Z-score Calculator

Z-score, otherwise known as the standard score, is the number of standard deviations by which a data point is above the mean. You can use our z-score calculator to determine this value for you. Read on to learn how to calculate the z-score and how to use the z-score table.

Z-score is a value used to describe the normal distribution. It is defined as the distance between the mean score and the experimental data point, expressed in terms of SD (standard deviation). In statistical data analysis, it is also called standard score, z value, standardized score, and normal score.

To find the z-score, you first need to calculate a data set's mean and standard deviation. Mean, denoted with the symbol μ, is the sum of all values in the data set divided by the number of data points. It can be written down as μ = ∑x / n. Standard deviation is found according to the expression:

σ = √[∑(x - μ)² / n],

where x stands for raw value and n for the number of data points. To explore the formula in more detail, we recommend checking the dedicated standard deviation calculator.

To find the z-score, you simply need to apply the following formula:

z = (x - μ) / σ

Let's assume the following task: during a test, four students scored 50, 53, 62, and 70 points. What is the z-score of the result 62?

1. Find the mean of the results. μ = (50 + 53 + 62 + 70) / 4 = 58.75. You can also use our average calculator to do it.
2. Calculate the individual values of (x - μ)² for each result:

• (50 - 58.75)² = 76.5625;
• (53 - 58.75)² = 33.0625;
• (62 - 58.75)² = 10.5625; and
• (70 - 58.75)² = 126.5625.

3. Calculate the standard deviation: √[(76.5625 + 33.0625 + 10.5625 + 126.5625) / 4] =√(246.75 / 4) = 7.854 .
4. Input these results to the z-score equation for x = 62: z = (62 - 58.75) / 7.854 = 0.41.
5. You just found the z-score of 62! You can also use the z-score calculator to find the mean or standard deviation if you know the z-score.

A z-score table is where you can find the area to the left of the given z-score under the standard distribution graph. The first column of the table is a list of z-values (accurate to one decimal place). In the first row, you can find the digit that is in the second decimal place of your z-score.

For example, we found the z-score of 62 in our example to be equal to 0.41. First, you need to find z = 0.4 in the first column; this value shows you in which row you need to seek. Then, find the value of 0.01 in the first row. It will determine the row in which you must look. The area under the standard distribution graph (to the left of our z-score) is equal to 0.6591. Remember that the total area under this graph is equal to 1. Hence, we can say that the probability of a student scoring 62 or lower on the test is equal to 0.6591, or 65.91%.

Knowing this area, you can also find the p-value - the probability that the score will be higher than 62. It is simply 1 - 0.6591 = 0.3409 or 34.09%. To learn more about this quantity, head to Omni's p-value calculator.

99.7% of the observations of a process that follows the normal distribution can be found within three standard deviations to the left and right of the mean. Hence, only 0.3% of this process's possible realizations will lay outside of the three sigma interval.

If you try to expand this interval and go six sigmas to the left and right, you will find out that 99.9999998027% of your data points fall into this principle. If this principle is successfully applied, you can expect to have 3.4 defects for every one million realizations of a process.

Such events may be considered as very unlikely: accidents and mishaps, on the one hand, and streaks of luck, on the other. Suppose you perform a repetitive task that can be described by the normal distribution (such as a production of a standardized good) in the long run. In that case, you may expect serious errors to happen so rarely that they become negligible.

This is the reason behind the quality control system based on the standard normal distribution, called the six sigma. Engineered at Motorola in the 1980s, the system uses statistical analysis to measure end eliminate errors.

There are five main elements to this process: a) define, b) measure, c) analyze, d) improve, and e) control. The basic notion is that a process requires a serious correction when it deviates more than three sigma from its mean. In other words, the main objective of your quality management and controls should be to have your production process outcome as close to the normal distribution as possible.

Because of the six sigma methodology, in the last three decades, the normal distribution has been used to enhance processes from manufacturing to transactions, both in factories and offices.