Z scores (Z value) is the number of standard deviations a score or a value (x) is away from the mean. In other words, the Z-score measures the dispersion of data. Technically, a Z-score tells you how many standard deviations value (x) are 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 the Z value is negative, it means the value (x) is below the mean.
What is a 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 has a mean of zero and has intervals that increase by 1. Each number on the horizontal line corresponds to the 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 a 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.
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A Z-score tells how much standard deviation a value or score is from the mean (µ). For example, if a Z-score is negative 3 means the value (x) is 3 standard deviations left of the mean. Similarly, if the Z-score is positive 2.5 means the value (x) is 2.5 standard deviations to the 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, the 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. The Z-score table tells the total quantity of area contained on the left side of any score or value (x).
Using the Z score, find the percentage by using the formula: 1-NORMSDIST(Z), where Z is your calculated Z Score.
How to Calculate a Z Score by Hand
There are 2 different situations you need to be aware of when calculating a z score:
Z score for a sample
Z score for a population
While the z-score equations look very similar, remember that calculating the standard deviation of a population is different than the way you calculate the standard deviation of a sample.
The formula for transforming a score or observation x from any normal distribution to a standard normal score is :
Calculating a Z Score (generic)
Calculating a Z Score for a Population
Z Score for a Sample
How many parts in a population will be longer or greater than some number?
Z score examples using standard deviation
Example 1: Longer than / Greater 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%
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 to the total population
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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?
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% .
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?
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. 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.
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. 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.
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?
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. 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.
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
can you sold it? im straggling. tried to do as the above example and o lack ;-\
Ami – what are you struggling with?
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. 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.
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
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.
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%
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.
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%
my understanding is that Sample standard deviation is = (population standard deviation)/(sqrt(n)
but this is contrary to what is used in e.g. 1.1 above.
What could be the reason that we are using, sample s.d = population s.d * sqrt(n) formula above
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 of X>175 and <225
should be 1-.0455 correct which changes the problem..please clarify
I think your first step is to choose an organization to be certified with. There are many of them. I would suggest getting started with ASQ or IASSC.
From there you would be able to a) see what belt you want to go for and b) be able to search for training.
Please remember that not all training covers all of the different certification organizations, so you want to make sure that whatever training you select is aligned with your desired organization. Prices for training and actual certification costs are are all different based on what you select.
Dear Ted, all,
If you want to write an equation in one single line, please take into account the operator precedence.
For instance, the following
Z= x- x̅/s =0.496-0.502/0.005= -1.2
should be written
Z= x- x̅/s =(0.496-0.502)/0.005= -1.2
or else, according to the order of operations, the result would be -99.904.
Comments (59)
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%
A
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?
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. 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.
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?
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. 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.
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%
🙂
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. 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.
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?
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. 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.
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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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,
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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.
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.
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.
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.
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?
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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%
Hi Ted,
It looks like the title of Example 4 should be “Percentage INSIDE/WITHIN the range” – as Example 3 dealt with outside the range.
Owen
Thanks, Owen!
Hi Ted,
my understanding is that Sample standard deviation is = (population standard deviation)/(sqrt(n)
but this is contrary to what is used in e.g. 1.1 above.
What could be the reason that we are using, sample s.d = population s.d * sqrt(n) formula above
Hello JEEVAN SINGH,
We have updated the article with detail explanation.
Thanks
Hello JEEVAN SINGH,
We have updated the article with detail explanation
Thanks
Confused.
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 of X>175 and <225
should be 1-.0455 correct which changes the problem..please clarify
Never mind, I get it!
I want to write all the certifications of 6 sigmas. How much are the tutorial costs and examination registrations?
Good question, John.
I think your first step is to choose an organization to be certified with. There are many of them. I would suggest getting started with ASQ or IASSC.
From there you would be able to a) see what belt you want to go for and b) be able to search for training.
Please remember that not all training covers all of the different certification organizations, so you want to make sure that whatever training you select is aligned with your desired organization. Prices for training and actual certification costs are are all different based on what you select.
Dear Ted, all,
If you want to write an equation in one single line, please take into account the operator precedence.
For instance, the following
Z= x- x̅/s =0.496-0.502/0.005= -1.2
should be written
Z= x- x̅/s =(0.496-0.502)/0.005= -1.2
or else, according to the order of operations, the result would be -99.904.
Thank you, Steve. I have made the necessary updates in all the relevant sections.
Thanks