# z-score

(noun)

## Definition of z-score

The standardized value of observation x from a distribution that has mean μ and standard deviation σ.

Source: Boundless Learning - CC BY-SA 3.0

## Examples of z-score in the following topics:

• ### Z-Scores and Location in a Distribution

• A z-score is the signed number of standard deviations an observation is above the mean of a distribution.
•  Thus, a positive z-score represents an observation above the mean, while a negative z-score represents an observation below the mean.
• We obtain a z-score through a conversion process known as standardizing or normalizing.Z-scores are also called standard scores, z-values, normal scores or standardized variables.
•  The z-score, in turn, provides an assessment of how off-target a process is operating.The conversion of a raw score, x, to a z-score can be performed using the following equation:$z=\frac { x-\mu }{ \sigma }$where μ is the mean of the population and σ is the standard deviation of the population.
• Z is negative when the raw score is below the mean and positive when the raw score is above the mean.A key point is that calculating z requires the population mean and the population standard deviation, not the sample mean nor sample deviation.
• z-score is the signed number of standard deviations an observation is above the mean of a distribution.
• ### Calculating a Normal Approximation

• This can be done by finding z-scores and using the z-score table, which can be seen in .
• If we did not have the normal area calculator, we could find the solution using a table of the standard normal distribution (a z-table) as follows:Find a Z score for 7.5 using the formula $Z=\frac { 7.5-5 }{ 1.5811 } =1.5811$Find the area below a Z of 1.58 = 0.943.Find a Z score for 8.5 using the formula $Z=\frac { 8.5-5 }{ 1.5811 } =2.21$Find the area below a Z of 2.21 = .987.Subtract the value in step 2 from the value in step 4 to get 0.044.The same logic applies when calculating the probability of a range of outcomes.
• ### The Standard Normal Curve

• Luckily, one can transform any normal distribution with a certain mean μ and standard deviation σ into a standard normal distribution, by the z-score conversion formula:$z=\frac { x-\mu }{ \sigma }$ Therefore, a z-score is the standardized value of observation x from a distribution that has mean μ and standard deviation σ (how many standard deviations you are away from zero).
• The z-score gets its name because of the denomination of the standard normal distribution as the "Z" distribution.
• Since x = 70.4, μ = 64 inches and σ = 2.5 inches, we need to calculate z: $z=\frac { 70.4-64 }{ 2.5 } =\frac { 6.4 }{ 2.5 } =2.56$ Therefore, the probability P(X>70.4) is equal to P(Z>2.56), where X is the normally distributed height with mean μ = 64 inches and standard deviation σ = 2.5 inches {X~N(64,2.5) for short}, and Z is a standard normal distribution {Z~N(0,1)}.
• The next step requires that we use what is known as the z-score table to calculate probabilities for the standard normal distribution.
• Therefore: P(60.3<X<65) = P(-1.48<Z<0.40) = P(Z<0.40)-P(Z<-1.48) = (0.5+0.1554)-(0.5-0694) = 0.6554-0.0694 = 0.5860.
• ### Finding the Areas under the Normal Curve

• In order to do this, we use a z-score table, as shown in (same as in the process of standardization discussed in the previous section).
• The following is another simple example: find P(Z≥1.17).
• Since all the probabilities must sum to 1: P(Z>1.17) = 1 − P(Z<1.17) = 0.121 As a final example: find P(−1.16≤Z≤1.32).
• P(−1.16≥Z≤1.32) = P(Z≤1.32) − P(Z≤−1.16) The difficulty arrises from the fact that our table of values does not allow us to directly calculate P(Z≤−1.16).
• However, we can use the symmetry of the distribution, as follows: P(Z≤−1.16) = 1 − P(Z≤1.16) = 0.1230 So, we can say that: P(−1.16≤Z≤1.32) = 0.9066 − 0.1230 = 0.7836.
• ### Change of Scale

• Thus, a positive standard score represents a datum above the mean, while a negative standard score represents a datum below the mean.
• The use of "Z" is because the normal distribution is also known as the "Z distribution".
• They are most frequently used to compare a sample to a standard normal deviate (standard normal distribution, with μ = 0 and σ = 1).The z-score is only defined if one knows the population parameters.
• If one only has a sample set, then the analogous computation with sample mean and sample standard deviation yields the Student's t-statistic.The standard score of a raw score x is:$z=\frac { x-\mu }{ \sigma }$,where μ is the mean of the population, and σ is the standard deviation of the population.
•  The absolute value of z represents the distance between the raw score and the population mean in units of the standard deviation. z is negative when the raw score is below the mean, positive when above.A key point is that calculating z requires the population mean and the population standard deviation, not the sample mean or sample deviation.
• ### Wilcoxon t-Test

