Business Statistics Unit 6

Normal Distributions

In this unit we will discuss the normal distribution.  The normal distribution is a bell shaped distribution that is often used in statistical research and sampling.

Introduction

The normal distribution is a continuous probability distribution.  This indicates that we can measure a variable using the degree of precision necessary for us to make a decision.  The normal distribution is represented by the normal curve, which is a symmetrical distribution following a bell-shaped curve.  The normal curve's shape is influenced by the standard deviation.  The smaller the standard deviation, the more pointed the normal curve is.

Figure 6.1.  A Normal Distribution Curve.

You notice that a normal curve has two tails:  one to the left and one to the right.  The mean is found in the middle of the curve at its highest point.  We can use the normal curve to help us find probabilities associated with specific data values. 

To calculate probabilities for a normal distribution, it is helpful to understand how we can visually relate probabilities to the normal distribution.  Often the standard normal distribution is used to help students understand how probabilities are determined.  Under the standard normal distribution, the mean is equal to 0 and each standard deviation above and below the mean has a distance of one. 

Figure 6.2.  The Standard Normal Curve.

The z Score and Normal Distributions

Every normal distribution has a unique mean and standard deviation.  Often textbooks discuss a normal distribution called the standard normal distribution where the mean equals 0 and the standard deviation equals 1.  The unique relationship that exists in a standard normal distribution is that the z score equals the standard deviation.  This means that if our standard deviation equals 1 our z score is either equal to 1.0 or -1.0 depending upon whether we are above or below the mean.

In the real world, normal distributions do not have means that are equal to 0 and standard deviations that are equal to 1.  Therefore we must use a simple formula to find the z score, which tells us for a particular value of x, how many standard deviations x is from the mean.  Therefore the z score measures distance from a particular point on the normal curve, and the mean.  In order to do this we use the following formula:

Formula Box 6.1.  The z Score.

The variable x refers to the specific data value or variable you are trying to find the z score for.  µ represents the mean of the set of data values, and σ represents the standard deviation of the set of data values.

For the standard normal distribution with a mean of 0 and a standard deviation of 1, if x equals 1, then our z score equals 1.  If x equals 2, our z score equals 2, and if x equals 3 our z score also equals 3.  If x equals -1, our z score is now equal to -1 and if x equals -2 our z score equals -2.  Finally when our value of x equals -3, our z score is also equal to -3.

Lets use some more realistic examples.  Lets say our variable x equals 5, and the mean equals 3 with a standard deviation of 1.5.  To find the z score we simply use the formula in formula box 6.1 and plug in the values.

z = (5 - 3) / 1.5

Therefore our z score equals 1.3333.

Another example.  Lets say our variable x is equal to 40, and the mean equals 50 with a standard deviation of 5.  To find the z score we use the formula and plug in the values.

z = (40 - 50) / 5

Therefore our z score equals -2.00.  Notice that we have a negative number.  That indicates that our variable x is to the left of the mean.  Ready for some practice?  Lets go!

Practice Exercise 6.1.  Finding the Z Score.

We are going to find the z score for three different problems in this practice exercise.

1.  If x is a normally distributed variable with a mean of 12 and a standard deviation of 2, what's the z score that represents an x value of 15?

To solve this problem we simply use formula 6.1 and plug in our values associated with each symbol.

z = (15 - 12) / 2

Therefore our z score equals 1.50.  A z score of 1.50 indicates that the value x of 15 is 1.50 standard deviations above the mean.

2.  If x is a normally distributed variable with a mean of 20 and a standard deviation of 5, what's the z score that represents an x value of 10?

To solve this problem we simply use formula 6.1 and plug in our values associated with each symbol.

z = (10 - 20) / 5

Therefore our z score equals -2.00.  A z score of -2.00 indicates that the value x of 10 is 2.00 standard deviations below the mean.

3.  If x is a normally distributed variable with a mean of 125 and a standard deviation of 4.5, what's the z score that represents an x value of 135?

To solve this problem we simply use formula 6.1 and plug in our values associated with each symbol.

z = (135 - 125) / 4.5

Therefore our z score equals 2.222.

Finding Probabilities Using the Z Score and the Normal Distribution Tables

NOTE:  Since we are going to use Excel to find our probabilities, it is not necessary you understand how to look up the probability values using tables in your textbook.

