Page 203 - DMTH404_STATISTICS

Unit 14: Theoretical Probability Distributions

Solution. Notes

The number of defective pins in a box is a Poisson variate with mean equal to 5. A box will meet
the guaranteed quality if r £ 4. Thus, the required probability is given by

4 5 r 5 é 25 125 625 ù 1569
P
+
(r  ) 4 = e  5 å = e  ê 1 5 + + + ú 0.00678= ´ = 0.44324.
r= 0 ! r ë 2 6 24 û 24
Lot Acceptance using Poisson Distribution

Example 21: Videocon company purchases heaters from Amar Electronics. Recently a
shipment of 1000 heaters arrived out of which 60 were tested. The shipment will be accepted if
not more than two heaters are defective. What is the probability that the shipment will be
accepted? From past experience, it is known that 5% of the heaters made by Amar Electronics are
defective.
Solution.
5
Mean number of defective items in a sample of 60 = 60 × = 3
100
2 e  3 .3 r
P(r £ 2) = å
r= 0 ! r
2
é 3 ù
+
–3
= e –3 ê 1 3 + ú = e .8.5 = 0.4232
ë 2! û
14.6.4 Poisson Approximation to Binomial

When n, the number of trials become large, the computation of probabilities by using the
binomial probability mass function becomes a cumbersome task. Usually, when n ³ 20 and p £
0.05, Poisson distribution can be used as an approximation to binomial with parameter m = np.

Example 22: Find the probability of 4 successes in 30 trials by using (i) binomial
distribution and (ii) Poisson distribution. The probability of success in each trial is given to be
0.02.
Solution.
(i) Here n = 30 and p = 0.02
a 30 a 4 a 26
 P r = 4f = C 0.02f 0.98f 27405 0.00000016 0.59 0.00259.= ´ ´ =
4
(ii) Here m = np = 30 × 0.02 = 0.6

´
e  0.6 (0.6 ) 4 0.5488 0.1296
P
 (r = ) 4 = = = 0.00296.
4! 24

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