Z scores (Z value) is the number of standard deviations a score or a value (x) away from the mean. In other words, Z-score measures the dispersion of data. Technically, Z-score tells a value (x) is how many standard deviations below or above the population mean (µ). If the Z value is positive, it indicates that the value or score (x) is above the mean. Similarly, if Z value is negative, it means the value (x) is below the mean.

**What is Standard normal distribution?**

A Normal Standard Distribution curve is a symmetric distribution where the area under the normal curve is 1 or 100%. The standard normal distribution is a type of special normal distribution with a mean (µ) of 0 and a standard deviation of 1.

A standard normal distribution always mean zero and has intervals that increase by 1. Each number on the horizontal line corresponds to z-score. Hence, use Z Scores to transform a given standard distribution into something that is easy to calculate probabilities on. As it can determine the likelihood of some event happening.

Any normal distribution with any value of mean (µ) and sigma can be transformed into the standard normal distribution, where the mean of zero and a standard deviation of 1. This is also called standardization.

A Z-scores tells how many standard deviation a value or score is from the mean (µ). For example if a Z-score negative 3 means the value (x) is 3 standard deviation left of the mean. Similarly, if the Z-score is positive 2.5 means the value (x) is 2.5 standard right of the mean (µ).

This is a common transformation, so there is a reference chart that allows us to look up values. Those values correlate to the value under the normal distribution curve – in other words, what’s the chance of an event happening. We use the Z table to find the percent chance.

**How to interpret the Z-score table**

Most importantly, Z-score helps to calculate how much area that specific Z-score is associated with. A Z-score table is also known as a standard normal table used to find the exact area. Z-score table tells the total quantity of area contained to the left side of any score or value (x).

There are two methods to read the Z-table:

**Case 1**: Use Z-table to see the area under the value (x)

In the Z-table top row and the first column corresponds to the Z-values and all the numbers in the middle corresponds to the areas.

For example a Z-score of -1.53 has an area of 0.0630 to the left of it. In other words, p(Z<-1.53) = 0.0630.

Standard normal table also used to determine the area to the right of any Z-value by subtracting the area on the left from 1. Simply, 1-Area_{Left} = Area_{right}

For example Z-score of 0.83 has an area of 0.7967 to the left of it. So, Area of right is 1-0.7967 = 0.2033.

**Case 2**: Use the Z-table to see what that score is associated with a specific area.

P(o<=Z<=x)

- Pick the right Z row by reading down the right column.
- Read across the top to find the decimal space.
- Finally, find the intersection and multiply by 100.

For example: Value of Z corresponds to an area of 0.9750 to the left of it is 1.96.

## How to create Z score in excel

Using Z score, find the percentage by using the formula: 1-NORMSDIST(Z), where Z is your calculated Z Score.

### What if the Z Score is off the Chart?

Z score tables sometimes only go up to 3. But depending on the spread of the population, z scores could go on for a while. A Z score of 3 refers to 3 standard deviations. That would mean that more than 99% of the population was covered by the z score. There’s not a lot left, but there is some. Use Excel to find the actual value if your table doesn’t go that high.

## How do you Calculate a Z Score?

The formula for transforming a score or observation x from any normal distribution to a standard normal score is :

### Calculating a Z Score for a Population

### Z Score for a Sample

Great example of using a Z score to determine how well you scored on a test (compared to the rest of the field) here.

How many parts in a population will be longer or greater than some number?

## Z score examples using standard deviation

### Example 1: Longer than

Hospital stays for admitted patients at a certain hospital are measured in hours and were found to be normally distributed with an average of 200 hours and a standard deviation of 75 hours. How many of these stays can be expected to **last for longer** than 300 hours?

- x=300
- x̅ = 200
- s=75

Z= x- x̅/s =300-200/75= 100/75= 1.33

Z score from the table for 1.33 = 0.9082

Since, we are looking for longer, solution is P(x>300) = P(Z>1.33) = 1- P(Z<1.33)= 1-0.9082 = 0.0918 = 9.18%

### Example 2: Less than

Hospital stays, for admitted patients at a certain hospital are measured in hours and were found to be normally distributed with an average of 200 hours and a standard deviation of 75 hours. How many of these stays can be expected to last less than 75 hours?

- x=75
- x̅ = 200
- s=75

Z= x- x̅/s =75-200/75= -125/75= -1.667

Z score from the table for -1.667 = 0.0475

Since, we are looking for less than, solution is = 4.75%

### Example 3: Both less than AND greater than. (Percentage outside the range)

The mean inside diameter of a sample of 200 washers produced by a machine is 0.502 inches and the standard deviation is 0.005 inches. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 inches, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.

