What is $P\left(|\bar{x}-\mu|\leq 1.5\frac{0.2}{\sqrt{16}}\right)$ equal to?

Question

I'm having a question regarding normalizing the distribution in 2-tailed test.

I have $\mu=3.3$, $\sigma = 0.2$ and $n=16$. I need to determine

$$ P\left(|\bar{x}-\mu|\leq 1.5\frac{0.2}{\sqrt{16}}\right). $$

My work:

$$ P\left(|\bar{x}-\mu|\leq 1.5\frac{0.2}{\sqrt{16}}\right)=P\left(-1.5\frac{0.2}{\sqrt{16}} \leq \bar{x}-\mu\leq 1.5\frac{0.2}{\sqrt{16}}\right)=2\Phi(1.5)-1. $$

My friend's work:

$$ P\left(|\bar{x}-\mu|\leq 1.5\frac{0.2}{\sqrt{16}}\right)=P(|Z|<1.5)=\Phi(1.5). $$

Which one do you think is correct? I think I have to expand the absolute sign, but my friend did not think like that. Any help is appreciated

Answer 1

Your friend's work is not correct for the reason that $$\Pr[Z \le 1.5] = \Phi(1.5),$$ but $$\Pr[|Z| \le 1.5] = \Phi(1.5) - \Phi(-1.5) = \Phi(1.5) - (1 - \Phi(1.5)) = 2\Phi(1.5) - 1.$$

Your computation is correct although perhaps unnecessarily complicated, since $$Z = \frac{\bar x - \mu}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1).$$ This assumes that the sample is in fact drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma$.

Answer 2

Hint:

For $X$ a normal random variable:

$$\mathbb{P}(|X|\leq x) = \mathbb{P}(-x\leq X \leq x) = 2\mathbb{P}(0\leq X \leq x)$$

Given the symmetry of the normal distribution.