Answer
424.5k+ views
Hint: We can solve the given question using Poisson distribution formula,
$ \Rightarrow $ \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\]
Where ‘m’ is the average number of events in a given time interval or mean or variance of distribution and ‘X=r’ is the number of events observed over a given time period. Put the given values in the formula and simplify it. Here, e is the base of the natural logarithm (also called Euler’s number).
Complete step by step answer:
Given that the Poisson distribution when X$ = 0$ is P=$0.2$
And we have to find the variance of the distribution.
Now, according to Poisson distribution,
$ \Rightarrow $ \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\] --- (i)
Where ‘m’ is the mean or variance of distribution or the average number of events in a given time interval and ‘X=r’ is the number of events observed over a given time period.
Here, e is the base of natural logarithm (also called Euler’s number).
Now we know the value of X$ = $ r$ = 0$
On putting this value in eq. (i), we get-
$ \Rightarrow {\text{P}}\left( {{\text{X = 0}}} \right) = \dfrac{{{{\text{e}}^{{\text{ - m}}}}{{\text{m}}^{\text{0}}}}}{{0!}}$
We know that $0! = 1$ and ${{\text{m}}^{\text{0}}} = 1$ . On putting these values in the above equation we get,
$ \Rightarrow {\text{P}}\left( {{\text{X = 0}}} \right) = \dfrac{{{{\text{e}}^{{\text{ - m}}}}.1}}{1}$
$ \Rightarrow {\text{P}}\left( {{\text{X = 0}}} \right) = {{\text{e}}^{{\text{ - m}}}}$
And it is given that P=$0.2$
So on putting the value of P in the equation we get,
$ \Rightarrow {{\text{e}}^{{\text{ - m}}}} = 0.2$
We can write $0.2 = \dfrac{2}{{10}}$ then the equation becomes,
$ \Rightarrow {e^{ - m}} = \dfrac{2}{{10}}$
$ \Rightarrow \dfrac{{10}}{2} = \dfrac{1}{{{e^{ - m}}}}$
$ \Rightarrow 5 = {e^m}$
On taking log both side we get,
$ \Rightarrow {\log _e}5 = {\log _e}{e^m}$
We know that ${\log _e}{x^a} = a{\log _e}x$
On applying this we get,
$ \Rightarrow {\log _e}5 = m{\log _e}e$
Since ${\log _e}e = 1$ , we get
$ \Rightarrow m = {\log _e}5$
Hence option C is correct.
Note: Poisson distribution is used to predict the probability of certain events from happening when you know how often the event has occurred. The conditions for Poisson distribution are
1. An event can occur any number of times during a time period.
2. Events occur independently. Example-for the number of phone calls an office would receive, there is no reason to expect a caller to affect the chances of another person calling.
3. The rate of occurrence is constant and not based on time.
4. The probability of an event occurring is proportional to the length of the time period
$ \Rightarrow $ \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\]
Where ‘m’ is the average number of events in a given time interval or mean or variance of distribution and ‘X=r’ is the number of events observed over a given time period. Put the given values in the formula and simplify it. Here, e is the base of the natural logarithm (also called Euler’s number).
Complete step by step answer:
Given that the Poisson distribution when X$ = 0$ is P=$0.2$
And we have to find the variance of the distribution.
Now, according to Poisson distribution,
$ \Rightarrow $ \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\] --- (i)
Where ‘m’ is the mean or variance of distribution or the average number of events in a given time interval and ‘X=r’ is the number of events observed over a given time period.
Here, e is the base of natural logarithm (also called Euler’s number).
Now we know the value of X$ = $ r$ = 0$
On putting this value in eq. (i), we get-
$ \Rightarrow {\text{P}}\left( {{\text{X = 0}}} \right) = \dfrac{{{{\text{e}}^{{\text{ - m}}}}{{\text{m}}^{\text{0}}}}}{{0!}}$
We know that $0! = 1$ and ${{\text{m}}^{\text{0}}} = 1$ . On putting these values in the above equation we get,
$ \Rightarrow {\text{P}}\left( {{\text{X = 0}}} \right) = \dfrac{{{{\text{e}}^{{\text{ - m}}}}.1}}{1}$
$ \Rightarrow {\text{P}}\left( {{\text{X = 0}}} \right) = {{\text{e}}^{{\text{ - m}}}}$
And it is given that P=$0.2$
So on putting the value of P in the equation we get,
$ \Rightarrow {{\text{e}}^{{\text{ - m}}}} = 0.2$
We can write $0.2 = \dfrac{2}{{10}}$ then the equation becomes,
$ \Rightarrow {e^{ - m}} = \dfrac{2}{{10}}$
$ \Rightarrow \dfrac{{10}}{2} = \dfrac{1}{{{e^{ - m}}}}$
$ \Rightarrow 5 = {e^m}$
On taking log both side we get,
$ \Rightarrow {\log _e}5 = {\log _e}{e^m}$
We know that ${\log _e}{x^a} = a{\log _e}x$
On applying this we get,
$ \Rightarrow {\log _e}5 = m{\log _e}e$
Since ${\log _e}e = 1$ , we get
$ \Rightarrow m = {\log _e}5$
Hence option C is correct.
Note: Poisson distribution is used to predict the probability of certain events from happening when you know how often the event has occurred. The conditions for Poisson distribution are
1. An event can occur any number of times during a time period.
2. Events occur independently. Example-for the number of phone calls an office would receive, there is no reason to expect a caller to affect the chances of another person calling.
3. The rate of occurrence is constant and not based on time.
4. The probability of an event occurring is proportional to the length of the time period
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
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![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)
Difference between Prokaryotic cell and Eukaryotic class 11 biology 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)
Write a letter to the principal requesting him to grant class 10 english CBSE
![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)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)