W06 - Examples

Normal distribution

Basic generalized normal calculation

Basic generalized normal calculation

Suppose $X\sim \mathcal{N}(-3,4)$. Find $P[X\ge -1.7]$.

Solution

First write $X$ as a linear transformation of $Z$:

$$X\sim 2Z-3$$

Then:

$$X\ge -1.7⟺Z\ge 0.65$$

Look in a table to find that $\Phi (0.65)\approx 0.74$ and therefore:

$$\begin{array}{c}P[Z\ge 0.65]\gg \gg 1-P[Z\le 0.65]\\ \\ \gg \gg \approx 1-0.74\gg \gg 0.26\end{array}$$ Link to original

Gaussian: interval of 2/3

Gaussian: interval of 2/3

Find the number $a$ such that $P[-a\le Z\le +a]=2/3$.

Solution

First convert the question:

$$\begin{array}{rl}P[-a\le Z\le +a]& \gg \gg F_Z(a)-F_Z(-a)\\ & \\ & \gg \gg \Phi (a)-\Phi (-a)\\ & \\ & \gg \gg 2\Phi (a)-1\end{array}$$

Solve for $a$ so that this value is $2/3$:

$$2\Phi (a)-1=2/3\gg \gg \Phi (a)=5/6\gg \gg a={\Phi }^{-1}(5/6)$$

Use a $\Phi$ table to conclude $a\approx 0.97$.

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Heights of American males

Heights of American males

Suppose that the height of an American male in inches follows the normal distribution $\mathcal{N}(71,6.25)$.

(a) What percent of American males are over 6 feet, 2 inches tall?

(b) What percent of those over 6 feet tall are also over 6 feet, 5 inches tall?

Solution

(a)
Let $H$ be a random variable measuring the height of American males in inches, so $H\sim \mathcal{N}(71,2.5^2)$. Thus $H\sim 2.5Z+71$, and:

$$\begin{array}{rl}P[H>74]& \gg \gg 1-P[H\le 74]\\ & \\ & \gg \gg 1-P[2.5Z+71\le 74]\\ & \\ & \gg \gg 1-P[Z\le 1.20]\\ & \\ & \gg \gg 1-0.8849\approx 11.5\%\end{array}$$

(b)
We seek $P[H>77\mid H>72]$ as the answer. Compute as follows:

$$\begin{array}{rl}P[H>77\mid H>72]& =\frac{P[H>77]}{P[H>72]}\\ & \\ & \gg \gg \frac{P[2.5Z+71>77]}{P[2.5Z+71>72]}\\ & \\ & \gg \gg \frac{1-P[Z\le 2.4]}{1-P[Z\le 0.4]}=\frac{1-0.9918}{1-0.6554}\approx 2.38\%\end{array}$$ Link to original

Variance of normal from CDF table

Variance of normal from CDF table

Suppose $X\sim \mathcal{N}(5,{\sigma }^2)$, and suppose you know $P[X>9]=0.2$.

Find the approximate value of $\sigma$ using a $\Phi$ table.

Solution

$$X\sim \mathcal{N}(5,{\sigma }^2)⟹X\sim \sigma Z+5$$

So $1-P[X\le 9]=0.2$ and thus $P[\sigma Z+5\le 9]=0.8$. Then:

$$P[\sigma Z+5\le 9]=P[Z\le 4/\sigma ]$$

so $P[Z\le 4/\sigma ]=0.8$.

Looking in the chart of $\Phi$ for the nearest inverse of 0.8, we obtain $4/\sigma =0.842$, hence $\sigma =4.75$.

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