082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

29.10 The binomial distribution 955
Example29.18 The probability that a component is acceptable is 0.93. Ten components are picked at
random. What is the probability that
(a) atleastnine areacceptable
(b) atmostthree areacceptable?
Solution (a) P(exactly 9 components are acceptable) =
Hence,
P(exactly 10 components areacceptable) =
( 10
9
)
(0.93) 9 (0.07) = 0.364
( ) 10
(0.93) 10 = 0.484
10
P(atleast9components areacceptable) = 0.364 +0.484 = 0.848
(b) We require the probability that none are acceptable, one is acceptable, two are acceptable
and threeare acceptable:
( ) 10
P(0are acceptable) = (0.93) 0 (0.07) 10 = 2.825 ×10 −12
0
( ) 10
P(1is acceptable) = (0.93) 1 (0.07) 9 = 3.753 ×10 −10
1
( ) 10
P(2are acceptable) = (0.93) 2 (0.07) 8 = 2.244 ×10 −8
2
( ) 10
P(3are acceptable) = (0.93) 3 (0.07) 7 = 7.949 ×10 −7
3
Hence,
P(atmost3are acceptable) = 2.825 ×10 −12 +3.753 ×10 −10
+2.244 ×10 −8 +7.949 ×10 −7
= 8.18 ×10 −7
that is, the probability that at most three components are acceptable is almost zero;
itisvirtually impossible.
29.10.2 Meanandstandarddeviationofabinomialdistribution
Let the probability of success in a single trial be p and let the number of trials be n.
The number of successes in n trials is a discrete random variable, x, with a binomial
distribution. Thenxcan have any value from {0,1,2,3,...,n}, although clearly some
valuesaremorelikelytooccurthanothers.Theexpectedvalueofxcanbeshowntobe
np. Thus,ifmany values ofxarerecorded, the mean of these will approachnp.
Expected value of the binomial distribution =np
The standard deviation of the binomial distributioncan alsobefound. This isgiven by
standard deviation of the binomial distribution = √ np(1−p)