The z score enables us to find the number of standard deviations a variable x is from the mean.  We can also use the z score to find probabilities associated with normal distributions.  It is important for you to understand that the z score represents an area from the mean to a specific variable x.  Take a look at the figure below.  Notice that the normal distribution has a mean of 10 and a standard deviation of 2.0.  For the variable x of 7.0, the z score is equal to -1.50.

Figure 6.3.  Area Represented by a z Score.

The area between the mean and any variable x is represented by a probability that can be found using the normal distribution table found in your textbook.  The only information needed is the z score.

Once we have a z score we can then use the table in the back of your book titled Areas Under the Normal Probability Distribution to find the probability.  Take a look at your table now.  Notice that there is a column marked z  containing a column of numbers from 0.0 to 4.0, and a row of numbers at the top from .00 to .09.  We take our z score value and use the table to find the probability.

Table 6.1.  Areas Under The Standard Normal Probability Distribution - Sample Table.

Now let's use an example to find a probability.  Suppose we are looking for the probability (or area) between our mean of 0 and a value of x of 1.0.  We must first use formula 6.1 to find the z score.  Since our mean equals 0 and the standard deviation for this problem equals 1.0, our z score equals 1.00.  Using the z score value of 1.00, we go into the normal tables by going down the first column until we find 1.0.  Then we go over to the .00 column and we find the probability equal to .3413.  The probability which is .3413 represents the area under the normal curve from our mean of 0 to our value of x which was 1.0.

Ready for another example?  Suppose we are looking for the probability of being between -1.0 and 1.2.  Both the -1.0 and 1.2 represent the values of x.  Therefore we will need to solve for each value of x.  Let's find the first probability associated with an x value of 1.2.  First we find the z score using formula 6.1, with a mean of 0 and a standard deviation of 1.0, our z score is 1.2.  Now using your normal table, find the probability for a z score of 1.2.  You should find the number .3849.  Now we have found the probability from 0 our mean to 1.2.  Next we must find the probability for a value of x of -1.0.

We must first use our formula to find the z score associated with the variable x of -1.0.  Using the formula, we find that our z score is -1.0.  Now notice that the normal tables do not give probabilities for negative values.  That's OK, you can use the probability for the positive number.  Why?  Because the probabilities are the same regardless of whether the z score is positive or negative.  That's an important characteristic of the normal distribution.  On each side of the mean, the probabilities add up to .50 for a total of 1.0.  So our probability for a z score of -1.0 equals .3413.

Finally to solve our problem, we must add up both of the probabilities that we have found.  Therefore .3413 + .3849 = .8185 our answer.

Figure 7.4.  The Normal Curve with x values of -1.0 and 1.2, mean = 0.

You should be comfortable by now finding the z score.  If not, please review the previous examples again and also review the examples given in your book. 

Practice Exercise 6.2.  Finding Probabilities Using the Normal Distribution Tables.

We will once again use three different examples to illustrate how to find probabilities using the z score.  It is important to remember that the total of all probabilities under the normal distribution equals 1.0 and that the total probability above the mean is equal to .50, and the total probability below the mean is equal to .50.

1.  If x is a normally distributed variable with a mean of 12 and a standard deviation of 3.0, find the z score and probability associated with an x value of 18.

Our first step to solve this problem is to find the z score.  Using the formula for the z score given in formula 6.1, we find that the z score equals:

z = (18 - 12) / 3.0 = 2.0

Now that we have the z score, we can use the normal probability tables in our book to find the probability.  Using the table, the probability is equal to .4772.

2.  If x is a normally distributed with a mean of 50 and a standard deviation of 10, find the probability of x being less than 35.

For this problem our first step remains the same:  find the z score for a value of x equal to 35.  Using the formula, we find the z score equals:

z = (35 - 50) / 10 = -1.50

Now that we have the z score we can use the normal probability table to find the probability.  Using the table, the probability equals .4332.  Our next step is to picture the normal distribution and understand what the probability in the table represents.  Our probability of .4332 represents all values from the mean of 50 to our value of x which is 35.  This is indicated by the pink shaded portion of the normal distribution pictured below.

Since we are looking for all values less than 35, we are looking for the probability of being below 35.  Since the total probability below the mean is equal to .50, we simply subtract .4332 from .50 and get .0668.  This is the probability of having a value of x less than 35.

3.  Suppose we have a normal distribution with a mean of 25, and a standard deviation of 2.0.  What is the probability of the value of x being between 24 to 26?

Our first step remains the same.  We need to find the z score.  However in this problem, we need to find two z scores.  A z score for our variable 24 and the z score for the variable 26.  The z score for 24 is:  (24 - 25) / 2.0 = -0.50.  Looking in the normal distribution table, the probability is equal to .1915.

