# Intuition behind $2\Phi(x)-1$

When using the central limit theorem to calculate for $$n$$ I come across having to use $$2\Phi(x)-1$$ to find this. However, I'm unsure why I have to use this and what it means and how it's derived. It just seems to be rather common as an qpproach to follow when trying to derive $$n$$, what's the explanation behind $$2\Phi(x)-1$$?

where symbol $$\Phi(x)$$ denotes the cumulative distribution function of a standard normal variable.

For example: Suppose that a random sample of size $$n$$ is to be taken from a distribution for which the mean is $$\mu$$ and the standard deviation is $$3$$. Use the central limit theorem to determine approximately the smallest value of $$n$$ for which the following relation will be satisfied: $$Pr(|\bar{X}_n-\mu|<0.3)\ge 0.95$$

By following the $$Z$$ distribution: $$Z = \frac{\sqrt(n)(\bar{X}_n-\mu)}{\sigma}$$ I can get: $$Pr\left(\frac{\sqrt(n)(\bar{X}_n-\mu)}{3}<0.3\right)=Pr(|\bar{X}_n-\mu|<0.1\sqrt{n})\ge 0.95$$

For the next step I do: $$Pr(|\bar{X}_n-\mu|<0.1\sqrt{n})\approx2\Phi(0.1\sqrt{n})-1\ge 0.95$$

Then re-arranging to find $$n$$.

So what's the intuition behind using $$2\Phi(x)-1$$?

• You seem to think $Pr\left(\frac{\sqrt{n}(\bar{X}_n-\mu)}{3}<0.3\right)=Pr(|\bar{X}_n-\mu|<0.1\sqrt{n}).$ Cam you explain why? Sep 3, 2021 at 15:08
• @BruceET Perhaps it makes more sense saying $P(|Z| < 0.1\sqrt{n})$ given that we already know $Pr(|\bar{X}_n-\mu|<0.3)$ Sep 3, 2021 at 15:46

For a standard normal random variable $$Z \sim N(0,1)$$ and a positive real number $$\alpha$$, \begin{align} P(|Z| \leq \alpha) &= P(-\alpha \leq Z \leq \alpha)\\ &= \Phi(\alpha) - \Phi(-\alpha)\tag{1}\\ &= \Phi(\alpha) - [1 - \Phi(\alpha)]\tag{2}\\ &= 2\Phi(\alpha) - 1 \tag{3} \end{align} where about the only place where any sort of intuition might have been used is in recognizing that symmetry of the standard normal pdf about $$0$$ allows us to recognize (or intuit) that $$\Phi(-\alpha)$$, the area under the pdf to the left of the point $$-\alpha$$, must equal $$1-\Phi(\alpha)$$, the area under the pdf to the right of the point $$\alpha$$, and so $$(2)$$ follows from $$(1)$$. In particular, note that tables of $$\Phi(\cdot)$$ show that $$\Phi(1.96) = 0.9750$$ and so $$P(|Z| \leq 1.96) = 2\times 0.9750 - 1 = 0.95.$$ Since the right side of $$(3)$$ is an increasing function of $$\alpha$$, choosing $$\alpha$$ to be larger that $$1.96$$ will certainly give us a probability larger than $$0.95$$.
Next, given a random sample $$X = (x_1, x_2, \ldots,x_n)$$ from a distribution with known mean $$\mu$$ and known standard deviation $$\sigma$$, it is known that the sample mean $$\bar{X}$$ is a random variable with mean $$\mu$$ and standard deviation $$\dfrac{\sigma}{\sqrt{n}}$$ and so genuflections in the general direction of the CLT allows us to claim that $$\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{\sqrt{n}(\bar{X} - \mu)} {\sigma} \sim Z \sim N(0,1)$$ or at least when $$n$$ is "large".