I'm given the following problem. Suppose that a random sample of size n is to be taken from the normal distribution with mean $\mu$ and variance 9.
determine the value n for which PR(|$\bar{X}_n – \mu| \le 1) \ge 0.95$)
so using the theorem $P_k = PR(|X – \mu| \le k\sigma) = PR(|Z| \le k$)
I find that
PR(|$\bar{X}_n – \mu| \le 1) = PR(|Z| \le {\sqrt(n)}/3) \ge 0.95$
Now what do I do from here to find the smallest value of n such that the above is satisfied?
Best Answer
We can use the fact that:
$$\frac{\bar{X}_n-\mu}{\frac{\sigma}{\sqrt{n}}}\sim N(0,1)$$
(as $\sigma$=3) So:
$$P(|\bar{X}_n-\mu|\leq1)=P(\frac{|\bar{X}_n-\mu|}{\frac{\sigma}{\sqrt{n}}}\leq\frac{\sqrt{n}}{\sigma})=P(|Z|\leq \frac{\sqrt{n}}{3})$$
And
$$P(|Z|\leq 1.9599)=0.95$$
So
$$\frac{\sqrt{n}}{3}=1.9599$$ $$n=(3*1.9599)^2=34.57$$
The smallest value of n is 35.