Answer
424.8k+ views
Hint: 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Guru Purnima speech in English in 100 words class 7 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Three liquids are given to you One is hydrochloric class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)