Solution assignment 02 Estimation

Return to Assignments Estimation

Rule of thumb
If the variable $X$ is normally distributed with mean $\mu$ and standard deviation $sigma$ , in brief $x\simN(\mu,\sigma)$ , sometimes the following rule is used: there is about $68$ % chance that an observation $x$ is in the interval $[\mu-\sigma,\mu+\sigma]$ and about $95$ % chance that it will lie in the interval $[\mu-2\sigma,\mu+2\sigma]$ .
How would you verify this rule?

Solution.
Just compute the probability:
$\displaystyle{P(-\sigma\leq{x}\leq\sigma)=P(-1\leq{Z}\leq1)=2*0.3413=68.26%\approx68}$ %, by using the standard normal variable $\displaystyle{Z=\frac{X-\mu}{\sigma}}$ and Table 3. A similar computation leads to $\displaystyle{P(-2\sigma\leq{x}\leq2\sigma)=P(-2\leq{Z}\leq2)\approx0.95}$ .

Return to Assignments Estimation

Solution assignment 02 Estimation

Contact

Tutor Math, Statistics or GMAT?

Language preference