Respuesta :
Answer:
(a) The distribution of [tex]X=\sum\limits^{n}_{i=1}{X_{i}}[/tex] is a Binomial distribution.
(b) The sampling distribution of the sample mean will be approximately normal.
(c) The value of [tex]P(\bar X>0.50)[/tex] is 0.50.
Step-by-step explanation:
It is provided that random variables [tex]X_{i}[/tex] are independent and identically distributed Bernoulli random variables with p = 0.50.
The random sample selected is of size, n = 100.
(a)
Theorem:
Let [tex]X_{1},\ X_{2},\ X_{3},...\ X_{n}[/tex] be independent Bernoulli random variables, each with parameter p, then the sum of of thee random variables, [tex]X=X_{1}+X_{2}+X_{3}...+X_{n}[/tex] is a Binomial random variable with parameter n and p.
Thus, the distribution of [tex]X=\sum\limits^{n}_{i=1}{X_{i}}[/tex] 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. Â
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,
[tex]\mu_{\bar x}=\mu=p=0.50[/tex]
And the standard deviation of the distribution of sample mean is given by,
[tex]\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05[/tex]
(c)
Compute the value of [tex]P(\bar X>0.50)[/tex] as follows:
[tex]P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\[/tex]
          [tex]=P(Z>0)\\=1-P(Z<0)\\=1-0.50\\=0.50[/tex]
*Use a z-table.
Thus, the value of [tex]P(\bar X>0.50)[/tex] is 0.50.