Now we need to find the second z score for the variable 26.  The z score for 26 is:  (26 - 25) / 2.0 = 0.50.  Looking in the normal distribution table, the probability is equal to .1915.  Did you notice that the probabilities are the same for both variables?  That is because they are the same distance from the mean.  Therefore they will have the same probability. 

Now we need to take our two probabilities and visualize the normal curve.  Both of the probabilities are for the distance from the mean of 25 to the variable 26, and from the mean of 25 to the variable 24.  Since this is the area we are looking for, we simply add the two probabilities together to find our answer.  .1915 + .1915 = .3830 which is the probability of being between 24 and 26 in our normal distribution with a mean of 25 and a standard deviation of 2.0.

Finding x When it is Unknown

Sometimes the value of x also known as the cutoff score, is not known.  To find the value of x, we change the formula in Formula Box 6.1 to solve for x.

Formula Box 6.2.  Finding x for Normally Distributed Variables.

or depending upon if the z score is positive or negative.

We have simply rearranged the formula so that we can find the value of x instead of solving for z.  For example, suppose we have a set of normally distributed data that has a mean of 10, a standard deviation of 2.0, and a z score of 1.56.  Find the x value.

Using our formula in Formula Box 6.2, we simply fill in the numbers:

x = 10 + 1.56(2.0) = 13.12.

Therefore our value of x equals 13.12 and -13.12.  remember we are working with a normal distribution so there will be two values for x:  a value above the mean and a value below the mean.  Since our problem did not specify which value of x was important, both values should be stated.  For a more practical application of finding the value of x, take a look at practice exercise 6.3 below.

Practice Exercise 6.3.  Finding x for Normally Distributed Variables.

Scores on a recent test in statistics are normally distributed with a mean of 42 points and a standard deviation of 3.0 points.  Starting at what test score do the top 5% of the students fall in?  In order to solve this problem we must first understand that we are looking for the value of x to the right of the mean where the top 5% of the test scores starts.  The second item we must understand is that although a z score is not given, we have enough information to find it.

Remember we are looking for the top 5% of students.  This means that 95% of the students will be below this score.  We can use the normal distributions tables and our knowledge of the normal distribution to find the z score.  Looking at the table values, remember that the tables provide value for z between the mean and the value of x.  We are looking for a value of z where 45% of the values are between the value of x and the mean, and 50% of the test scores are below the mean.  We will need to find the value of z that is closest to .4500.  What's different here?  Well we are using the tables backwards.  We have the probability and are trying to find z instead of knowing z and finding the probability.

Using the normal distribution table in your book, find the value of z that is closest to a probability of .4500.  You should notice that there are two probabilities that come close to our probability of .4500:  .4495 and .4505.  The z score for .4495 is 1.64 and the z score for .4505 is 1.65.  Which one do you use?  Well normally we would use the closest probability and choose that z score.  Since both of the probabilities are the same distance from our probability of .4500, we simply take the two z scores (1.64 and 1.65), add them together and divide by 2.  This gives us a z score of 1.645.

Now we are finally ready to solve our problem using formula 6.2.

x = 42 + 1.645(3.0) = 46.935 or for a test we should use 47 points.  This means that the top 5% of the students scored 47 points or more on the test.  Notice that we were not concerned about the x value below the mean since we were only interested in the top values.  The x value below the mean would be equal to:

x= 42 - 1.645(3.0) = 37.065.

Using Excel to Find Normal Probabilities        IMPORTANT INFORMATION!

We can use Excel to find normal probabilities and skip looking up values in the table.  In fact, Excel will provide a more accurate probability since many values in the normal distribution table are rounded.

Lets take a look at a normal distribution with a mean of 25 and a standard deviation of 2.5.  What is the probability of being less than 21 (x = 21)?

Excel 2007

1. Open a blank worksheet.

2. From the Home tab, click on fx, the Insert Function button.

3. The Insert Function box appears.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.
   
   
    

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 21.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 25.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 2.5.  Then click in the Cumulative box.

12. In the Cumulative box, type in True or the number 1.  Then click OK.
    
    

The calculated probability is equal to .054799.  Notice that Excel provides you with the probability of being less than our value of x which is 21.  The normal distribution tables provide you with a probability between our value of x and the mean.  This is an important difference to note.

Excel 2003

1. Open a blank Excel worksheet.

2. From the top menu choices, select Insert.

3. Now select Function.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 21.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 25.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 2.5.  Then click in the Cumulative box.

