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GENERATE WORK

GENERATE WORK

**Input Data : **

Random Value (X) = 9.25

Mean (μ) = 9

Standard Deviation (σ) = 0.73

**Objective : **

Find what is the normalized score of random member X?

**Formula **

z-score = (x - μ)σ

**Solution :**

z-score = (9.25 - 9)0.73

= 0.250.73

z-score = 0.3425

**Z-score calculator** calculates the standardized score for any raw value of $X$. It's an online statistics and probability tool requires an unstandardized raw value, the population or sample mean, and the *standard deviation*. The result will describe the measure of the *number of standard deviations between a value and the mean*.
It is necessary to follow the next steps:

- Enter an unstandardized raw value, the population mean, and the standard deviation of the population in the box. These values must be real numbers and may be separated by commas. The values can be copied from a text document or a spreadsheet.
- Press the "
**GENERATE WORK**" button to make the computation. - z-score calculator will give the standard score for a data point.

z-score of a population data is determined by the formula $$z=\frac{x-\mu}{\sigma}$$ where $x$ is a random member, $\mu$ is an expected mean of population and $\sigma$ is the standard deviation of an entire population.

z-score of a sample data is determined by the formula $$z=\frac{x-\bar X}{s_X}$$ where $x$ is each value in the data set, $\bar{X}$ is the sample mean and $s_X$ is the sample standard deviation.

In statistics and probability, it is also called standard score, z-value, standardized score or normal score. A z-score measures the distance between an observation and the mean, measured in units of standard deviation. In other words, z-score is the number of standard deviations there are between a given value and the mean of the set.
If a z-score is zero, then the data point's score is identical to the mean. If a z-score is 1, then it represents a value that is one standard deviation from the mean. z-score may be positive or negative. A positive value represents the score above the mean and a negative score represents the score below the mean. It is used to describe __the normal distribution.__

The standard normal distribution is the normal distribution with mean $0$ and standard deviation $1$. It is denoted as $N(0,1)$.
A random variable with the standard normal distribution is called a __standard normal variable__. Specific values of a standard normal variable are called z-values.
If $X$ is a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$, then the distribution of $$Z =\frac{X-\mu}{\sigma}$$ is the standard normal distribution.
For a specific value $x$ of $X$, $\frac{x-\mu}{\sigma}$ is called the z-value. The z-value is a specific value of $Z$.

To calculate a z-score, divide the deviation by the standard deviation, i.e. $$z=\frac{\mbox{Deviation}}{\mbox{Standard Deviation}}$$ Since the deviation is the observed value of the variable, subtracted by the mean value, then the z-score is determined by the formula $$z=\frac{x-\mu}{\sigma}$$ In the following, we will give a stepwise guide for calculation the z-score for the given population.

- Find the mean of the population;
- Find the standard deviation of the population;
- Chose a random sample from the population;
- Find the difference between the value of random variable and population mean;
- Find the ratio between the difference and population standard deviation.

The class of five students scored $68, 75, 81, 87,$ and $90$. Find the normalized or z-score of $75$.

Following the previously exposed steps, we obtain

- $\mu =\frac{68 + 75 + 81 + 87 + 90}{5}= 80.2$
- $\sigma=\sqrt{\frac{(68 - 80.2)^2 + ( 75 - 80.2)^2 + ( 81 - 80.2)^2 + ( 87 - 80.2)^2 + ( 90 - 80.2)^2)}{5}}= 7.98498$
- By selecting $75$ as a random member from the population of $68, 75, 81, 87$, and $90$;
- The difference between the value of random variable and population mean is
$$75 - 80.2= -5.2$$
- The ratio between the difference and population standard deviation is
$$z-\mbox{score} =\frac{x -\mu}{\sigma}= \frac{-5.2}{7.98}= -0.6516$$

$$\mbox{Sample Standard Deviation}=s_X=\sqrt{\frac{1}{n-1}\sum_{i=1}^{i=n}(x_i-\bar{X})^2},$$

where $\bar{X}$ is the sample mean.
$$\mbox{Population Standard Deviation}=\sigma_X=\sqrt{\frac{1}{N}\sum_{i=1}^{i=N}(x_i-\mu)^2},$$

where $\mu$ is the population mean.
The z-score of a sample is determined by the formula
$$z=\frac{x-\bar{X}}{s_X}$$
where $\bar{X}$ is the sample mean and $s_X$ is the sample standard deviation.The z-Score calculator, formulas, solved example with step by step calculation to find the normalized, standard or relative standing value of a random variable of the normal distribution, calculated from the population of $68, 75, 81, 87$ and $90$. For any other values of random member, mean and standard deviation, just supply three real numbers and click on the "GENERATE WORK" button. The grade school students may use this Z-score calculator to generate the work, verify the results derived by hand or do their homework problems efficiently.

The standard score transformation is useful to compare the relative standings between the members of the distribution with population mean and standard deviation. In other words, z-score determines how many standard deviations $\sigma$ a raw score of a random variable of the population above or below the population mean. Standard score is an important factor in statistics and probability to identify which random member in the normal distribution performing good, bad or moderate.

The graph of a standard normal distribution is called the __standard normal curve__. z-scores can be used to calculate probability by comparing the location of the z-score to the area under a normal curve either to the left or right. A standard normal distribution has the following properties:

- The normal curve is bell shaped and is symmetric about the mean;
- The total area under the normal curve is equal to one.

**Practice Problem 1:**

Find the z-score of a value of $35$, if a mean of data set is $25$, and a standard deviation is $2$.

**Practice Problem 2:**

Find the z-score of the price of a ball that cost $\$27$, if the mean ball price is $\$25.2$, with a standard deviation of $\$1.6$.

**Practice Problem 3:**

Given a normal distribution with a mean of $237$ and standard deviation of $56$, find a value with a z-score of $1.94.$

**Practice Problem 4:**

In a population that is normally distributed with mean $7$ and standard deviation $11$, the bottom $90\%$ of the values are those less than $x$. Find the value of $x$.

**Practice Problem 5:**

Find z-values correspond to the middle $60\%$ of the standard normal distribution.