556 STATISTICS AND PROBABILITY(b) three employees will be absent on a
particular day. [(a) 0.7514 (b) 0.0019]- A manufacturer estimates that 3% of his
output of a small item is defective. Find the
probabilities that in a sample of 10 items
(a) less than two and (b) more than two items
will be defective. [(a) 0.9655 (b) 0.0028] - Five coins are tossed simultaneously. Deter-
mine the probabilities of having 0, 1, 2, 3, 4
and 5 heads upwards, and draw a histogram
depicting the results.
⎡
⎢
⎢
⎣Vertical adjacent rectangles,
whose heights are proportional to
0.0313, 0.1563, 0.3125, 0.3125,
0.1563 and 0.0313⎤⎥
⎥
⎦- If the probability of rain falling during a par-
ticular period is 2/5, find the probabilities of
having 0, 1, 2, 3, 4, 5, 6 and 7 wet days in a
week. Show these results on a histogram.
⎡ ⎢ ⎢ ⎢ ⎢ ⎣
Vertical adjacent rectangles,
whose heights are proportional
to 0.0280, 0.1306, 0.2613,
0.2903, 0.1935, 0.0774,
0.0172 and 0.0016
⎤ ⎥ ⎥ ⎥ ⎥ ⎦- An automatic machine produces, on average,
10% of its components outside of the toler-
ance required. In a sample of 10 components
from this machine, determine the probabil-
ity of having three components outside of the
tolerance required by assuming a binomial
distribution. [0.0574]
57.2 The Poisson distribution
When the number of trials,n, in a binomial distri-
bution becomes large (usually taken as larger than
10), the calculations associated with determining the
values of the terms becomes laborious. Ifnis large
andpis small, and the productnpis less than 5, a
very good approximation to a binomial distribution
is given by the corresponding Poisson distribution,
in which calculations are usually simpler.
The Poisson approximation to a binomial distri-
bution may be defined as follows:
‘the probabilities that an event will happen 0, 1, 2, 3, ...,
ntimes inntrials are given by the successive terms of the
expressione−λ(
1 +λ+λ^2
2!+λ^3
3!+···)taken from left to right’.The symbolλis the expectation of an event happen-
ing and is equal tonp.Problem 6. If 3% of the gearwheels produced
by a company are defective, determine the
probabilities that in a sample of 80 gearwheels
(a) two and (b) more than two will be defective.The sample number,n, is large, the probability of a
defective gearwheel,p, is small and the productnp
is 80× 0 .03, i.e. 2.4, which is less than 5.
Hence a Poisson approximation to a binomial dis-
tribution may be used. The expectation of a defective
gearwheel,λ=np= 2 .4.
The probabilities of 0, 1, 2,...defective gear-
wheels are given by the successive terms of the
expressione−λ(
1 +λ+λ^2
2!+λ^3
3!+···)taken from left to right, i.e. bye−λ,λe−λ,λ^2 e−λ
2!,...Thus probability of no defective gearwheels ise−λ=e−^2.^4 = 0. 0907probability of 1 defective gearwheel isλe−λ= 2 .4e−^2.^4 = 0. 2177probability of 2 defective gearwheels isλ^2 e−λ
2!=2. 42 e−^2.^4
2 × 1= 0. 2613(a) The probability of having 2 defective gearwheels
is0.2613.
(b) The probability of having more than 2 defective
gearwheels is 1−(the sum of the probabilities