1. 6 Normal Curves and Sampling Distributions
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2. Areas Under Any Normal Curve
Section
6.3
Areas Under Any Normal Curve
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3. Focus Points Compute the probability of “standardized events.”
Find a z score from a given normal probability (inverse normal).
Use the inverse normal to solve guarantee problems.

4. Normal Distribution Areas

5. Normal Distribution Areas
In many applied situations, the original normal curve is not the standard normal curve.
Generally, there will not be a table of areas available for the original normal curve.
This does not mean that we cannot find the probability that a measurement x will fall into an interval from a to b. What we must do is convert the original measurements x, a, and b to z values.

6. Normal Distribution Areas
Procedure:

7. Example 7 – Normal distribution probability
Let x have a normal distribution with  = 10 and  = 2. Find the probability that an x value selected at random from this distribution is between 11 and 14. In symbols, find P(11  x  14).
Solution:
Since probabilities correspond to areas under the distribution curve, we want to find the area under the x curve above the interval from x = 11 to x = 14.
To do so, we will convert the x values to standard z values and then use Table 5 of Appendix II to find the corresponding area under the standard curve.

8. Example 7 – Solution We use the formula
cont’d
We use the formula
to convert the given x interval to a z interval.
(Use x = 11,  = 10,  = 2.)
(Use x = 14,  = 10,  = 2.)

9. Example 7 – Solution
cont’d
The corresponding areas under the x and z curves are shown in Figure 6-23.
Corresponding Areas Under the x Curve and z Curve
Figure 6-23

10. Example 7 – Solution From Figure 6-23 we see that
cont’d
From Figure 6-23 we see that
P(11  x  14) = P(0.50  z  2.00)
= P(z  2.00) – P(z  0.50)
= – =
Interpretation The probability is that an x value selected at random from a normal distribution with mean 10 and standard deviation 2 lies between 11 and 14.

11. Inverse Normal Distribution

12. Inverse Normal Distribution
Sometimes we need to find z or x values that correspond to a given area under the normal curve.
This situation arises when we want to specify a guarantee period such that a given percentage of the total products produced by a company last at least as long as the duration of the guarantee period. In such cases, we use the standard normal distribution table “in reverse.”
When we look up an area and find the corresponding z value, we are using the inverse normal probability distribution.

13. Example 8 – Find x, given probability
Magic Video Games, Inc., sells an expensive video games package. Because the package is so expensive, the company wants to advertise an impressive guarantee for the life expectancy of its computer control system.
The guarantee policy will refund the full purchase price if the computer fails during the guarantee period.
The research department has done tests that show that the mean life for the computer is 30 months, with standard deviation of 4 months.

14. Example 8 – Find x, given probability
cont’d
The computer life is normally distributed. How long can the guarantee period be if management does not want to refund the purchase price on more than 7% of the Magic Video packages?

15. Example 8 – Solution
Let us look at the distribution of lifetimes for the computer control system, and shade the portion of the distribution in which the computer lasts fewer months than the guarantee period. (See Figure 6-26.)
7% of the Computers Have a Lifetime Less Than the Guarantee Period
Figure 6-26

16. Example 8 – Solution
cont’d
If a computer system lasts fewer months than the guarantee period, a full-price refund will have to be made.
The lifetimes requiring a refund are in the shaded region in Figure This region represents 7% of the total area under the curve.
We can use Table 5 of Appendix II to find the z value such that 7% of the total area under the standard normal curve lies to the left of the z value.

17. Example 8 – Solution
cont’d
Then we convert the z value to its corresponding x value to find the guarantee period. We want to find the z value with 7% of the area under the standard normal curve to the left of z.
Since we are given the area in a left tail, we can use Table 5 of Appendix II directly to find z. The area value is

18. Example 8 – Solution
cont’d
However, this area is not in our table, so we use the closest area, which is , and the corresponding z value of z = –1.48(see Table 6-5).
Excerpt from Table 5 of Appendix II
Table 6-5

19. Example 8 – Solution
cont’d
To translate this value back to an x value (in months), we use the formula
x = z + 
= –1.48(4) (Use  = 4 months and  = 30 months.)
= months
Interpretation The company can guarantee the Magic Video Games package for x = 24 months. For this guarantee period, they expect to refund the purchase price of no more than 7% of the video games packages.

20. Inverse Normal Distribution
Procedure:

21. Inverse Normal Distribution
cont’d

22. Example 10 – Assessing normality
Consider the following data, which are rounded to the nearest integer.

23. Example 10(a) – Assessing normality
Look at the histogram and box-and-whisker plot generated by Minitab in Figure 6-30 and comment about normality of the data from these indicators.
Solution: Note that the histogram is approximately normal. The box-and whisker plot shows just one outlier. Both of these graphs indicate normality.
Histogram and Box-and-Whisker Plot
Figure 6-30

24. Example 10(b) – Assessing normality
Use Pearson’s index to check for skewness.
Solution:
Summary statistics from Minitab:
We see that x = 19.46,median = 19.5, and s = 2.705
Pearson’s index
Since the index is between –1 and 1, we detect no skewness. The data appear to be symmetric.

25. Example 10(c) – Assessing normality
Look at the normal quantile plot in Figure 6-31 and comment on normality.
Solution:
The data fall close to a straight line, so the data appear to come from a normal distribution.
Normal Quantile Plot
Figure 6-31

26. Example 10(d) – Assessing normality
Interpretation Interpret the results.
Solution:
The histogram is roughly bell-shaped, there is only one outlier, Pearson’s index does not indicate skewness, and the points on the normal quantile plot lie fairly close to a straight line.
It appears that the data are likely from a distribution that is approximately normal.