Question

Mr. Karma receives 6 telephone calls between 6 am and 8 am. What is the probability that Karma receives more than one call in the next 15 minutes?

Select one:

a. 0.7334

b. 0.1734

c. 0.1503

d. 0.1900

​

Answer 1

Given : Mr. Karma receives 6 telephone calls between 6 am and 8 am.

To Find : probability that Karma receives more than one call in the next 15 minutes

a. 0.7334

b. 0.1734

c. 0.1503

d. 0.1900

Solution:

Mr. Karma receives 6 telephone calls between 6 am and 8 am

=> 6 calls in 2 hrs

=> 6/8 calls in 15 minutes

= 0.75 calls

Poisson distribution

P(x)  = { λˣ  $e^{-\lambda}$) } / x!

λ  = Mean  = 0.75

probability of more than one call in the next 15 minutes

= 1 - ( P(0)  + P(1)  )

P(0)  = (0.75)⁰ $e^{-0.75}$ / 0!  =  0.47237

P(1)  =  (0.75)¹ $e^{-0.75}$ / 1!  =    0.35427

probability of more than one call in the next 15 minutes   =  1 - (0.47237 + 0.35427 )

= 1 -  0.82664

= 0.17336

=0.1734

probability that Karma receives more than one call in the next 15 minutes =0.1734

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