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I'm doing my homework and i dont have the answers, can someone say if its correct?

"Let the population X, defined by X~Ber(p);

Find The sampling distribution of an RV. X¯=1n∑ni=1Xi where Xi is the obtained value from an extraction with reposition of this population X. Use N=5 And P=0.3." $$T=\sum_{i=1}^{n}Xi$$ $$E[T] = n*p$$

$$E[\bar{X}] = E[T]/n$$ $$E[\bar{X}] = \frac{0.3*5}{5} = 0.3$$

Based on histogram, i find that the "truth" mean expectation is 1.5.

Finding the $P(|e| < 0.2)$ Where e = sampling error of the mean = $1.5-0.3 = 1.2$ so, $P(1.2 < 0.2) = 0$

Is that correct?

PlayMa256
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1 Answers1

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As you found, the random variable $\bar{X}$ has mean $0.3$. The sampling error of the mean is $\bar{X}-0.3$. We are asked for the probability that the sampling error has absolute value $\lt 0.2$, so we want to find $\Pr(|\bar{X}-0.3|\lt 0.2)$. Note that $$ \Pr(|\bar{X}-0.3|\lt 0.2)=\Pr(0.1 \lt \bar{X}\lt 0.5).\tag{1}$$ Since $\bar{X}=\frac{X_1+X_2+X_3+X_4+X_5}{5}=\frac{T}{5}$, the probability in (1) is equal to
$$\Pr(0.5\lt T\lt 2.5).\tag{2}$$ But $T$ only takes on integer values, so we want $\Pr(T=1)+\Pr(T=2)$.

Recall that $T$ has Binomial distribution with parameters $n=5$ and $p=0.3$. Thus $\Pr(T=1)=\binom{5}{1}(0.3)^1(0.7)^4$. You can find a similar expression for $\Pr(T=2)$. Add.

André Nicolas
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  • and to know the squared mean error? – PlayMa256 Jun 18 '14 at 19:06
  • You may have intended mean squared error, aka variance. Depends whether it is of $T$ or of $\bar{X}$. If it is of $\bar{X}$, it will be $\frac{p(1-p)}{n}$, where $p=0.3$ and $n=5$. Comments are too short for full discussion. – André Nicolas Jun 18 '14 at 19:14
  • in the question is MSE($\bar{X},\mu$), so, MSE = E($\bar{X} - \mu)^2)$ – PlayMa256 Jun 18 '14 at 19:19
  • Yes, they are are asking for the variance of $\bar{X}$. – André Nicolas Jun 18 '14 at 19:22
  • i dont know if i found correctly, based on bernoulli distribution i find the variance = 0,21 and the mean = 0,3, on binomial distribution looking on the histogram and calculatin the mean = 1,5 and variance = 1,05. Which one is correctly? – PlayMa256 Jun 18 '14 at 20:48
  • The mean of $\bar{X}$ is $0.3$, as you knew earlier. The variance of $\bar{X}$ is $0.042$. The mean of $T$ is $1.5$, and the variance of $T$ is $0.21$. – André Nicolas Jun 18 '14 at 21:30