Standardization of Normal Variables (Z-values)
In the study of normal distributions, standardization allows us to compare and analyze different normal distributions on a common scale. This is achieved through the use of z-values, also known as z-scores or standard scores.
A z-value represents the number of standard deviations a data point is from the mean of its distribution. The formula for calculating a z-value is:
$$ z = \frac{x - \mu}{\sigma} $$
Where:
- $x$ is the raw score (data point)
- $\mu$ is the mean of the distribution
- $\sigma$ is the standard deviation of the distribution
Suppose we have a normal distribution with a mean of 100 and a standard deviation of 15. If we want to find the z-score for a value of 130, we would calculate:
$z = \frac{130 - 100}{15} = 2$
This means that 130 is 2 standard deviations above the mean.
Z-scores can be positive or negative. A positive z-score indicates that the value is above the mean, while a negative z-score indicates that the value is below the mean.
Interpreting Z-scores
Z-scores provide a standardized way to understand how far a data point is from the mean in terms of standard deviations. This is particularly useful when comparing values from different normal distributions.
- A z-score of 0 means the data point is exactly at the mean.
- A z-score of 1 means the data point is one standard deviation above the mean.
- A z-score of -1 means the data point is one standard deviation below the mean.
Students often forget that z-scores represent distances from the mean in terms of standard deviations, not raw units. A z-score of 2 doesn't mean "2 units above the mean," but rather "2 standard deviations above the mean."
Finding Probabilities
To find the probability that a value falls below a certain point in a normal distribution, we use the cumulative distribution function (CDF). On most graphing calculators, this is represented by the normalcdf function.
Using a normal distribution with mean 70 and standard deviation 5, find the probability that a randomly selected value is less than 75.
On a TI-84 calculator: normalcdf(-99999, 75, 70, 5) ≈ 0.8413
This means there's about an 84.13% chance that a randomly selected value is less than 75.
Finding Values (Inverse Normal)
The inverse normal function is used to find a value in a normal distribution given a probability. On most graphing calculators, this is represented by the invNorm function.
Find the value that is at the 90th percentile of a normal distribution with mean 100 and standard deviation 15.
On a TI-84 calculator: invNorm(0.90, 100, 15) ≈ 119.2
This means that 90% of the values in this distribution are below 119.2.
Inverse Normal Calculations with Unknown Parameters
Sometimes, we need to perform inverse normal calculations when the mean and standard deviation are not explicitly given. In these cases, we can use z-scores and percentiles to determine the unknown parameters.
Suppose we know that in a normal distribution, the 16th percentile is 42 and the 84th percentile is 58. Find the mean and standard deviation of this distribution.
- The 16th and 84th percentiles correspond to z-scores of -1 and 1 respectively.
- Let $\mu$ be the mean and $\sigma$ be the standard deviation.
- We can set up two equations: $42 = \mu - \sigma$ (16th percentile) $58 = \mu + \sigma$ (84th percentile)
- Adding these equations: $100 = 2\mu$ $\mu = 50$
- Subtracting the first equation from the second: $16 = 2\sigma$ $\sigma = 8$
Therefore, the mean is 50 and the standard deviation is 8.
Using Z-values to Calculate Unknown Means and Standard Deviations
Z-values can be particularly useful when we need to find unknown means or standard deviations in a normal distribution.
Finding Unknown Mean
If we know a z-score, the corresponding x-value, and the standard deviation, we can find the mean:
$$ \mu = x - z\sigma $$
If a z-score of 1.5 corresponds to a value of 85 in a normal distribution with a standard deviation of 10, what is the mean?
$\mu = 85 - (1.5 \times 10) = 85 - 15 = 70$
The mean of the distribution is 70.
Finding Unknown Standard Deviation
If we know a z-score, the corresponding x-value, and the mean, we can find the standard deviation:
$$ \sigma = \frac{x - \mu}{z} $$
If a z-score of -2 corresponds to a value of 60 in a normal distribution with a mean of 80, what is the standard deviation?
$\sigma = \frac{60 - 80}{-2} = \frac{-20}{-2} = 10$
The standard deviation of the distribution is 10.
When solving problems involving z-scores and normal distributions, always start by clearly identifying what information you're given and what you need to find. Then, choose the appropriate formula or method based on the available information.
