The binomial and Poisson distributions 561

0.28

0.24

Probability of having an accident

0.20

0.16

0.12

0.08

0.04

(^0012345)
Number of people
6
Figure 57.2
The average occurrence of the event,λ,is
7500 × 0 .0003, i.e. 2.25
The probability of no people having anaccident is
e−λ=e−^2.^25 = 0. 1054
The probability of 1 person having anaccident is
λe−λ= 2 .25e−^2.^25 = 0. 2371
The probability of 2 people having anaccident is
λ^2 e−λ
2!

2. 
   

   252 e−^2.^25
   2!
   = 0. 2668
   and so on, giving probabilities of 0.2001, 0.1126,
   0.0506 and 0.0190 for 3, 4, 5 and 6 respectively hav-
   ing an accident. The histogram for these probabilities is
   shown in Fig. 57.2.
   Now try the following exercise
   Exercise 214 Further problems on the
   Poisson distribution

   
   


3. 
   

   In problem 7 of Exercise 213, page 559,
   determinetheprobabilityof having threecom-
   ponents outside of the required tolerance using
   the Poisson distribution. [0.0613]

   
   


4. 
   

   The probability that an employee will go to
   hospital in a certain period of time is 0.0015.

   
   

Use a Poisson distribution to determine the

probabilityof more than two employees going

to hospital during this period of time if there

are 2000 employees on the payroll.

[0.5768]

3. When packaging a product, a manufacturer
   finds that one packet in twenty is underweight.
   Determine the probabilities that in a box of
   72 packets (a) two and (b) less than four will
   be underweight.
   [(a) 0.1771 (b) 0.5153]


4. A manufacturer estimates that 0.25% of his
   outputof a component are defective. The com-
   ponents are marketed in packets of 200. Deter-
   mine the probability of a packet containing
   less than three defective components.
   [0.9856]


5. The demand for a particular tool from a store
   is, on average, five times a day and the demand
   follows a Poisson distribution. How many of
   these tools should be kept in the stores so that
   the probability of there being one available
   when required is greater than 10%?
   ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎢⎢

⎢

⎢

⎢

⎣

The probabilities of the demand

for 0, 1, 2,...tools are

0.0067, 0.0337, 0.0842, 0.1404,

0.1755, 0.1755, 0.1462, 0.1044,

0.0653,...This shows that the

probability of wanting a tool

8 times a day is 0.0653, i.e.

less than 10%. Hence 7 should

be kept in the store

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎥⎥

⎥

⎥

⎥

⎦

6. Failure of a group of particular machine
   tools follows a Poisson distribution with a
   mean value of 0.7. Determine the probabili-
   tiesof0,1,2,3,4and5failuresinaweekand
   present these results on a histogram.
   ⎡
   ⎢
   ⎢
   ⎢
   ⎣

Vertical adjacent rectangles

having heights proportional

to 0.4966, 0.3476, 0.1217,

0.0284, 0.0050 and 0.0007

⎤

⎥

⎥

⎥

⎦