12. In the Cumulative box, type in True.  Then click OK.

13. The calculated probability is equal to .054799.  Notice that Excel provides you with the probability of being less than our value of x which is 21.  The normal distribution tables provide you with a probability between our value of x and the mean.  This is an important difference to note.

Another example.  Suppose on a recent statistics test the average grade was 90 points with a standard deviation of 5 points.  What is the probability of having a grade of 85 points or more?

Excel 2007

1. Open a blank worksheet.

2. From the Home tab, click on fx, the Insert Function button.

3. The Insert Function box appears.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 85.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 90.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 5.  Then click in the Cumulative box.

12. In the Cumulative box, type in True or the number 1.  Then click OK.

13. The calculated probability is equal to .1587.  This represents the probability from 0 to 85.  But we were looking for the probability for 85 or more.  What do we do?  Since the total probability in a normal distribution equals 1, we subtract .1587 from 1 and we get .8413.  Therefore the probability of having a grade of 85 points or more is equal to .8413.

Excel 2003

1. Open a blank Excel worksheet.

2. From the top menu choices, select Insert.

3. Now select Function.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 85.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 90.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 5.  Then click in the Cumulative box.

12. In the Cumulative box, type in True.  Then click OK.

13. The calculated probability is equal to .1587.  This represents the probability from 0 to 85.  But we were looking for the probability for 85 or more.  What do we do?  Since the total probability in a normal distribution equals 1, we subtract .1587 from 1 and we get .8413.  Therefore the probability of having a grade of 85 points or more is equal to .8413.

You want another example?  OK!

At a popular local restaurant, the average wait time to get seated for dinner is 40 minutes with a standard deviation of 7 minutes.  What is the probability the wait time will be between 35 and 50 minutes?  To solve this problem, we will need to run Excel twice for the two different values of x.

Excel 2007

1. Open a blank worksheet.

2. From the Home tab, click on fx, the Insert Function button.

3. The Insert Function box appears.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 35.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 40.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 7.  Then click in the Cumulative box.

12. In the Cumulative box, type in True or the number 1.  Then click OK.

13. The calculated probability is equal to .2375.  This represents the probability for 0 to 35 minutes.  Now, let's run Excel again using the x value of 50 minutes.

1. Select another cell on your open Excel worksheet.

2. From the Home tab, click on fx, the Insert Function button.

3. The Insert Function box appears.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 50.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 40.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 7.  Then click in the Cumulative box.

12. In the Cumulative box, type in True or the number 1.  Then click OK.

13. The calculated probability is equal to .9234.  This represents the probability of 0 to 50 minutes.  What were we looking for?  The probability our wait is between 35 to 50 minutes.  Here's what we do:  take the 0 to 50 minutes probability of .9234 and subtract from it the 0 to 35 minute probability of .2375.  You get .6859 which is the answer for the probability that our wait time will be between 35 to 50 minutes.

Excel 2003

1. Open a blank Excel worksheet.

2. From the top menu choices, select Insert.

3. Now select Function.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 35 for the first value of x.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 40.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 7.  Then click in the Cumulative box.

12. In the Cumulative box, type in True.  Then click OK.

13. The calculated probability is equal to .2375.  This represents the probability for 0 to 35 minutes.  Now, let's run Excel again using the x value of 50 minutes.

1. Select another cell on your open Excel worksheet.

2. From the top menu choices, select Insert.

3. Now select Function.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMDIST.  Then click OK.

8. The Function Arguments box appears.

9. In the X box, enter the value of x.  Here it is 50.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 40.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 7.  Then click in the Cumulative box.

12. In the Cumulative box, type in True.  Then click OK.

13. The calculated probability is equal to .9234.  This represents the probability of 0 to 50 minutes.  What were we looking for?  The probability our wait is between 35 to 50 minutes.  Here's what we do:  take the 0 to 50 minutes probability of .9234 and subtract from it the 0 to 35 minute probability of .2375.  You get .6859 which is the answer for the probability that our wait time will be between 35 to 50 minutes.

We can also use Excel to find the value of x when we know our mean, standard deviation, and either the z score or a probability.  Here is how we do it.

Lets take a look at some normally distributed data where the mean is equal to 15, the standard deviation is 2.0 and the probability equals 95%.  What is the value of x?

