BINOMIAL DISTRIBUTION
Is an extension to discrete random variables and in order to apply
binomial distribution the following conditions are applicable
1 The data must be discrete
2 There must be
fixed number of tries int
3 There must only be 2 possible outcomes
a
Probability of eachoutcome is fixed
s Probability of eachoutcome is independent
6 Probability of favourable outcome is p
7 Probability of unfavourable outcome is q
8
9 1 1
9 The parameter of Binomial distribution with parametres n and p
x
B n p
Binomialdistribution within parametres in
to apply Binomial distribution is as follows
mtyyow
w.to
her 191
a
pit
where
n
and p
n is the number of tries
p is the probability of favourable outcome
q is the probabilityof unfavourableoutcome
OC n
11 Expectation or mean is Echl
12 Variance
or
var in
npq
up
n
Q
12 bulbs are checked for defeat The probability that a bulb is
detected is 0.35 Find the probability that
cil Exactly 5 bulbs are defected
pin 51
s 10.65712 510.3515
0.204
Iiil At most 2 bulbs are defected
10
P 2
PII
12co 10.65
10.3570 0.00569
ypppigspfpp.gs
0.151
pipps
iii more than 3 bulbs are defected
1 P 10
P n 3
1
1
1101 P 1
7121 P 3
12co 10.65
10.3570 0.00569
12C 10.65
10.35
0.0368
12C 10.6571010.3572 0.109
0 195
C 10.65
10.35
1
0.00569
1
10.346
0.654
0.0368
0.109 0.195
3
P 4 312
iv At least 10 bulbs are defected
10
P n
0.35
0.000769
Cii 0.65 x 0.35
0.0000 753
Cio 10.65
2
x
0.35 12
12 0.65
54M 0 000848
0.0000033g
101 Less than a bulbs are defected
PIN
9
1
P a
t P 10
p ll
1112
0.00476
Cio 10.65 0.35
0.000769
2cal o.gs
0.35
0.0000 753
12C12 0.65
0.35
0.00000338
Ivil Expected number of defectedbulbs
0 35
4 2
will variance ofdefectedbulbs
12 0.35 0 65
HW
Pg168
11117
12cg 0.6513 0.3579
0.994
12
P
1
18
2.73
112
P3
P6
PU
P7
Pls
7181