Question

a) Let the random variable X have the following distribution:
P( and X = )0 = P(X = )2 = p
P(X = )1 =1− 2 p where
2
1
0 ≤ p ≤
for what value of p is the ) var(X a maximum? Justify. (3)
b) In a binomial distribution consisting of 5 independent trials, probabilities of 1 and 2
successes are 0.4096 and 0.2048 respectively. Find the parameter ‘ p ’ of the
distribution. (3)
c) Fit a Poisson distribution to the following data

Answer 1

Answer on Question #60649 – Math – Statistics and Probability

Question

a) Let the random variable $X$ have the following distribution:

$$P(X = 0) = P(X = 2) = p$$

$P(X = 1) = 1 - 2p$, where $0 \leq p \leq 1$.

for what value of $p$ is the var(X) a maximum? Justify.

Solution

$$\begin{aligned} \operatorname{Var}(X) &= M(X^2) - (M(X))^2 = \sum_{i=1}^{3} x_i^2 p_i - \left( \sum_{i=1}^{3} x_i p_i \right)^2 = \\ &= (0^2 \cdot p + 1^2 \cdot (1 - 2p) + 2^2 \cdot p) - (0 \cdot p + 1 \cdot (1 - 2p) + 2 \cdot p)^2 \\ &= 1 - 2p + 4p - 1 = 2p \end{aligned}$$ $$\max \operatorname{Var}(X) = \max \left\{ 2p, 0 \leq p \leq \frac{1}{2} \right\} = 1$$

$\operatorname{Var}X$ is maximum for $p = \frac{1}{2}$.

Question

b) In a binomial distribution consisting of 5 independent trials, probabilities of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the parameter 'p' of the distribution.

Solution

For a binomial distribution, the probability of $X$ successes in $n$ trials is given by

$$binom{n}{x} p^x q^{n-x}.$$

Probability of one success in 5 independent trials is

$$\binom{5}{1} p q^4 = 0.4096.$$

Probability of two successes in 5 independent trials is

$$\binom{5}{2} p^2 q^2 = 0.2048.$$

Now find:

$$\frac{\binom{5}{1} p q^4}{\binom{5}{2} p^2 q^2} = \frac{0.4096}{0.2048}$$ $$\frac{5q}{10p} = 2$$

On the other hand,

$$p + q = 1$$

Then

$$4p = q = 1 - p \rightarrow 4p + p = 1 \rightarrow 5p = 1 \rightarrow p = \frac{1}{5}.$$

Answer: 0.200.

Question

c) Fit a Poisson distribution to the following data:

$$\begin{array}{c c c c c c} x_i & 0 & 1 & 2 & 3 & 4 \end{array}$$ $$p_i \frac{109}{200} \quad \frac{65}{200} \quad \frac{22}{200} \quad \frac{3}{200} \quad \frac{1}{200}$$

Solution

Calculating the mean and variance:

$$E(N) = (M(X)) = \sum_{i=1}^{5} x_i p_i = \frac{1}{200} \left(0 \cdot 109 + 1 \cdot 65 + 2 \cdot 22 + 3 \cdot 3 + 4 \cdot 1 = 0.61 \right)$$ $$\begin{array}{l} \operatorname{Var}(X) = M(X^2) - (M(X))^2 = \sum_{i=1}^{5} x_i^2 p_i - \left( \sum_{i=1}^{5} x_i p_i \right)^2 \\ = \frac{1}{200} \left(0^2 \cdot 109 + 1^2 \cdot 65 + 2^2 \cdot 22 + 3^2 \cdot 3 + 4^2 \cdot 1 \right) - 0.61^2 = 0.98 - 0.37 \\ = 0.61 \end{array}$$

Since $E(N) = \operatorname{Var}(X)$, this can be modeled as a Poisson distribution because in this distribution

$\operatorname{E}(N) = \operatorname{Var}(X) = \lambda$, and here we can assign $\lambda = 0.61$:

$$\Pr(N = k) = \frac{\lambda^k}{k!} e^{-\lambda} = \frac{0.61^k}{k!} e^{-0.61}, \text{ for } k \geq 0.$$

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