Excel 2007

1. Open a blank worksheet.

2. From the Home tab, click on fx, the Insert Function button.

3. The Insert Function box appears.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMINV.  Then click OK.
   
   

8. The Function Arguments box appears.

9. In the Probability box, enter the probability.  Here it is .95.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 15.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 2.0.  Then click OK.
    
    

12. The calculated value of x is 18.2897. This is our answer since it clearly represents a position right of the mean.  Ninety-five percent of all the values are below the x value of 18.2897, and 5% of the values are above our x value of 18.2897.

Excel 2003

1. Open a blank Excel worksheet.

2. From the top menu choices, select Insert.

3. Now select Function.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMINV.  Then click OK.

8. The Function Arguments box appears.

9. In the Probability box, enter the probability.  For this problem our probability is equal to .95.

10. In the Mean box, enter the value of the mean.  Here it is 15.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 2.0.  Then click OK.

12. The calculated value of x is 18.2897. This is our answer since it clearly represents a position right of the mean.  Ninety-five percent of all the values are below the x value of 18.2897, and 5% of the values are above our x value of 18.2897.

What?  You want another example!

Recent research has shown that the average number of jobs a person will have with a different employer over his/her lifetime is 13 with a standard deviation of 1.5.  Find the range of the number of employers where 90% of the people will fall in their lifetime.  Yuch!  This is difficult!  Hmm...A normal distribution with a mean of 13, and we are trying to find the minimum number of employers and the maximum number of employers where 90% of the people are expected to fall.  Since I know this is one of those "backwards" problems where we are trying to find two x values, I need to figure out where my first x value is.  In Excel both probabilities and percentages start from the left side where my minimum value is.  Using 90%, that means that 45% of the employers are on either side of the mean.  So if I use 5% as my first percentage, I should find my first x value.

Excel 2007

1. Open a blank worksheet.

2. From the Home tab, click on fx, the Insert Function button.

3. The Insert Function box appears.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMINV.  Then click OK.

8. The Function Arguments box appears.

9. In the Probability box, enter the probability.  Here it is .05.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 13.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 1.5.  Then click OK.

12. The calculated value of x is 10.53.   This becomes my lower number for the number of employers.  To find my upper number I simply rerun NORMINV again and use 95% as my probability.  Notice I am NOT using 90% since I must consider how Excel calculates probabilities starting from the left side.  I need to be 90% away from from 5% location, therefore I use the 95% probability.

1. Move to a different cell in your worksheet.

2. From the Home tab, click on fx, the Insert Function button.

3. The Insert Function box appears.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMINV.  Then click OK.

8. The Function Arguments box appears.

9. In the Probability box, enter the probability.  Here it is .95.  Then click in the Mean box.

10. In the Mean box, enter the value of the mean.  Here it is 13.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 1.5.  Then click OK.

12. The calculated value of x is 15.47.   This becomes my upper number for the number of employers.  Therefore the answer to this problem is 10.53 to 15.47 years.

Excel 2003

1. Open a blank Excel worksheet.

2. From the top menu choices, select Insert.

3. Now select Function.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMINV.  Then click OK.

8. The Function Arguments box appears.

9. In the Probability box, enter the probability.  For this problem our probability is equal to .05.

10. In the Mean box, enter the value of the mean.  Here it is 13.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 1.5.  Then click OK.

12. The calculated value of x is 10.53.   This becomes my lower number for the number of employers.  To find my upper number I simply rerun NORMINV again and use 95% as my probability.  Notice I am NOT using 90% since I must consider how Excel calculates probabilities starting from the left side.  I need to be 90% away from from 5% location, therefore I use the 95% probability.

1. Move to a different cell in your worksheet.

2. From the top menu choices, select Insert.

3. Now select Function.

4. Under the Select a category box, scroll until you find Statistical.

5. Select Statistical by highlighting it with you mouse.

6. You will see new options listed in the Select a function box.

7. Scroll down until you find NORMINV.  Then click OK.

8. The Function Arguments box appears.

9. In the Probability box, enter the probability.  For this problem our probability is equal to .95.

10. In the Mean box, enter the value of the mean.  Here it is 13.  Then click in the Standard_dev box.

11. In the Standard_dev box, enter the standard deviation.  Here it is 1.5.  Then click OK.

12. The calculated value of x is 15.47.   This becomes my upper number for the number of employers.  Therefore the answer to this problem is 10.53 to 15.47 years.

Sampling Basics - Applying the Normal Distribution to Sample Data

Why do we sample?  In a previous unit we learned that sampling has several advantages over surveying an entire population.  Those advantages include a lower cost and obtaining information faster.  If a sample is conducted correctly, it can be almost as accurate as a census of the population.

