Z-Score

Tags

PAS

Cegep/2

Word count

273 words

Reading time

2 minutes

Distance of a measurement from the mean in standard deviations
Measure of relative standing
Sym. $Z$ (as a random variable)

Sample z-score:

$$z=\frac{x-\bar{x}}{s}$$

Population z-score:

$$z=\frac{x-\mu }{\sigma }$$

==Follows standard normal distribution:==

$$\begin{array}{rl}Z& \sim N(0,1)\\ E(Z)& =0\\ Var(Z)& =1\end{array}$$

[!abstract] $z_a$
$z$ such that an area of $a$ lies to its right, i.e. $P(Z>z_a)=a$

Examples

Suppose the scores, X, on a college entrance exam are normally distributed with mean of 550 and standard deviation of 100. A prestigious university will only consider admitting applicants whose scores exceed the 90th percentile of the distribution. Find the minimum score an applicant must achieve in order to receive consideration for admission to the university.

$$\begin{array}{rl}P(X\le x_0)& =0.9\\ ⟺P(\frac{X-\mu }{\sigma }\le \frac{x_0-\mu }{\sigma })& =0.9\\ ⟺P(Z\le \frac{x_0-550}{100})& =0.9\\ ⟺P(Z>\frac{x_0-550}{100})& =0.1\\ ⟺P(Z<-\frac{x_0-550}{100})& =0.1\\ -\frac{x_0-550}{100}& =-1.28\\ x_0& =678\end{array}$$

Using z-score, derive a $1-a$ confidence interval for $p$ by taking $n$ random samples.

$$\begin{array}{rl}P(-z_{\frac{\alpha }{2}}<Z<z_{\frac{\alpha }{2}})& =1-\alpha \\ ⟺P(-z_{\frac{\alpha }{2}}<\frac{\hat{P}-p}{\sqrt{\frac{p(1-p)}{n}}}<z_{\frac{\alpha }{2}})& =1-\alpha \\ ⟺P(\hat{P}-z_{\frac{\alpha }{2}}\sqrt{\frac{p(1-p)}{2}}<p<\hat{P}+z_{\frac{\alpha }{2}}\sqrt{\frac{p(1-p)}{n}})& =1-\alpha \\ P(\hat{P}-z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}<p<\hat{P}+z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}})& \approx 1-\alpha \end{array}$$

Contributors

Ajitani Hifumi

Changelog

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