# Modeling data distribution
### **z-score** - how many $$\sigma$$away from the mean $$\mu$$
example value 65
$$\frac{value - \mu}{\sigma} \Longrightarrow \frac{65-81}{6.3}=-2.54$$
A z-score measures exactly how many standard deviations above or below the mean a data point is. Here's the formula for calculating a z-score:
$$z = \frac{data point−mean}{standard deviation} \Longrightarrow z = \frac{x - \mu}{\sigma}$$
Here are some important facts about z-scores:
* A positive z-score says the data point is above average.
* A negative z-score says the data point is below average.
* A z-score close to 000 says the data point is close to average.
* A data point can be considered unusual if its z-score is above 333 or below -3−3minus, 3.
### **Standard Deviation and IQR change only with** multiplication and devision, but not with addition and subtraction. The mean and Median do change either way.
### Normal distribution: Empirical Rule (68-95-99.7%)
### Standard normal distribution:
$$\mu = 0 (\text{mean})\newline \sigma = 1 (\text{standard deviation})$$
### What is a normal distribution?
Early statisticians noticed the same shape coming up over and over again in different distributions—so they named it the normal distribution.
Normal distributions have the following features:
* symmetric bell shape
* mean and median are equal; both located at the center of the distribution
* ≈ 68% of the data falls within 1 standard deviation of the mean
* ≈ 95% of the data falls within 2 standard deviations of the mean
* ≈ 99.7% of the data falls within 3 standard deviations of the mean
### Quiz
A set of average city temperatures in August are normally distributed with a mean of $$21.25 ^\circ$$C and a standard deviation of $$2 ^\circ$$C.
$$\large \frac{value - \mu}{\sigma} = \text{z-score}$$
**What proportion of temperatures are between** $$19.63^\circ$$ **C and** $$20.53^\circ$$ **C?**\
\&#xNAN;*You may round your answer to four decimal places.*
1. Let's find the z-score for$$19.63^\circ$$C and $$20.53^\circ$$C:
$$z\_1 = \frac{19.63 - 21.25}{2} = \frac{-1.62}{2} = -0.81$$
$$z\_2 = \frac{20.53 - 21.25}{2} = \frac{-0.72}{2} = -0.36$$
2. We want to find the proportion of temperatures between these two z-scores:
$$z\_1z\_2$$
3. Looking up $$z\_1 = -0.81$$ on the z-table, we see that $$0.2090$$ of temperatures are **below** $$\blueD{19.63}^\circ$$C:
$$z\_1$$
4. Looking up $$z\_2 = -0.36$$ on the z-table, we see that $$0.3594$$ of temperatures are **below** $$\goldD{20.53}^\circ$$C:
$$z\_2$$
5. To find the area between $$z\_1$$ and $$z\_2$$we can subtract the area below $$z\_1$$ from the area below $$z\_2$$
$$0.3594 - 0.2090 = \greenD{0.1504}$$
$$z\_1z\_2$$
6. The answer: $$\greenD{0.1504}$$
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