Section 7.2

1. Section 7.2 • Standard Units and Areas • Under the Standard Normal Distribution

2. Standard Score • The z value or z score tells the number of standard deviations between the original measurement x and the mean μ of the x distribution. • The z value is in standard units.

3. x Values and Corresponding z Values

4. Calculating z scores The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. a. Convert 21 minutes to a z score. b. Convert 29.7 minutes to a z score. Solution a. b.

5. Raw Score • A raw score is the result of converting from standard units (z scores) back to original measurements, x values. • From • We get the formula: x = z s + m

6. Interpreting z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Interpret a z score of 1.60. The delivery time is 28.2 minutes.

7. Standard Normal Distribution is a normal distribution with μ = 0 and σ = 1 m= 0 s= 1 Any x values are converted to z scores. x = z s + m = z * 1 + 0 = z

8. Importance of the Standard Normal Distribution: For μ = 0 and σ = 1 Any Normal Curve: Areas will be equal.

9. Areas of a Standard Normal Distribution We have seen how to convert any normal distribution to the standard normal distribution. We can change any x value to a z value and back again. But what is the is the advantage of all that work? The advantage is that there are extensive tables that show the area under the standard normal curve for almost any interval along the z axis. The areas are important because each area is equal to the probability that the measurement on an item selected at random falls in that interval Thus the standard normal distribution can be a tremendously helpful tool.

10. Areas of a Standard Normal Distribution • Appendix • Table 3 • Pages A6 - A7

11. Use of the Normal Probability Table Appendix Table 3 is a left-tail style table. Entries give the cumulative areas to the left of a specified z.

12. To Find the area to the Left of a Given z score • Find the row associated with the sign, units and tenths portion of z in the left column of Table 3. • Move across the selected row to the column headed by the hundredths digit of the given z.

13. Find the area to the left of z = – 2.84

14. To find the area to the left of z = – 2.84

15. To find the area to the left of z = – 2.84

16. The area to the left of z = – 2.84 is .0023

17. To Find the Area to the Left of a Given Negativez Value: Use Table 3 of the Appendix directly.

18. To Find the Area to the Leftof a Given Positive z Value: Use Table 3 of the Appendix directly.

19. To Find the Area to the Rightof a Given z Value: Subtract the area to the left of z from 1.0000.

20. Alternate Way To Find the Area to the Right of a Given Positive z Value: Use the symmetry of the normal distribution. Area to the right of z = area to left of –z.

21. To Find the Area Between Two z Values Subtract area to left of z1 from area to left of z2 . (When z2 > z1.)

22. Convention for Using Table 3 • Treat any area to the left of a z value smaller than 3.49 as 0.000 • Treat any area to the left of a z value greater than 3.49 as 1.000

23. Example Use Table 3 of the Appendix to find the specified areas. • Find the area between z = 1.00 and z = 2.70. • Find the area to the right of z = 0.94. Solution • (Area between 1.00 and 2.70) = = (Area left to 2.70) – (Area left to 1.00) = 0.9965 – 0.8413 = 0.1552 • (Area to the right of 0.94) = = (Area under entire curve) – (Area to the left of 0.94) = 1.000 – 0.8264 = 0.1736 Alternatively: (Area to the right of 0.94) = (Area to the left of - 0.94) = 0.1736.

24. Use of the Normal Probability Table a. P( z < 1.64 ) = __________ b. P( z < - 2.71 ) = __________ .9495 .0034 .3925 c. P(0 < z < 1.24) = ______ d. P(0 < z < 1.60) = _______ e. P( 2.37 < z < 0) = ______ .4452 .4911

25. Use of the Normal Probability Table .9974 f. P( 3 < z < 3 ) = _______ g. P( 2.34 < z < 1.57 ) = _____ h. P( 1.24 < z < 1.88 ) = _______ .9322 .0774 .2254 i. P( 2.44 < z < 0.73 ) = _______ j . P( z > 2.39 ) = ________ k. P( z > 1.43 ) = __________ .0084 .9236 Assignment 14

26. Example:Graph and investigate the normal distribution curve where the mean is 0 and the standard deviation is 1. For graphing the normal distribution, choose normalpdf. The normalpdf (normal probability density function) is found under DISTR (2nd VARS) #1normalpdf(. Go to the Y = menu. The parameters will be (variable, mean, standard deviation). Adjust the WINDOW.You will have to set your own window.  Guideline is: Xmin = mean - 3 SDXmax = mean + 3 SDXscl = SDYmin = 0Ymax = 1/(2 SD)Yscl = 0 GRAPH.  Using TRACE, simply type the desired x value and the point will be plotted. 

27. Example cont. Investigate:  What happens to the curve as the standard deviation increases? Double the standard deviation and see what happens to the graph.When graphing 2 normal curves, the window will need to be adjusted. Xmin = mean - 3 (largest SD)Xmax = mean + 3 (largest SD)Xscl = largest SDYmin = 0Ymax = 1/(2 smallest SD)Yscl = 0 Observe that as the standard deviation increases, the morespread out the graph becomes.  

28. Example cont. Now, the area under the curve between particular values represents the probabilities of events occurring within that specific range.  This area can be seen using the command ShadeNorm(. To find ShadeNorm(, go to DISTR and right arrow to DRAW.   Choose #1:ShadeNorm(.ShadeNorm (lower bound, upperbound, mean, standard deviation)By entering parameters -1,1 you will see the area, indicating approximately 68% probability of a score falling within 1 standard deviation from the mean in a normally distributed set of values.  Since the calculator defaults to a mean of 0 and standard deviation of 1, it was not necessary to enter these values in this example, but is is a good idea to get in the habit of entering all 4 parameters.  Notice how this answer supports the percentage listed in the chart of the standard normal distribution

29. Example cont.