A poisson variate x is such that `P(x=2)=9P(x=4)+90.P(x=6)` then mean and standard deviation are

Options

A. 1, 1
B. 1, 2
C. 2, 2
D. `2,sqrt(2)`

Answer

To solve the problem, we need to find the mean and standard deviation of a Poisson random variable \(X\) given the equation:

\[ P(X=2) = 9P(X=4) + 90P(X=6) \]

### Step 1: Write the Poisson Probability Formula
The probability mass function for a Poisson distribution is given by:

\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

where \(\lambda\) is the mean (and variance) of the distribution.

### Step 2: Substitute the Values into the Equation
Using the formula, we can express \(P(X=2)\), \(P(X=4)\), and \(P(X=6)\):

\[ P(X=2) = \frac{e^{-\lambda} \lambda^2}{2!} \]
\[ P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} \]
\[ P(X=6) = \frac{e^{-\lambda} \lambda^6}{6!} \]

Substituting these into the original equation gives:

\[ \frac{e^{-\lambda} \lambda^2}{2} = 9 \cdot \frac{e^{-\lambda} \lambda^4}{24} + 90 \cdot \frac{e^{-\lambda} \lambda^6}{720} \]

### Step 3: Simplify the Equation
We can cancel \(e^{-\lambda}\) from both sides (assuming \(\lambda > 0\)):

\[ \frac{\lambda^2}{2} = 9 \cdot \frac{\lambda^4}{24} + 90 \cdot \frac{\lambda^6}{720} \]

Now simplify the right-hand side:

\[ \frac{\lambda^2}{2} = \frac{9\lambda^4}{24} + \frac{90\lambda^6}{720} \]

Calculating the fractions:

\[ \frac{90\lambda^6}{720} = \frac{\lambda^6}{8} \]

Thus, the equation becomes:

\[ \frac{\lambda^2}{2} = \frac{3\lambda^4}{8} + \frac{\lambda^6}{8} \]

### Step 4: Multiply Through by 8 to Eliminate Denominators
Multiply the entire equation by 8:

\[ 4\lambda^2 = 3\lambda^4 + \lambda^6 \]

### Step 5: Rearrange the Equation
Rearranging gives:

\[ \lambda^6 + 3\lambda^4 - 4\lambda^2 = 0 \]

### Step 6: Factor the Equation
Let \(y = \lambda^2\). Then the equation becomes:

\[ y^3 + 3y^2 - 4y = 0 \]

Factoring out \(y\):

\[ y(y^2 + 3y - 4) = 0 \]

The quadratic can be solved using the quadratic formula:

\[ y = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 4}}{2} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]

This gives:

\[ y = 1 \quad \text{(since } y = 0 \text{ is not valid for } \lambda^2\text{)} \]

### Step 7: Solve for \(\lambda\)
Since \(y = \lambda^2\), we have:

\[ \lambda^2 = 1 \implies \lambda = 1 \]

### Step 8: Find Mean and Standard Deviation
For a Poisson distribution, the mean \(\mu\) and variance \(\sigma^2\) are both equal to \(\lambda\):

\[ \text{Mean} = \lambda = 1 \]
\[ \text{Standard Deviation} = \sqrt{\lambda} = \sqrt{1} = 1 \]

### Final Answer
The mean and standard deviation are:

\[ \text{Mean} = 1, \quad \text{Standard Deviation} = 1 \]