The normal distribution can be used as a model to solve many problems about variables that are approximately normally distributed. Since each variable has its own mean and standard deviation, statisticians use what is called thestandard normal distribution to solve the problems.
The standard normal distribution has all the properties of a normal dis- tribution, but the mean is zero and the standard deviation is one. See Figure 9-10.
Fig. 9-9. Fig. 9-8.
A value for any variable that is approximately normally distributed can be transformed into a standard normal value by using the following formula:
z¼ valuemean
standard deviation
The standard normal values are calledz values or z scores.
EXAMPLE: Find the corresponding z value for a value of 18 if the mean of a variable is 12 and the standard deviation is 4.
SOLUTION: z¼ valuemean standard deviation¼ 1812 4 ¼ 6 4¼1:5
Hence thezvalue of 1.5 corresponds to a value of 18 for an approximately normal distribution which has a mean of 12 and a standard deviation of 4.
z values are negative for values of variables that are below the mean.
EXAMPLE:Find the correspondingzvalue for a value of 9 if the mean of a variable is 12 and the standard deviation is 4.
SOLUTION: z¼ valuemean standard deviation¼ 912 4 ¼ 3 4¼ 0:75
Hence in this case a value of 9 is equivalent to a zvalue of 0.75. In addition to finding probabilities for values that are between zero, one, two, and three standard deviations of the mean, probabilities for other values
Fig. 9-10.
CHAPTER 9
The Normal Distribution
can be found by converting them to zvalues and using the standard normal distribution.
Areas between any two given z values under the standard normal distribution curve can be found by using calculus instead; however, tables for specificz values can be found in any statistics textbook. An abbreviated table of areas is shown in Table 9-1.
Table 9-1 Approximate Cumulative Areas for the Standard Normal Distribution
z Area z Area z Area z Area
3.0 .001 1.5 .067 0.0 .500 1.5 .933 2.9 .002 1.4 .081 0.1 .540 1.6 .945 2.8 .003 1.3 .097 0.2 .579 1.7 .955 2.7 .004 1.2 .115 0.3 .618 1.8 .964 2.6 .005 1.1 .136 0.4 .655 1.9 .971 2.5 .006 1.0 .159 0.5 .692 2.0 .977 2.4 .008 0.9 .184 0.6 .726 2.1 .982 2.3 .011 .08 .212 0.7 .758 2.2 .986 2.2 .014 0.7 .242 0.8 .788 2.3 .989 2.1 .018 0.6 .274 0.9 .816 2.4 .992 2.0 .023 0.5 .309 1.0 .841 2.5 .994 1.9 .029 0.4 .345 1.1 .864 2.6 .995 1.8 .036 0.3 .382 1.2 .885 2.7 .997 1.7 .045 0.2 .421 1.3 .903 2.8 .997 1.6 .055 0.1 .460 1.4 .919 2.9 .998 3.0 .999
This table gives the approximate cumulative areas for z values between
3 and þ3. The next three examples will show how to find the area (and corresponding probability in decimal form).
EXAMPLE: Find the area under the standard normal distribution curve to the left of z¼1.3.
SOLUTION:
The area is shown in Figure 9-11.
In order to find the area under the standard normal distribution curve to the left of any givenzvalue, just look it up directly in Table 9-1. The area is 0.903 or 90.3%.
EXAMPLE: Find the area under the standard normal distribution curve between z¼ 1.6 andz¼0.8.
SOLUTION:
The area is shown in Figure 9-12.
Fig. 9-11.
Fig. 9-12.
CHAPTER 9
The Normal Distribution
To find the area under the standard normal distribution curve between any given two z values, look up the areas in Table 9-1 and subtract the smaller area from the larger. In this case the area corresponding toz¼ 1.6 is 0.055, and the area corresponding to z¼0.8 is 0.788, so the area betweenz¼ 1.6 and z¼0.8 is 0.7880.055¼0.733¼73.3%. In other words, 73.3% of the area under the standard normal distribution curve is between z¼ 1.6 and
z¼0.8.
EXAMPLE: Find the area under the standard normal distribution curve to the right ofz¼ 0.5.
SOLUTION:
The area is shown in Figure 9-13.
To find the area under the standard normal distribution curve to the right of any given z value, look up the area in the table and subtract that from 1. The area corresponding to z¼0.5 is 0.309. Hence 10.309¼0.691. The area to the right of z¼0.5 is 0.691. In other words, 69.1% of the area under the standard normal distribution curve lies to the right of
z¼ 0.5.
