1. Advanced Algebra
The Normal Curve
Ernesto Diaz 1

2. Copyright © Cengage Learning. All rights reserved.
14.3
The Normal Curve
Copyright © Cengage Learning. All rights reserved.

3. Bell-Shaped Curves

4. Bell-Shaped Curves
Suppose we survey the results of 20 children’s scores on an IQ test.
The scores (rounded to the nearest 5 points) are 115, 90, 100, 95, 105, 95, 105, 105, 95, 125, 120, 110, 100, 100, 90, 110, 100, 115, 105, and 80.
We can find = 103 and s 

5. Bell-Shaped Curves
A frequency graph of these data is shown in part a of Figure If we consider 10,000 scores instead of only 20, we might obtain the frequency distribution shown in part b of Figure
a. IQs of 20 children
b. IQs of 10,000 children
Frequency distributions for IQ scores
Figure 14.34

6. The Normal Distribution: as mathematical function (pdf)
This is a bell shaped curve with different centers and spreads depending on  and 
Note constants: = e=

7. **The beauty of the normal curve:
No matter what  and  are, the area between - and + is about 68%; the area between -2 and +2 is about 95%; and the area between -3 and +3 is about 99.7%. Almost all values fall within 3 standard deviations.

8. 68-95-99.7 Rule 68% of the data 95% of the data 99.7% of the data
SAY: within 1 standard deviation either way of the mean
within 2 standard deviations of the mean
within 3 standard deviations either way of the mean
WORKS FOR ALL NORMAL CURVES NO MATTER HOW SKINNY OR FAT
95% of the data
99.7% of the data

9. Bell-Shaped Curves
The data illustrated in Figure approximate a commonly used curve called a normal frequency curve, or simply a normal curve. (See Figure )
A normal curve
Figure 14.35

10. Bell-Shaped Curves
If we obtain the frequency distribution of a large number of measurements (as with IQ), the corresponding graph tends to look normal, or bell-shaped. The normal curve has some interesting properties.
In it, the mean, the median, and the mode all have the same value, and all occur exactly at the center of the distribution; we denote this value by the Greek letter mu (). The standard deviation for this distribution is  (sigma).

11. Bell-Shaped Curves
The normal distribution is a continuous (rather than a discrete) distribution, and it extends indefinitely in both directions, never touching the x-axis.
It is symmetric about a vertical line drawn through the mean, . Graphs of this curve for several choices of  are shown in Figure
Variations of normal curves
Figure 14.36

12. Example 2 – Find a normal distribution
Predict the distribution of IQ scores of 1,000 people if we assume that IQ scores are normally distributed, with a mean of 100 and a standard deviation of 15.
Solution:
First, find the breaking points around the mean For  = 100 and  = 15:
 +  = =  –  = 100 – 15 = 85
 + 2 = (15) =  – 2 = 100 – 2(15) = 70
 + 3 = (15) =  – 3 = 100 – 3(15) = 55

13. Example 2 – Solution
cont’d
We use Figure to find that 34.1% of the scores will be between 100 and 115 (i.e.,between  and  + 1) :
0.341  1,000 = 341
A normal curve
Figure 14.35

14. Example 2 – Solution
cont’d
About 13.6% will be between 115 and 130 (between  + 1 and  + 2):
0.136  1,000 = 136
About 2.2% will be between 130 and 145 (between  + 2 and  + 3):
0.022  1,000 = 22
About 0.1% will be above 145 (more than  + 3):
0.001  1,000 = 1

15. Example 2 – Solution
cont’d
The distribution for intervals below the mean is identical, since the normal curve is the same to the left and to the right of the mean. The distribution is shown below.

16. z-Scores

17. z-Scores
First, we introduce some terminology. We use z-scores (sometimes called standard scores) to determine how far, in terms of standard deviations, a given score is from the mean of the distribution.
We use the z-score to translate any normal curve into a standard normal curve (the particular normal curve with a mean of 0 and a standard deviation of 1) by using the definition.

