# Z confidence interval

$$W_1, \ldots, W_5$$ be iid $$Normal(3.1, 5.2^2)$$. What is the probability that $$\left(\frac{\bar W - 3.1}{\hat \sigma / \sqrt{5}} \geq 2.1\right)$$?

I am not sure how to start. Is this equivalent to the probability that $$Z \leq 2.1$$?

You're close, but it's actually $$1-P[t \le 2.1]$$.

The distribution of $$\bar{W} = \frac{1}{5}\sum_i^5 {W_i}$$ is

$$\bar{W} \sim \mathscr{N} \left( 3.1, \frac{5.2^2}{5} \right) .$$

Normally, $$(\bar{W} - 3.1)/\left( \frac{5.2^2}{\sqrt{5}}\right)$$ would be the standardization of $$\bar{W}$$ into a standard normal $$Z \sim \mathscr{N}(0,1)$$. Because of the use of $$\hat{\sigma}$$ in the question, this is presumably coming from a sample of $$5$$ observations of $$W$$, and we are using a $$t$$-distribution with $$4$$ degrees of freedom, not a standard normal.

So, we can rewrite the probability we want in terms of $$t$$:

$$P \left[ \frac{\bar{W} - 3.1}{\frac{\sigma}{\sqrt{5}}} \ge 2.1 \right]$$ $$= 1 - P \left[ \frac{\bar{W} - 3.1}{\frac{\sigma}{\sqrt{5}}} \le 2.1 \right]$$ $$= 1 - P \left[ t \le 2.1 \right] .$$ $$\approx 0.052 .$$

• Can you explain how you got 0.052 from the second to last step. (1-P[t <= 2.1]) – Jeremy Feb 25 at 2:43
• You can't calculate it by hand, you have to use a numeric solver (available in any statistical software or Excel) or a t-table. Here it is using WolframAlpha. – Amaan M Feb 25 at 18:59