# Determining a sample size - Normal Distributions

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?

• when you write "Suppose that a random variable of size n", should it be "suppose a sample size n"? Commented Dec 17, 2016 at 1:00
• yea i made a mistake Commented Dec 17, 2016 at 1:02

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.

• Thaks for the observation Commented Dec 17, 2016 at 2:54