Question

If the number of telephones calls an operator receives between 10.00 pm to 10.10 pm following Poisson distribution with mean 3. The probability that the operator receives one call during the interval the next day is 0.149.

Answer 1

The number of telephones an operator receives is between 10.00 pm to 10.10 pm following Poisson distribution so use the Poisson distribution.
Poisson distribution formula is given by:
$P(x,\mu ) = \dfrac{{{e^{ - \mu }}{\mu ^x}}}{{x!}}$
Where$\mu$ is mean, x is the actual number of successes that result from the experiment and e is approximation equal to 2.718.

Complete step-by-step answer :
Operators receives telephone calls following the Poisson distribution the formula for Poisson
Distribution is:
$P(x,\mu ) = \dfrac{{{e^{ - \mu }}{\mu ^x}}}{{x!}}$
Where$\mu$= 3, that is an average telephone call receives operator between 10.00 pm to 10.10 pm
$x = 1$, the operator receives one call during the interval the next day
$e = 2.718$, that is exponential constant
$P(x,\mu ) = \dfrac{{{e^{ - \mu }}{\mu ^x}}}{{x!}}$
In above equation put the value of x=1 and $\mu$= 3
$P(1,3) = \dfrac{{{e^{ - 3}}{3^1}}}{{1!}}$
Taking the inverse value to the denominator to remove the inverse and substituting the value of 1!=1 in the denominator.
$P(1,3) = \dfrac{{{3^1}}}{{{e^3}}}$
$= \dfrac{{{3^1}}}{{20.0855}}$
In above equation put the value of e = 2.718
Dividing the numerator from the denominator we solve the value.
$= 0.149$

The probability to receive one call during that interval on the next day is 0.149.

Note :
Students are likely to have little difference in the final answer because it depends on students substituting the final value of e up to which decimal value they substitute, students are advised to use the value of e given in the question for the answer.

Poisson distribution formula is given by:
$P(x,\mu ) = \dfrac{{{e^{ - \mu }}{\mu ^x}}}{{x!}}$
Where$\mu$is mean, x is the actual number of successes that result from the experiment and e is approximation equal to 2.718.