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# If for a Poisson distribution P(X=$0$) =$0.2$ then the variance of the distribution isA.$5$ B.${\log _{10}}5$ C.${\log _e}5$ D.${\log _5}e$

Last updated date: 25th Jun 2024
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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).

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