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: (4)

5
No. of mistakes per
page:
0 1 2 3 4 Total
Number of pages: 109 65 22 3 1 200

Answer 1

Answer on Question #60655 - Math - Statistics and Probability

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

$P(X = 0) = P(X = 2) = p$ and $P(X = 1) = 1 - 2p$ where

$0 \leq p \leq 1/2$

for what value of $p$ is the var(X) a maximum? Justify. (3)

Solution

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

Answer: $\operatorname{Max}\{Var(X)\} = 1$

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)

Solution

Probability mass function of a binomial distribution:

$$P_n(k) = \binom{n}{k} p^k (1 - p)^{n - k}, \text{ where } \binom{n}{k} = \frac{n!}{k! (n - k)!}$$

by task:

$$n = 5$$ $$P_5(1) = 0.4096 = \frac{5!}{1! (5 - 1)!} p^1 (1 - p)^{5 - 1} = 5p (1 - p)^4$$ $$P_5(2) = 0.2048 = \frac{5!}{2! (5 - 2)!} p^2 (1 - p)^{5 - 2} = 10p^2 (1 - p)^3$$

To find the parameter $p$ we need to solve the appropriate system of equations:

$$\left\{ \begin{array}{l} 5p (1 - p)^4 = 0.4096 \\ 10p^2 (1 - p)^3 = 0.2048 \end{array} \right.$$

To solve this system divide the first equation by the second:

$$\begin{array}{l} \frac{5p(1 - p)^4}{10p^2(1 - p)^3} = \frac{0.4096}{0.2048} \\ \frac{1 - p}{2p} = 2 \\ 1 - p = 4p \\ 5p = 1 \\ p = \frac{1}{5} = 0.2 \\ \end{array}$$

Answer: $p = 0.2$

c) Fit a Poisson distribution to the following data: (4)

Solution

Calculating the mean and variance:

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

Since $E(N) = Var(X)$, this can be modeled as a Poisson distribution because in this distribution $E(N) = Var(X) = \lambda$, and here we can assign $\lambda = 0.61$:

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

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

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