Question

We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variables with p = 0.5 (E(Xi)=p and var(Xi)=p(1-p)). a. What distribution does Pn i=1 Xi follow exactly (sum of independent Bernoulli random variables)? State the type of distribution and what the parameter is. b. Using the central limit theorem, what approximate distribution does X¯ = 1 n Pn i=1 Xi follow? State i the type and the mean and variance. c. What is the approximate probability that X¯ = 1 n Pn i=1 Xi is greater than 0.5? Write down the integral using proper notation and use empirical rule or pnorm function in R to obtain numeric answer.

Answer 1

Answer:

(a) The distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of $P(\bar X>0.50)$ is 0.50.

Step-by-step explanation:

It is provided that random variables $X_{i}$ are independent and identically distributed Bernoulli random variables with p = 0.50.

The random sample selected is of size, n = 100.

(a)

Theorem:

Let $X_{1},\ X_{2},\ X_{3},...\ X_{n}$ be independent Bernoulli random variables, each with parameter p, then the sum of of thee random variables, $X=X_{1}+X_{2}+X_{3}...+X_{n}$ is a Binomial random variable with parameter n and p.

Thus, the distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. n = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu=p=0.50$

And the standard deviation of the distribution of sample mean is given by,

$\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05$

(c)

Compute the value of $P(\bar X>0.50)$ as follows:

$P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})$

                    $=P(Z>0)\\=1-P(Z$

*Use a z-table.

Thus, the value of $P(\bar X>0.50)$ is 0.50.