Lesson 5

Discover

 

In Lesson 4 you learned that z-scores could be used to determine how many standard deviations a data point was away from the mean. You used the z-score and the area under the standard normal curve to determine the percent of data that was less than or greater than a given value. Z-scores and the area under the standard normal curve can also be used to determine the percent of data that is within two data values.

 

Consider the following situation.

 

One of the best-selling items from OJ Juice Company is its 500-mL bottle of orange juice. When the production process is running smoothly, the volume of juice in the bottles is normally distributed with a mean of 500 mL and a standard deviation of 0.75 mL. The company’s quality control department has set a range of volumes that each bottle leaving the plant must meet. Each bottle of orange juice must be between 498.0 mL and 502.0 mL.

 

Try This 1

 

Use the “Areas Under the Normal Distribution (x and z-values)” applet to determine the z-scores and the area under the curve that corresponds to a lower limit of 498.0 mL and an upper limit of 502.0 mL for the volumes in OJ Juice Company’s 500-mL bottles.

 

 

Follow the guidelines for using the applet.

 

First, you need to enter the mean and the standard deviation by clicking on the mean and standard deviation buttons that are located in the top left corner of the applet. Enter the appropriate value, and then click OK. (If reading the values in this multimedia piece is a challenge, you may find it helpful to increase the text size in your web browser by selecting “Ctrl” and “+.”)

 

Now you can move the blue and green sliders to the corresponding data points (i.e., x = 498.0 mL and x = 502.0 mL). As you move the sliders, the applet will use the z-score formula to calculate the corresponding z-score for each value of x. The calculations are shown to the right of the curve. To show the z-score calculation for the lower limit, click on the blue z-score box. To show the z-score calculation for the upper limit, click on the green z-score box.

 

Note: You may want to verify the z-score calculations manually as the applet may have rounding errors.

 

The area under the curve is shown in red. The percentage of data that is within the two z-scores (and corresponding x-values) is shown to the left of the curve.

1. What percentage of OJ Juice Company’s bottles contain between 498.0 mL and 502.0 mL?
   
   
2. If 15 000 bottles of orange juice are produced each day, how many bottles would meet the quality standard of volumes between 498.0 mL and 502.0 mL?
   
   
3. Suppose the management at OJ Juice Company wants to shorten the acceptable range of volumes to a minimum of 499.0 mL and a maximum of 500.5 mL. What percentage of bottles will meet this new quality standard?
   
   
4. Of the 15 000 bottles of orange juice produced each day, how many bottles would not meet this new quality standard?

Answer

 

Share 1

 

Compare your results from Try This 1 with another student or appropriate partner. Once you have come to an agreement on the correct answers, discuss the following statement and question:

 

Putting product back through a production line or discarding product that doesn’t meet certain specifications is expensive. What action might the OJ Juice Company take to ensure more bottles meet the new quality standard?