We are familiar with sampling.  Have you ever gone to an electronics store and viewed different TVs or listened to different stereo systems?  That's sampling.  Have you ever gone to the grocery store and tried a new product that was being demonstrated at the store?  That's sampling too.

We sample in order to use the sample data to make inferences about the population.  Our sample enables us to understand characteristics of a population without examining the entire population.  In order to be accurate do we need a large sample?  No.  In many cases a sample size can be very small and provide us with accurate information.  Even if we wanted to get the opinion of voters in the entire United States, a population size of perhaps almost 100 million people, we can use a sample size of only about 1000 voters and get an accurate sample.

Sampling Distribution of Means - The Application of the Central Limit Theorem

When we obtain a sample and calculate the mean, one of the first concepts you should understand is that the sample mean's value may not be the same as the population mean's value.  A sample mean will usually be very close to a population mean, but it is normal for it to not be identical to the population mean.

For example, take a look at the data in table 6.1.  The table provides information on the number of miles driven annually by sales representatives for an electronics company.  Assume that the miles driven by sales representatives are normally distributed.

Table 6.1.  Annual Miles Driven by Sales Representatives - Midwest Sales District.

Let's calculate the population mean for the mileage data in table 6.1.  If you add all of the miles driven, you should arrive at a total of 516,430 miles.  Therefore our population mean is equal to:

516,430 / 20 = 25,821.5

Now let's suppose we decide to take a random sample of six sales representatives from our population and calculate the sample mean.  For our sample the following sales representatives and miles were selected:

Table 6.2.  Random Sample of Sales Representatives.

Now we will calculate the sample mean based upon our random sample of six sales representatives.  The total number of miles driven in our sample is equal to 139,495.  Our sample mean is equal to:

139,495 / 6 = 23,249.1667

Notice that the sample mean is different than the population mean.  If we obtained another random sample of six sales representatives from our population and calculated the sample mean, we would likely obtain a different sample mean value again.  This sample mean would likely have a different value that our first sample mean and it would be different than the population mean.

How many different samples are possible when you have 20 different data values and you are looking for a sample of six? Well the best way to determine the number of different samples that could be obtained would be to use the formula for a combination that was discussed in unit 5.  If we plug our current numbers into the combination formula from unit 5 we would get:

Number of different samples where n = 20 and r = 6:

20! / 6!(20 - 6)! = 38,760

There are 38,760 different possible sample combinations available when you pick six variables out of 20 variables.  Therefore it is likely that you could have 38,760 different sample means!

Now here things get interesting.  If you take all 38,760 different sample means, put them into a frequency distribution and add them together and then calculate the mean of the 38,760 sample means, guess what?  The mean of the sample means is equal to the population mean!  We call all of the sample means placed into a frequency distribution the sampling distribution of means.

It is also important to note that our symbols for the mean of the sampling distribution of means is shown as:

The variance and standard deviation also have slightly different symbols:

Variance: 

Standard Deviation (Standard Error): 

Some other characteristics of the sampling distribution of means.  If the population is normally distributed the sampling distribution of means will also be normally distributed.  In fact even if the population is not normally distribution, the sample distribution of means will be normally distributed if the sample size is greater than 30.  Therefore if the normal distribution can be used to represent the sampling distribution of means, we can also use the normal distribution to determine the probability of a single sample mean being within a specified number of standard deviations from the population mean.  This is the basic concept of the Central Limit Theorem.  As long as the sample size is greater than 30, the sampling distribution approximates the normal probability distribution.

Standard Deviation of the Sampling Distribution of Means (Standard Error)

To calculate the extent at which the sample mean can differ from the population mean, we use the standard deviation of the sampling distribution.  For a sampling distribution, the standard deviation is called the standard error of the mean.

The formula for the standard error of the mean is given in formula box 6.3.

Formula Box 6.3.  The Standard Error of the Mean.

In order to solve for the standard error of the mean you will need to know the population standard deviation and the sample size.  On rare occasions a different formula is used to find the standard error of the mean.  This formula contains the finite correction factor which is used when the sample size represents more than 5 percent of the population size.  The formula with the finite correction factor is given in Formula Box 6.4.

Formula Box 8.2.  The Standard Error of the Mean - Finite Correction Factor

You should notice that the only difference between the two formulas is the additional step of taking the square root of the population size minus the sample size, divided by the the population size minus 1.  For purposes of this course we will not be using the finite correction factor for any calculations or problems.

