Page 295 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Quantitative Techniques – I

Notes Solution:
Let p be the probability of making a mistake in typing a word.
1. Let the random variable r denote the number of mistakes per letter. Since 20 letters are
typed, r will follow Poisson distribution with mean = 20 × p.
Since less than 1% of the letters are rejected, it implies that the probability of making at
least one mistake is less than 0.01, i.e.,

P(r 1) 0.01 or 1 – P(r = 0) 0.01

1 – e -20p 0.01 or e -20p 0.99
Taking log of both sides

– 20p.log 2.72 log 0.99

20 0.4346 p 1.9956

0.0044
– 8.692p – 0.0044 or p 0.00051.
8.692
2. In this case r is a Poisson variate which denotes the number of mistakes per day. Since the
typist has to type 20 × 200 = 4000 words per day, the mean number of mistakes = 4000p.

It is given that there is no mistake on 90% of the days, i.e.,
P(r = 0) = 0.90 or e -4000p = 0.90
Taking log of both sides, we have

– 4000p log 2.72 = log 0.90 or 4000 0.4346p 1.9542 0.0458

0.0458
p 0.000026.
4000 0.4346

Lot Acceptance using Poisson Distribution

Example: 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
3
2 e .3 r
P(r 2)
r 0 r!
3 2
= e –3 1 3 = e .8.5 = 0.4232
–3
2!

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