Using Table 9-1 and the formula for transforming values for variables that are approximately normally distributed, you can find the probabilities of various events.
EXAMPLE: The scores on a national achievement exam are normally dis- tributed with a mean of 500 and a standard deviation of 100. If a student is selected at random, find the probability that the student scored below 680.
Fig. 9-13.
SOLUTION:
Find the zvalue for 680:
z¼ valuemean standard deviation¼ 680500 100 ¼ 180 100¼1:8 Draw a figure and shade the area. See Figure 9-14.
Look up 1.8 in Table 9-1 and find the area. It is 0.964. Hence the probability that a randomly selected student scores below 680 is 0.964 or 96.4%.
EXAMPLE:The average life of a certain brand of automobile tires is 24,000 miles under normal driving conditions. The standard deviation is 2000 miles, and the variable is approximately normally distributed. For a randomly selected tire, find the probability that it will last between 21,800 miles and 25,400 miles.
SOLUTION:
Find the twoz values using the formulaz¼ valuemean
standard deviation. The z value for 21,800 miles is
z¼21ó80024ó000
2000 ¼
2200
2000 ¼ 1:1
The corresponding area from Table 9-1 for 1.1 is 0.136. The z value for 25,400 miles is
z¼25ó40024ó000
2000 ¼0:7
The corresponding area from Table 9-1 for 0.7 is 0.758.
Fig. 9-14.
CHAPTER 9
The Normal Distribution
Draw and label the normal distribution curve. Shade in the area. See Figure 9-15.
Subtract the smaller area from the larger area, 0.7580.136¼0.622. Hence the probability that a randomly selected tire will last between 21,800 and 25,400 miles is 0.622 or 62.2%.
EXAMPLE: The average time it takes college freshmen to complete a reasoning skills test is 24 minutes. The standard deviation is 5 minutes. If a randomly selected freshman takes the exam, find the probability that he or she takes more than 32 minutes to complete the test. Assume the variable is normally distributed.
SOLUTION:
Find thezvalue for 32 minutes:
z¼ valuemean standard deviation¼ 3224 5 ¼ 8 5¼1:6
Draw the standard normal distribution and label it as shown. Shade the appropriate area. See Figure 9-16.
Fig. 9-16. Fig. 9-15.
The area for z¼1.6 from Table 9-1 is 0.945. Subtract the area from 1. 1.000.945¼0.055. Hence the probability that it will take a randomly selected student longer than 32 minutes to complete the test is 0.055 or 5.5%.
PRACTICE
1. In order to qualify for a position, an applicant must score 86 or above on a skills test. If the test scores are normally distributed with a mean of 80 and a standard deviation of 4, find the probability that a ran- domly selected applicant will qualify for the position.
2. If a brisk walk at 4 miles per hour burns an average of 300 calories per hour, find the probability that a person will burn between 260 and 290 calories if the person walks briskly for one hour. Assume the standard deviation is 20 and the variable is approximately normally distributed.
3. The average count for snow per year that a city receives is 40 inches. The standard deviation is 10 inches. Find the probability that next year the city will get less than 53 inches. Assume the variable is normally distributed.
4. If the average systolic blood pressure is 120 and the standard devia- tion is 10, find the probability that a randomly selected person will have a blood pressure less than 108. Assume the variable is normally distributed.
5. A survey found that on average adults watch 2.5 hours of television per day. The standard deviation is 0.5 hours. Find the probability that a randomly selected adult will watch between 2.2 and 2.8 hours per day. Assume the variable is normally distributed.
ANSWERS
1. z¼8680
4 ¼
6 4¼1:5
The required area is shown in Figure 9-17.
The area for z¼1.5 is 0.933. Since we are looking for the area greater than z¼1.5, subtract the table value from 1: 10.933¼
0.067. Hence the probability is 0.067 or 6.7%.
CHAPTER 9
The Normal Distribution
2. z¼290300 20 ¼ 10 20¼ 0:5 z¼260300 20 ¼ 40 20¼ 2
The required area is shown in Figure 9-18.
The area for0.5 is 0.309. The area for2 is 0.023. Since we are looking for the area betweenz¼ 0.5 andz¼ 2, subtract the areas. Hence the probability that a person burns between 260 and 290 calories is 0.3090.023¼0.286 or 28.6%.
3. z¼5340
10 ¼ 13 10¼1:3
The required area is shown in Figure 9-19.
Fig. 9-18. Fig. 9-17.