It is important for you to understand that as our sample size gets larger, the standard error of the mean gets smaller.  This is simply because the increased sample size reduces the number of values a sample mean may have which reduces the dispersion.

Practice Exercise 6.4.  Calculating the Standard Error of the Mean.

A local restaurant has determined that the mean time customers spend at the restaurant is 50 minutes with a standard deviation of 5 minutes.  The restaurant manager conducts a random sample of eating times during the most recent one week period.  The sample size was 60 customers.  Find the mean of the sampling distribution and the standard error.

To find the mean of the sampling distribution, we must remember that the mean of the sampling distribution is equal to the population mean which is 50 minutes.  That's it!

To find the standard error of the mean, we use formula 8.1.

= 5/√60

Using Excel to solve:  =5/sqrt(60)

Our answer is .6455.

What if we have a problem where we need to calculate the z score?  Are there any changes to the formula for the z score when we use a sampling distribution?  The answer is yes!  The formula for the z score changes to reflect that we are working with a sample.  The new formula is:

Therefore z equals the sample mean minus the mean of the sampling distribution, divided by the standard error of the mean.

Practice Exercise 6.5.  Sampling Distributions of Means

The ages of students in a statistics class are normally distributed with a mean of 27 and a standard deviation of 3 years.  A sample of 15 students is selected at random.  Find the probability that the mean age is less than 25 years.

We solve this problem using the normal distribution and the concepts discussed in the previous unit, along with the z score formula for a sample.  To solve for the z score we must first find the standard error of the mean.  This is equal to the standard deviation divided by the square root of the sample size.  Therefore we have 3 / √15 = .7746.  Now we use the formula (or Excel using NORMDIST) for the z score for a sample to solve the problem.

z = 25 - 27 / .7746 = -2.58

Now we use the normal tables (or Excel using NORMDIST) to find the probability for a z value of -2.58.  The probability from the table is equal to .4951.  Since we are looking for the probability being less than 25 years old, and the tables give us the probability from a value of x to the mean, we need to subtract .4951 from .50 and we get .0049 which is our answer.

Using Excel 2007:

1. Open a blank worksheet.

2. To find the standard error in an Excel spreadsheet, you will need to enter the value of the population standard deviation, and divide it by the square root of the sample size.  For this problem the format is:  = 3 / sqrt(15).  Your answer is .774597.

3. Now you can use the NORMDIST function to find your probability.

4. Open a blank worksheet.

5. From the Home tab, click on fx, the Insert Function button.

6. The Insert Function box appears.

7. Under the Select a category box, scroll until you find Statistical.

8. Select Statistical by highlighting it with you mouse.

9. You will see new options listed in the Select a function box.

10. Scroll down until you find NORMDIST.  Then click OK.

11. The Function Arguments box appears.

12. In the X box, enter the value of the sample mean.  Here it is 25.  Then click in the Mean box.

13. In the Mean box, enter the value of the mean.  Here it is 27.  Then click in the Standard_dev box.

14. In the Standard_dev box, enter the standard error you calculated.  Here it is .774597.  Then click in the Cumulative box.

15. In the Cumulative box, type in True or the number 1.  Then click OK.

16. Your probability is .004912.  Since we are looking for the probability for the area below 25, and Excel provides us with the probability for the area below the value of x, we have our answer.

Using Excel 2003:

1. Open a blank worksheet.

2. To find the standard error in an Excel spreadsheet, you will need to enter the value of the population standard deviation, and divide it by the square root of the sample size.  For this problem the format is:  = 3 / sqrt(15).  Your answer is .774597.

3. Now you can use the NORMDIST function to find your probability.

4. In Excel, select Insert.

5. Now select Function.

6. Under the Select a Category box, scroll until you find Statistical.

7. Select Statistical by highlighting it with your mouse.

8. You will see new options listed in the Select a function box.

9. Scroll down until you find NORMDIST.  Then click OK.

10. The Function Arguments box appears.

11. In the X box, enter the value of the sample mean.  Here it is 25.  Then click in the Mean box.

12. In the Mean box, enter the value of the population mean.  Here it is 27.  Then click in the Standard_dev box.

13. In the Standard_dev box, enter the standard error you calculated.  Here it is .774597.  Then click in the Cumulative box.

14. In the Cumulative box, type in True or 1.  The click OK.

15. Your probability is .004912.  Since we are looking for the probability for the area below 25, and Excel provides us with the probability for the area below the value of x, we have our answer.

