We use Z Scores to transform a given standard distribution into something that is easy for us to calculate probabilities on. Why? So we can determine the likelihood of some event happening.

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.

Sometimes, depending on the wording of the problem, we look for different things. I find it’s easiest if we draw the problem and decide what exactly we are looking for.

## There are 3 types of Z Score Questions:

- The percent chance of an event happening beyond a certain point.
- This is the number under the curve beyond the z value.

- The percent chance of an event happening under a certain point.
- This is the number under the curve up to the z value.

- The percent chance of an event happening between two points.
- This could be the number under the curve bounded by two points.
- Sometimes one of those points is the mean – or the center of the distribution.
- In this case, Z scores are used to determine how far off a particular point in a distribution is from the mean.

- This could be the number under the curve bounded by two points.

Depending on the information given in the problem, there are also 2 different ways you go about solving the questions asked. One is by using the Z table to find the Z score. The other is by calculating the Z score. These are really the same thing just done in reverse. We will cover both types.

As you might guess, there are a number of ways of asking these kinds question. We will cover the ones I’ve seen most often. If I miss one, just let me know in the comments.

Before we get into how to solve problems, let’s cover the Z table.

## The Normalized Z Table

The Z table (a table based on Normal Standard Distribution) is used to solve how far off a point in a distribution is likely to be from the center (mean).

A common statistical way of standardizing data on one scale so a comparison can take place is using a z-score. The z-score is like a common yard stick for all types of data. – from here.

Both z and t distributions are symmetric and bell-shaped, and both have a mean of zero.

By using the Z transformation, we can convert any normal distribution into a normal distribution with a mean of 0 and a standard deviation of 1. Thus, we can use a single normal table to find probabilities. – Pyzdeck

## How to Create Your Own Z Table

Don’t have a reference chart handy? Have Excel? Great!

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

## How to Use a Z Table to find a Z Score

Be careful. Pay attention to what side of the z you should be on.

P(o<=Z<=a)

Step 1: Pick the right Z row by reading down the right column

Step 2: Read across the top to find the decimal space.

Step 3: Find the intersection and multiply by 100.

Ex This means that 6.18% of the normal curve resides to the right of Z. This also means that 93.82% resides to the left of the normal curve.

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

Don’t panic. Z score tables sometimes only go up to 3. But depending on the spread of your 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. You can use Excel to find the actual value if your table doesn’t go that high.

## How do you Calculate a Z Score?

### Calculating a Z Score for a Population

### Calculating a 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? <from book>

### 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?

### 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.

## 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.

## Example 5: What is the Area Under the Normal Curve?

A common Z value test question is to find the area under a normal curve.

What is the area under the curve between +.7 and +1.3 Standard deviations?

Ans. Show example curve. Look up both on Z table. Subtract the 2. The remainder is the area.

## Additional Notes:

## 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

With an average voltage of 60, you’d expect the total to be 540v. This ends up being a standard deviation of a sample problem.

First, we find the standard deviation of the sample. To do this we use the variance equation:

Thus, Z = 530 – 540 / Sqrt(9*4^2) = -10 / Sqrt(144) = -10/12 = -0.833

Using the Z table, we find the area to the right of the Z is 0.7976. **So, there is a 79.76%.**

#### 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%.

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

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

Eric Park says

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%

Six Sigma Study Guide says

Mean : 758

SD : 19.4

USL : 800

LSL : 700

Out Of Limit ;

Z1 = 700-758 / 19.4 : 2.989 , Z Table : 0.0014

Z2 = 800-758 / 19.4 : 2.164 , Z Table : 0.0154

So, P(X< =Z1)+P(X>=Z2) : 0.0014 + 0.0154 = 1.68%

Ell Mar says

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

Amy says

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.

Sofi says

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?

Rems says

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

Six Sigma Study Guide says

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

chen says

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?

Julioskij says

Problem 1, solution is 9.18% 🙂

Julioskij says

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%

🙂

Julioskij says

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?

Tharik Mansoor says

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

Six Sigma Study Guide says

What have you tried so far, Tharik?

Tharik Mansoor says

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

Tharik Mansoor says

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

Tharik Mansoor says

Unsatisfied customers/month : 450*2.97%=13.81