18. z-Scores

19. Example 5 – Find probabilities in a normal distribution
The Eureka Lightbulb Company tested a new line of lightbulbs and found their lifetimes to be normally distributed, with a mean life of 98 hours and a standard deviation of 13 hours.
a. What percentage of bulbs will last less than 72 hours?
b. What percentage of bulbs will last less than 100 hours?
c. What is the probability that a bulb selected at random will last longer than 111 hours?
d. What is the probability that a bulb will last between and 120 hours?

20. Example 5 – Solution
cont’d
Draw a normal curve with mean 98 and standard deviation 13, as shown in Figure
Lightbulb lifetimes are normally distributed
Figure 14.38

21. Example 5 – Solution
cont’d
For this example, we are given  = 98 and  = 13 .
a. For x = 72, z =
= –2.
This is 2 standard deviations below the mean.

22. Example 5 – Solution
cont’d
The percentage we seek is shown in blue in Figure 14.39a.
About 2.3% (2.2% + 0.1% = 2.3%) will last less than 72 hours. We can also use the z-score to find the percentage.
2 standard deviations below the mean
Using z -scores
Figure 14.39(a)

23. Example 5 – Solution
cont’d
From Table 14.10, the area to the left of z = –2 is 0.5 – = , so the probability is about 2.3%.
Standard Normal Distribution
Table 14.10

24. Example 5 – Solution
cont’d
Standard Normal Distribution
Table 14.10

25. Example 5 – Solution  0.15. b. For x = 100, z =
cont’d
b. For x = 100, z =
This is 0.15 standard deviation above the mean. From Table 14.10, we find (this is shown in green in Figure 14.39b).
 0.15.
0.15 standard deviations above the mean
Using z-scores
Figure 14.39(b)

26. Example 5 – Solution
cont’d
Since we want the percent of values less than 100, we must add 50% for the numbers below the mean (shown in blue). The percentage we seek is
= or about 56.0%.

27. Example 5 – Solution c. For x = 111, z = = 1.
cont’d
c. For x = 111, z = = 1.
This is one standard deviation above the mean, which is the same as the z-score. The percentage we seek is shown in blue (see Figure 14.39c).
1 standard deviations above the mean
Using z-scores
Figure 14.39(c)

28. Example 5 – Solution
cont’d
We know that about 15.9% (13.6% + 2.2% + 0.1% = 15.9%) of the bulbs will last longer than 111 hours, so
P(bulb life > 111 hours)  0.159
We can also use Table For z = 1.00, the table entry is , and we are looking for the area to the right, so we compute
– =

29. Example 5 – Solution
cont’d
d. We first find the z-scores using  = 98 and  = 13. For
x = 106, z =  0.62.
From Table 14.10, the area between the z-score and the mean is (shown in green).
For x = 120, z =  1.69.

30. Example 5 – Solution
cont’d
From Table 14.10, the area between this z-score and the mean is The desired answer (shown in yellow) is approximately
– =
Since percent and probability are the same, we see the probability that the life of the bulb is between 106 and hours is about 22.2%.

31. z-Scores

32. z-Scores
Sometimes data do not fall into a normal distribution, but are skewed, which means their distribution has more tail on one side or the other. For example, Figure 14.40a shows that the 1941 scores on the SAT exam (when the test was first used) were normally distributed.
1941 SAT scale
Distribution of SAT scores
Figure 14.40(a)

33. z-Scores
However, by 1990, the scale had become skewed to the left, as shown in Figure 14.40b.
1990 SAT scale
Distribution of SAT scores
Figure 14.40(b)

34. z-Scores
In a normal distribution, the mean, median, and mode all have the same value, but if the distribution is skewed, the relative positions of the mean, median, and mode would be as shown in Figure
a. Skewed to the right
(positive skew)
b. Normal distribution
c. Skewed to the left
(negative skew)
Comparison of three distributions
Figure 14.41