Mathematics, 28.05.2020 04:00 gabw2758

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.

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Answer from: lucypackham9

(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.

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 $P(\bar X0.50)$ as follows:

$P(\bar X0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}\frac{0.50-0.50}{0.05})\\$

$=P(Z0)\\=1-P(Z$

*Use a z-table.

Thus, the value of $P(\bar X0.50)$ is 0.50.

Answer from: Quest

answer: the answer is one of

$f=\dfrac{1}{2},~\dfrac{1}{4},~\dfrac{3}{4},~\dfrac{1}{6},~\dfrac{5}{6},~\dfrac{1}{8},~\dfrac{3}{8},~\dfrac{5}{8},~\dfrac{7}{8}.$

step-by-step explanation: we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

$f=\dfrac{1}{2},~\dfrac{1}{4},~\dfrac{3}{4},~\dfrac{1}{6},~\dfrac{5}{6},~\dfrac{1}{8},~\dfrac{3}{8},~\dfrac{5}{8},~\dfrac{7}{8}.$

Answer from: Quest

The year before the 10s so it was like when caveman were alive

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