Finding standard deviation from a normal distribution

Question

This is for a homework assignment, I'm really only confused about one step within the problem though.

Question: Suppose $X \sim N(400, \sigma^2)$ where $\sigma$ is the standard deviation. If we have $P(380 \leq X \leq 420) = 0.99$, what is the standard deviation $\sigma$? (Hint: Use standardization)

So the class came up with this answer:

1) $z = \frac{x - 400}{\sigma} \sim N(0,1)$

2) $P(\frac{-20}{\sigma} \leq z \leq \frac{20}{\sigma}) = 0.99$

3) $1 - 0.99 = 0.01$

4) $\frac{0.01}{2} = 0.005$

5) $1 - 0.005 = 0.995$

6) $\frac{20}{\sigma} = z_{0.995} = 2.576$

7) Solving for $\sigma$ gives you $7.76$

I understand everything except for steps 3, 4, and 5. Mainly why do we decide to subtract $1 - 0.99$ and then why do we randomly decide to divide by $2$? Finally, why would we want to subtract it by $1$ again?

Answer 1

Q-function is defined as $Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty \exp\left(-\frac{u^2}{2}\right) \, du$

$P(−\frac{20}{σ}≤z≤\frac{20}{σ})=0.99$

It implies $\frac{1}{\sqrt{2\pi}} \int_{−\frac{20}{σ}}^{\frac{20}{σ}} \exp\left(-\frac{u^2}{2}\right) \, du = 0.99$

As normal distribution is symmetric, $\frac{1}{\sqrt{2\pi}} \int_{0}^{\frac{20}{σ}} \exp\left(-\frac{u^2}{2}\right) \, du = 0.99 / 2 = 0.495$

Also $\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} \exp\left(-\frac{u^2}{2}\right) \, du = 0.5$

So $\frac{1}{\sqrt{2\pi}} \int_{\frac{20}{σ}}^{\infty} \exp\left(-\frac{u^2}{2}\right) \, du = 0.5-0.495 = 0.005$

Then from table of Q-function you can find $σ$.

Hope my answer clarifies your doubt.

Answer 2

I understand everything except for steps 3, 4, and 5. Mainly why do we decide to subtract 1−0.99 and then why do we randomly decide to divide by 2? Finally, why would we want to subtract it by 1 again?

Standard Normal Tables (or Z-Tables) look up the the cumulative probability ($\alpha$) and report the value $z_\alpha$, such that:

$$\mathsf P(Z\leq z_\alpha)=\alpha$$

So the manipulation in steps 3,4,5 is to obtain the $\alpha$ value you need to look up to find the required $z_\alpha$ value.

Basically you have: $$\begin{align}\mathsf P(-z_\alpha\leq Z\leq z_\alpha)=\beta \tag{given} \\ \mathsf P(Z\leq -z_\alpha)+\mathsf P(-z_\alpha\leq Z\leq z_\alpha)+\mathsf P(Z\geq z_\alpha) &= 1 \tag{total probability} \\\mathsf P(Z\leq -z_\alpha)&=\mathsf P(Z\geq z_\alpha)\tag{symmetry} \\[1ex] \hline & \hline \\[0ex] \therefore \mathsf P(Z\leq z_\alpha) & = 1-\frac{1-\beta}{2} \\[2ex] \beta =0.99 & \implies \alpha = 0.995 \\[2ex] \text{Find}\quad & z_{0.995} \end{align}$$