The area forz¼1.3 is 0.903. Since we are looking for an area less thanz¼1.3, use the value found in the table. Hence the probability that the city receives less than 53 inches of snow is 0.903 or 90.3%.
4. z¼108120
10 ¼
12
10¼ 1:2
The required area is shown in Figure 9-20.
The area forz¼ 1.2 is 0.115. Since we are looking for the area less thanz¼ 1.2, use the value in the table. Hence the probability that a person has a systolic blood pressure less than 108 is 0.115 or 11.5%.
5. z¼2:82:5 0:5 ¼ 0:3 0:5¼0:6 z¼2:22:5 0:5 ¼ 0:3 0:5¼ 0:6
The required area is shown in Figure 9-21.
Fig. 9-20. Fig. 9-19.
CHAPTER 9
The Normal Distribution
The area for z¼0.6 is 0.726. The area for z¼ 0.6 is 0.274. Since we are looking for the area between0.6 and 0.6, subtract the areas: 0.7260.274¼0.452 or 45.2%. Hence the probability that an adult will watch between 2.2 and 2.8 hours of television per day is 0.452 or 45.2%.
Summary
Statistics is a branch of mathematics that uses probability. Statistics uses data to analyze, summarize, make inferences, and draw conclusions from data. There are three commonly used measures of average. They are the mean, median, and mode. The mean is the sum of the data values divided by the number of data values. The median is the midpoint of the data values when they are arranged in numerical order. The mode is the data value that occurs most often.
There are two commonly used measures of variability. They are the range and standard deviation. The range is the difference between the smallest data value and the largest data value. The standard deviation is the square root of the average of the squares of the differences of each value from the mean.
Many variables are approximately normally distributed and the standard normal distribution can be used to find probabilities for various situations involving values of these variables.
The standard normal distribution is a continuous, bell-shaped curve such that the mean, median, and mode are at its center. It is also symmetrical about the mean. The mean is equal to zero and the standard deviation is equal to one. About 68% of the area under the standard normal distribution lies within one standard deviation of the mean, about 95% within two standard deviations, and about 100% within three standard deviations of the mean.
Fig. 9-21.
CHAPTER QUIZ
1. What is the mean of 18, 24, 6, 12, 15, and 12? a. 12
b. 14.5 c. 87 d. 10.6
2. What is the median of 25, 15, 32, 43, 15, and 6? a. 20
b. 15 c. 22.8 d. 23.5
3. What is the mode of 17, 19, 26, 43, 26, 15, and 14? a. 22.9
b. 26 c. 17 d. None
4. What is the range of 32, 50, 14, 26, 18 and 25? a. 36
b. 25 c. 18 d. 43
5. What is the standard deviation of 12, 18, 14, 22, and 21? a. 3.9
b. 18.8 c. 4.3 d. 17.4
6. Which isnot a property of the normal distribution? a. It is bell-shaped.
b. It is continuous.
c. The mean is at the center.
d. It is not symmetrical about the mean.
CHAPTER 9
The Normal Distribution
7. The area under the standard normal distribution is a. 1 or 100%
b. 0.5 or 50% c. Unknown d. Infinite
8. For the standard normal distribution,
a. The mean¼1 and the standard deviation¼0. b. The mean¼1 and the standard deviation¼2. c. The mean¼0 and the standard deviation¼1. d. The mean and standard deviation can both vary.
9. What percent of the area under the standard normal distribution lies within two standard deviations of the mean?
a. 68% b. 95% c. 100% d. Unknown
10. When the value of a variable is transformed into a standard normal variable, the new value is called a(n)
a. xvalue b. y value c. 0 value d. z value
Use Table 9-1 to answer questions 11 through 15.
11. The area under the standard normal distribution to the right of
z¼0.9 is a. 90.1% b. 18.4% c. 81.6% d. 10.2 %
12. The area under the standard normal distribution to the left of
z¼ 1.2 is a. 88.5% b. 62.3% c. 48.7% d. 11.5%
13. The area under the standard normal distribution between z¼ 1.7 and z¼0.5 is a. 69.2% b. 4.5% c. 35.6% d. 64.7%
14. An exam which is approximately normally distributed has a mean of 200 and a standard deviation of 20. If a person who took the exam is selected at random, find the probability that the person scored above 230.
a. 93.3% b. 70.2% c. 6.7% d. 30.8%
15. The average height for adult females is 64 inches. Assume the variable is normally distributed with a standard deviation of 2 inches. If a female is randomly selected, find the probability that her height is between 62 and 66.8 inches.
a. 92% b. 76% c. 32% d. 16%