• Thus,For ${ N }_{ r }\ge 10$, a z-score can be calculated as ).If z > zcritical then reject H0.For Nr < 10, W is compared to a critical value from a reference table.If $W\ge { W }_{ critical,{ N }_{ r } }$ then reject H0.Alternatively, a p-value can be calculated from enumeration of all possible combinations of W given Nr.
• ### Using the Normal Curve

• Z-ValueThe functional form for a normal distribution is a bit complicated.
• The table only gives the probabilities to the left of the z-value.
• For example, if we wanted to find out the probability that a variable is more than 0.51 sigmas above the mean, P(z>0.51), we just need to calculate 1-P(z<0.51)=1-0.6950=0.3050, or 30.5%.There is another note of caution to take into consideration when using the table: The table provided only gives values for positive z-values, which correspond to values above the mean.
• What if we wished instead to find out the probability that a value falls below a z-value of -0.51, or 0.51 standard deviations below the mean?
• We must remember that the standard normal curve is symmetrical , meaning that P(z<-0.51)=P(z>0.51), which we calculate above to be 30.5%.We may even wish to find the probability that a variable is between two z-values, such as between 0.50 and 1.50, or P(0.50).68-95-99.7 RuleAlthough we can always use the z-score table to find probabilities, the 68-95-99.7 rule helps for quick calculations.
• ### Comparing Two Independent Population Proportions

• To conduct the test, we use a pooled proportion, pc.The pooled proportion is calculated as follows:${ p }_{ c }=\frac { { x }_{ A }+{ x }_{ B } }{ { n }_{ A }+{ n }_{ B } }$The distribution for the differences is:${ P }_{ A }^{ ' }-{ P }_{ B }^{ ' }\sim N\left[ 0,\sqrt { { p }_{ c }\cdot (1-{ p }_{ c })\cdot \left( \frac { 1 }{ { n }_{ A } } +\frac { 1 }{ { n }_{ B } } \right) } \right]$.The test statistic (z-score) is:$z=\frac { { (p }_{ A }^{ ' }-{ p }_{ B }^{ ' })-({ p }_{ A }-{ p }_{ B }) }{ \sqrt { { p }_{ c }\cdot (1-{ p }_{ c })\cdot \left( \frac { 1 }{ { n }_{ A } } +\frac { 1 }{ { n }_{ B } } \right) } }$.ExampleTwo types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. 20 out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. 12 out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication.
• ### Estimates and Sample Size

• To calculate E, we need to know the desired confidence level (${ Z }_{ \frac { \alpha }{ 2 } }$) and the population standard deviation, $\sigma$.
• $E={ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } }$To change the size of the error (E), two variables in the formula could be changed: the level of confidence (${ Z }_{ \frac { \alpha }{ 2 } }$) or the sample size (n).
• To determine this, begin by solving the equation for the E in terms of n:$n={ \left( \frac { { Z }_{ \frac { \alpha }{ 2 } }\sigma }{ E } \right) }^{ 2 }$where ${ Z }_{ \frac { \alpha }{ 2 } }$ is the critical z score based on the desired confidence level, E is the desired margin of error, and $\sigma$ is the population standard deviation.Since the population standard deviation is often unknown, the sample standard deviation from a previous sample of size n ≥ 30 may be used as an approximation to s.
•  In this case, the margin of error, E, is found using the formula:$E={ Z }_{ \frac { \alpha }{ 2 } }\sqrt { \frac { p'q' }{ n } }$where:p' = x/n is the point estimate for the population proportionx is the number of successes in the samplen is the number in the sample; andq' = 1 - p'Then, solving for the minimum sample size n needed to estimate p:$n=p'q'\left( \frac { { Z }_{ \frac { \alpha }{ 2 } } }{ E } \right) ^{ 2 }$ExampleThe Mesa College mathematics department has noticed that a number of students place in a non-transfer level course and only need a 6 week refresher rather than an entire semester long course.