Page 201 - DMTH404_STATISTICS

Unit 14: Theoretical Probability Distributions

Solution. Notes

Since n is large and p is small, Poisson distribution is applicable. The random variable is the
number of defective toys with mean m = np = 300 × 0.01 = 3. The required probability is given by

´
e  3 .3 5 0.04979 243
P (r = ) 5 = = = 0.10082.
5! 120

Example 16: In a town, on an average 10 accidents occur in a span of 50 days. Assuming
that the number of accidents per day follow Poisson distribution, find the probability that there
will be three or more accidents in a day.

Solution.
10
The random variable denotes the number accidents per day. Thus, we have m = =0.2 . The
50
required probability is given by

2
é (0.2 ù
)
P (r  ) 3 = 1 P (r  ) 2 = 1 e  0.2 ê 1 0.2 + ú 1 0.8187 1.22=  ´ = 0.00119.
+


ê ë 2! ú û
Example 17: A car hire firm has two cars which it hire out every day. The number of
demands for a car on each day is distributed as a Poisson variate with mean 1.5. Calculate the
proportion of days on which neither car is used and the proportion of days on which some
-1.5
demand is refused. [ e = 0.2231]
Solution.
When both car are not used, r = 0
 (r = ) 0 = e  1.5 = 0.2231. Hence the proportion of days on which neither car is used is 22.31%.
P
Further, some demand is refused when more than 2 cars are demanded, i.e., r > 2

2 e  1.5 ( ) r é ( ) 2
1.5 ù
1.5
 (r > ) 2 = 1 P (r  ) 2 =  å 1 0.2231 1 1.5=  ê + + ú = 0.1913.

1
P
r= 0 ! r ê ë 2! ú û
Hence the proportion of days is 19.13%.
Example 18: A firm produces articles of which 0.1 percent are usually defective. It packs
them in cases each containing 500 articles. If a wholesaler purchases 100 such cases, how many
cases are expected to be free of defective items and how many are expected to contain one
defective item?
Solution.

The Poisson variate is number of defective items with mean

1
=
m = ´ 500 0.5.
1000

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