**Less than**

- x=0.496
- x̅ = 0.502
- s=0.005

Z= x- x̅/s =0.496-0.502/0.005= -1.2

Z score from the table for -1.2 = 0.1151

Since, we are looking for less than, solution for lower bound = 11.51%

**More than**

- x=0.508
- x̅ = 0.502
- s=0.005

Z= x- x̅/s =0.508-0.502/0.005= 1.2

Z score from the table for 1.2 = 0.8849

Since, we are looking for More than, solution is 1-0.8849 = 0.1151 = 11.51%

So, total is 23.02%

## Example 4: Both upper and lower bound. (Percentage OUTSIDE the range.)

The weights of 500 American men were taken and the sample mean was found to be 194 pounds with a standard deviation of 11.2 pounds. What percentages of men have weights between 175 and 225 pounds.

**P(X<175)**

- x=175
- x̅ = 194
- s=11.2

Z= x- x̅/s =175-194/11.2= -1.6964

Z score from the table for -1.6964= 0.0455

P(X<175) = 4.55%

**P(X<225)**

- x=225
- x̅ =194
- s=11.2

Z= x- x̅/s =225-194/11.2= 2.7678

Z score from the table for 2.7678 = 0.9971=99.71%

Since, we are looking for weights between 175 and 225, P(175<x<225) = 99.71%-4.55% = 95.16%

## Additional Notes:

## Helpful Videos

## Z Transformations

### Z Transformation Example with Standard Deviation

#### Question

A batch of batteries with an average of 60v and a Standard Deviation of 4v. If 9 batteries are selected at random, what is the probability that the total voltage of the batteries is greater than 530?

#### Answer

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OR

Enroll in Pass Your Six Sigma Exam

#### Question

What is the probability that the average voltage is less than 62?

#### Answer

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Enroll in Pass Your Six Sigma Exam

## Z Test for Two Proportion

## What do you do when the sample size is less than 20?

Great question! You’d apply student t-scores.

## Six Sigma Green Belt Z Score Questions

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**Question:** This formula Z = (X – μ)/σ is used to calculate a Z score that, with the appropriate table, can tell a Belt what ____________________________________.

A) Ratio the area under the curve is to the total population

B) Number of Standard Deviations are between X and μ

C) The Median of the sample population is

D) Proportion of the data is between X and μ

**Question:** A battery manufacturer was considering changing suppliers for a particular part. The purchasing manager required that the average cost of the part be less than or equal to $32 in order to stay within budget. A sample of the 32 initial deliveries had a Mean of the new product upgrade price of $28 with an estimated Standard Deviation of $3. Based on the data provided, the Z value for the data assuming a Normal Distribution is?

A) 0.67

B) 1.33

C) 2.67

D) 4.33

## Want a List of Free Z Score Questions and Z Table downloads to help study?

### Z Score Practice Questions and Z Charts

Practice makes perfect! Download a list of top sites for **free Z Score practice questions and Z table charts.**

## Comments (47)

A process measurement has a mean of 758 and a standard deviation of 19.4. If the specification limits are 700 and 800, what percent of product can be expected to be of limits assuming a normal distribution?

A. 1.7%

B. 7.1%

C. 3.4%

D. 2.9%

I’ve put this answer and the complete question walkthrough in the Pass Your Six Sigma Green Belt paid course. Members can find it there.

Hello,

From the above example, Z value of 2.164 s 98.46; instead of 0.0154. Your value of 0.0154 would be correct if the mean value of the specifications was 716 instead of 800.

Further, if the data crosses 2 standard devations it should be greater than 95% .

Pl. correct if i am wrong.

Regards,

Hari Kiran

Hi all,

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2.9%

In example 4, why did you look for z of (-2.76) while it is a positive number? please explain. thank you.

I’ve put this answer and the complete question walkthrough in the Pass Your Six Sigma Green Belt paid course. Members can find it there.

Thank you for these examples. In example 4, I calculated (175-194)/11.2 = -1.696 for the first limit, and rounded up to -1.7, so the Z table number I got was .0446. You did not round up and referenced -1.69 and the Z table number you got was .0455. Is there guidance on rounding when using these tables? I don’t recall anything like this when I went through the Villanova SSBB class. Thanks.

Thanks for flagging this! I was just wondering the exact same thing. Example 4 appears to be rounded incorrectly. I got the answer of 95.26%. Did you get a response?

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an IQ score has z score of 175. What is the score if the mean is 100 and the standard deviation is 15.

How have you thought about trying to solve this question? Where do you get stuck?

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i need to find more math & statistic problem to practice, seems i come to the right place! thank you! is there any other ways i can find on the internet?

Hi Chen,

I offer 1000 questions in my my Green Belt Study Guide program here.

