5, it is known that 647
a tions exceed 16.25, What is the total no of observations in the pop-
ulation:
[Ans: 647)
‘an imteligence test is administered to 1,000 children. The average score is,
2 igmdSD=24
( Find the number of children exceeding the score 60.
Gi Find the number of children with score lying between 20 and 40. (As-
sume the normal distribution).
[Ans: (i) 227, (ii) 289]
= 10, find P(15 < x < 40).
[Ans: 0.6687]
4, The normal distribution 1 = 20 and S.D
4, The average seasonal rainfall in a place is 16 inches with S.D of 4 inches.
Matis the probability that in a year the rainfall in that place will be be-
tween 20 and 24 inches ?
[Ans: 0.1359]
5, Ina distribution exactly normal, 7% of items are under 35 and 89% are
under 63, What are the mean and S.D of the distribution.
[Ans: Mean = 20.29, S.D = 10.33]
6. Inacertain examination the percentage of passes and distributions were 46
and 9 respectively. Estimate the average marks obtained by the candidates,
the minimum pass and distinction marks being 40 and 75 respectively (As-
sume the distribution of marks to be normal).
[Ans: Mean = 36.4; $.D = 28.2]
1. The mini ae
saan height is to be prescribed for eligibility to government ser-
at 60% of the young men will have a fair chance to coming
UP to the :
with iar The weights of the young men are normally distributed
.6" and 8. 2.55”. Determine the minimum specifications.
[Ans: 59.9”]
Boece 1.262 Probability and Random Processes
8. The monthly income of a group of 10,000 persons are found to
distributed with mean Rs. 750 and S.D Rs. 50. Show that ony
group, about 95% has income exceeding Rs. 668 and only 5% hia hy
exceeding Rs. 832. What is the lowest income among the riches, iron,
(AMER
“9
‘The mean 1.Q. of a large number of children of the age 14 was 100 ang
16. Assuming the distribution was normal, find. Sp
(a) What % of children has I.Q under 80.
(b) Between what limits the 1.Q’s of the middle 40% of the children jn
(c) What % of the children are within the range 41 + 1.96. ‘.
[Ans: (a) 10.56%, (b) 91.6, 1084, () 995
10. Assume the mean heights of soldiers to be 68.22 inches with a variance g
10.8(inches)?. How many soldiers in a regiment of 1,000 would you expt
to be over 6 feet tall.
[Ans: 129
11. The height measurements of 600 adult males are arranged in an ascenting
” order and it is observed that 180" and 450" entries are 64.2" and 618"
respectively. Assuming that the sample of height is drawn from a nom
population, estimate the mean and S.D of the population.
[Ans: 67.78",3]
12. Given X is normally distributed. P[X < 45] = 0.31 and P[X > 64]=
0.08. Find the mean and S.D
[Ans: 50,10)
13. A company finds that the time taken by one of its engineers to compltt
a repair job has a normal distribution with mean 40 minutes and sD:
minutes. State what proportion of jobs take: (i) Less than 35 minutes.
More than 48 minutes.
[Ans: (9 0.159, Gi) 008
2)
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