Practice Exercise 6.6.  Sampling Distributions of Means.  The Normal Distribution Applied to a Sample.

A cracker company fills boxes of crackers using special packaging machines.  The fill amounts are normally distributed with a mean of 2 pounds and a standard deviation of 0.3 pound.  A sample of 10 boxes is selected at random.  What is the probability that the mean weight of the sample is more than 2.05 pounds?

We solve this problem using the normal distribution and the concepts discussed in the previous unit, along with the z score formula for a sample.  To solve for the z score we must first find the standard error of the mean.  This is equal to the standard deviation divided by the square root of the sample size.  Therefore we have 0.3 / √10 = .0949.  Now we use the formula (or Excel using NORMDIST) for the z score for a sample to solve the problem.

z = 2.05 - 2.00 / .0949 = .5269 or .53

Now we use the normal tables (or Excel using NORMDIST) to find the probability for a z value of .53.  The probability from the table is equal to .2019.  Since we are looking for the probability being greater than 2.05 pounds, we subtract .2019 from .50 and we get .2981 which is our answer.

Using Excel 2007:

1. Open a blank worksheet.

2. To find the standard error in an Excel spreadsheet, you will need to enter the value of the population standard deviation, and divide it by the square root of the sample size.  For this problem the format is:  =.3 / sqrt(10).  Your answer is .094868.

3. Now you can use the NORMDIST function to find your probability.

4. Open a blank worksheet.

5. From the Home tab, click on fx, the Insert Function button.

6. The Insert Function box appears.

7. Under the Select a category box, scroll until you find Statistical.

8. Select Statistical by highlighting it with you mouse.

9. You will see new options listed in the Select a function box.

10. Scroll down until you find NORMDIST.  Then click OK.

11. The Function Arguments box appears.

12. In the X box, enter the value of the sample mean.  Here it is 2.05.  Then click in the Mean box.

13. In the Mean box, enter the value of the mean.  Here it is 2.00.  Then click in the Standard_dev box.

14. In the Standard_dev box, enter the standard error you calculated.  Here it is .094868.  Then click in the Cumulative box.

15. In the Cumulative box, type in True or the number 1.  Then click OK.

16. Your probability is .70092.  Since we are looking for the probability for the area above 2.05, we need to subtract .70092 from 1.  Your answer is .29908.

Using Excel 2003:

1. Open a blank worksheet.

2. To find the standard error in an Excel spreadsheet, you will need to enter the value of the population standard deviation, and divide it by the square root of the sample size.  For this problem the format is:  =.3 / sqrt(10).  Your answer is .094868.

3. Now you can use the NORMDIST function to find your probability.

4. In Excel, select Insert.

5. Now select Function.

6. Under the Select a Category box, scroll until you find Statistical.

7. Select Statistical by highlighting it with your mouse.

8. You will see new options listed in the Select a function box.

9. Scroll down until you find NORMDIST.  Then click OK.

10. The Function Arguments box appears.

11. In the X box, enter the value of the sample mean.  Here it is 2.05.  Then click in the Mean box.

12. In the Mean box, enter the value of the population mean.  Here it is 2.00.  Then click in the Standard_dev box.

13. In the Standard_dev box, enter the standard error you calculated.  Here it is .094868.  Then click in the Cumulative box.

14. In the Cumulative box, type in True or 1.  The click OK.

15. Your probability is .70092.  Since we are looking for the probability for the area above 2.05, we need to subtract .70092 from 1.  Your answer is .29908.

IMPORTANT INFORMATION!

The most frequent question asked about this unit is when do you need to first find the standard error to solve a normal distribution problem?  You must always calculate the standard error first whenever you have sample data.  Sample data usually contains a reference to a sample in the problem.  It might actually say "sample" or it might say "eight units were selected", or similar language.  If a problem indicates one item or unit was used, that is not a sample.

Assignment

The homework problems for unit 6 can be found by going to the Course Documents link in Blackboard, and clicking on the link for Unit 6.  Look for the Unit 6 - Assignment link.  Once you have completed your homework assignment, you will need to post your answers on Blackboard®.  Once you have signed into Blackboard, simply go to the Unit 6 - Assignment 6 - Post Answers Here link and post your answers.  Immediate feedback is provided once you have completed the posting of all of your answers and clicked on submit.  Make sure you print the entire submitted homework assignment to assist you with quizzes and tests.

©2008, 2007 by E.H. McKay, III.  Version 5.0

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