What kind of statistics problems are you looking for?

Problem 1, solution is 9.18% 🙂

Hi, z-transf #1:

Using the Z table, we find the area to the right of the Z is 0.7976. So, there is a 79.76%. In the table it´s 79.67%

🙂

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Hi, I really don´t get the batteries problem? Sigma Xbar is a measure of the error of the sample distribution and should be = 4/3. So the 62v part of the question is clear. I don´t get why the first part is true than, could you help me?

What is the Ppk of a process with a spread of 24 units, an average of 68, an upper limit of 82 and a lower limit

of 54?

A. 1.68

B. 2.00

C. 4.00

D. 4.42

I’ve tried solving this a couple of times to no avail.

Edrezz- can you tell me the steps you took to try to solve this?

can you sold it? im straggling. tried to do as the above example and o lack ;-\

Ami – what are you struggling with?

Can you help me to resolve this problem…

Using this partial Z Table, how many units from a month’s production run are expected to not satisfy customer requirements for the following process?

Upper specification limit: 7.2 Lower specification limit: 4.3 Mean of the process: 5.9 Standard Deviation: 0.65 Monthly production: 450 units

Please choose the correct answer.

Response:

3

7

10

12

What have you tried so far, Tharik?

Mean : 5.9

SD : .65

USL : 7.2

LSL : 4.3

Out Of Limit ;

Z1 = 7.2-5.9 / 0.65 : 2 , Z Table : .0228

Z2 = 5.9-4.3/ 0.65 : 2.46, Z Table : .0069

So, P(X=Z2) : 0.0228+ 0.0069 = 2.97%

Total unsatisfied customer requirement : 450*2.97%=13.81

Hi Thanrik,

I moved this question into the Pass Your Six Sigma Green Belt paid program.

If you’d like to learn how to solve problems like this – and thousands of others – consider joining my Green Belt Study Guide program here.

Mean : 5.9

SD : .65

USL : 7.2

LSL : 4.3

Out Of Limit ;

Z1 = 7.2-5.9 / 0.65 : 2 , Z Table : .0228

Z2 = 5.9-4.3/ 0.65 : 2.46, Z Table : .0069

So, P(X=Z2) : 0.0228+ 0.0069 = 2.97%

Total unsatisfied customer requirement : 450*2.97%=13.81

Unsatisfied customers/month : 450*2.97%=13.81

Hi Thanrik,

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If you’d like to learn how to solve problems like this – and thousands of others – consider joining my Green Belt Study Guide program here.

How can I find the z-score normalization for a set of data in excel format?

Hi Rocky,

I offer one (and dozens of other helpful items, thousands of questions, and much more) in my consider joining my Green Belt Study Guide program here.

Which of the following describes the 95% confidence

interval of a 20% absentee rate in a department with

30 people?

(A) 6% to 34%

(B) 8% to 32%

(C) 13% to 27%

(D) 17% to 23%

The answer to this question is covered here.

Is it possible to have a z-score of + 300 or – 300?

Hi Priya,

I moved this question into the Pass Your Six Sigma Green Belt paid program.

How to calculate Z value in below question?

Question:

When σ = 10, what sample size is needed to specify a 95% confidence interval of ±3 units from the mean?

Here, Sample size = ((Z*σ )/(margin of error))^2.

In the formula, Z value need to find from Z- table. But how it can be found?

Hi Arpan,

I moved this question into the Pass Your Six Sigma Green Belt paid program.

Here I m wanting to know about how to use z table & how to calculate the rejection in PC’s & percentage, my question is –

A lot of 1000 pivot pins, specification length of pin is 15.0+/-0.2 mm, sample of 50 PC’s were inspected mean & std. Deviation was to be 15.05 & 0.1 mm respectively. Calculate how many PCs are likely to be recjeted from the lot. Use z table & pls give pictorial view.

What have you tried so far, Maninder?

I moved this question into the Pass Your Six Sigma Green Belt paid program.

In a distribution exactly normal, 7% of the items are under 35 and 79% are under 63. What is the mean and standard deviation of the distribution

What have you tried so far, Shadrack?

I moved this question into the Pass Your Six Sigma Green Belt paid program.

I have a question with regards to the following:

Question

What is the probability that the average voltage is less than 62?

Answer

The expected value would be 60.

Z = 62-60 / (4 / Sqrt(9)) = 2 / (4/3) = 3/2 = 1.5.

The area to the left of Z is 1 – Z. Thus 1 – 0.0668 = 0.9332. Or 93.32%.

My comment: Z should be -1.5

following the hypothesis test : Ha: xa or= 60; hence since Z=-1.5; Alpha= 0.06881; the question is related to the value less than 60 (left side) hence it should be only 6.881%