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Quantitative Techniques – I




                    Notes          Solution:
                                   It is  given that  m =  2.  Let the  number  of  arrivals  per minute  be denoted  by the  random
                                   variable r. The required probability is given by
                                                 2  3
                                                e .2  0.13534 8
                                   1.   P r  3                   0.18045
                                                 3!       6
                                                2   2  r
                                                  e .2     2      4
                                   2.   P r  2      r!   e  1 2   2   0.13534 5  0.6767.
                                                r 0
                                                             2  0
                                                            e .2
                                   3.   P r 1  1 P r  0  1        1 0.13534  0.86464.
                                                             0!

                                          Example: An executive makes, on an average, 5 telephone calls per hour at a cost which
                                   may be taken as   2 per call. Determine the probability that in any hour the telephone calls’ cost
                                   (i) exceeds   6, (ii) remains less than   10.

                                   Solution:
                                   The number of telephone calls per hour is a random variable with mean = 5. The required
                                   probability is given by
                                                             3  5  r
                                                               e .5
                                   1.   P r  3  1 P r  3  1
                                                                r!
                                                            r 0
                                                 5      25  125            236
                                                1 e  1 5        1 0.00678      0.7349.
                                                        2   6               6
                                                4   5  r
                                                  e .5     5     25 125   625          1569
                                   2.   P r  4      r!   e  1 5   2   6   24   0.00678       0.44324.
                                                r 0                                     24


                                          Example: A company makes electric toys. The probability that an electric toy is defective
                                   is 0.01. What is the probability that a shipment of 300 toys will contain exactly 5 defectives?
                                   Solution:

                                   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

                                                           3 5
                                                         e .3   0.04979 243
                                                 P r  5                     0.10082.
                                                           5!       120

                                          Example: 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.











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