Some Material Used Was Contributed by Tom Koesters East Forsyth High School
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Some material used was contributed by Tom Koesters – East Forsyth High School Section 16.2 – Normal Distributions Skewed data is described based on location of tail (longer “half” of distribution) Positively Skewed (Right) Negatively Skewed (Left) “Mean is to Right of Median” “Mean is to Left of Median
DISTRIBUTIONS
Median Mean Mean Median Normal Curves and Normal Distributions A set of data whose distribution has a bell-shaped curve is said to have a NORMAL DISTRIBUTION and the curve is called a NORMAL CURVE. The area under the curve is equal to 1.0 probability or 100%.
SHAPE: Symmetry: A normal curve is symmetrical along a vertical line (median) splitting the bell in half. CENTER: Mean equals Median: μ = M In a perfectly normal distribution, the line of symmetry is the median M and the mean μ of the data. o 50% of the data are greater than or equal to the mean o 50% are less than or equal to the mean.
Standard Deviation and Point of Inflection. A point of inflection P on a curve is the point where the concavity of the curve P P changes, and is one standard deviation σ away from the center (mean or median) of the data. σ P = μ ± σ μ–σ μ+σ
Example #1: Find the mean, median and standard deviation of each normal distribution shown. The point of inflection of the curve is P is the point of inflection of the curve.
a. μ = ___78_____ b. μ = ___82_____ c. μ = __87______M = __78_____ M = __82_____ M = __87_____ σ = _96 – 78 = 18 σ = _99 – 82 = 17 σ = _93 – 87 = 6 Some material used was contributed by Tom Koesters – East Forsyth High School Example #2: Consider the following normal distributions where P and P′ are the points of inflection of the curve. Find the mean and standard deviation. (round to the nearest tenth).
Total Distance = 12 half = 6 2a. μ = _70 = (76+64)/2_ 2b. μ = _79 = (86+72)/2_ 2c. μ = 79 = (87+71)/2_ σ = __6______σ = __7______σ = __8______
Section 16.3 – Standardizing Normal Data
Standardized Value, Z-Value, or Z-Score: To standardize the comparison of data in we refer to data values based on the number of standard deviations they are above or below the mean. This standardized value is often called the z-value or z-score . EXAMPLE: Given mean is 8 and standard deviation is 3 11 has z-value = 1.0 because 11 is 3 above 8 or one standard deviation above the mean 5 has z-value = -1.0 because 5 is 3 below 8 or one standard deviation below the mean
Z – Score Interpretation:
A. Z-value of 2.5 describes data that is __2.5___ standard deviation(s) ___above____ the mean?
B. Z-value of -2 describes data that is ___2___ standard deviation(s) ___below____ the mean?
C. Z-value of -0.3 describes data that is __0.3__ standard deviation(s) ___below___ the mean?
D. Z-value of 1.6 describes data that is __1.6___ standard deviation(s) ___above___ the mean?
z-score of zero (z = 0) is the mean of the data. (Zero means Not above or below) X z-value Calculation: z The z-value calculation is the difference of the data value (X) and mean (μ) all over the standard deviation (σ). Some material used was contributed by Tom Koesters – East Forsyth High School Example #1: Consider a normally distributed data set with a mean μ = 45 ft and a standard deviation σ = 10 ft. Find the z-value for the given data values.
1a) X = 55 ft. 55 45 1c) X = 50 ft. z 1.0 50 45 10 z 0.5 55 ft. is one σ above the mean 10 50 ft. is one-half σ above the mean
1b) X = 20 ft. 20 45 1d) X = 21.58 ft. z 2.5 21.58 45 10 z 2.342 35 ft. is two and a half σ below the mean 10 21.58 ft. is 2.342 σ below the mean
Example #2: Consider a normally distributed data set with μ = 18.5 and σ = 1.2. Find the z-values. a) X = 22.1 b) X = 16.34 c) X = 19.3 22.1 18.5 19.3 18.5 z 3.0 16.34 18.5 z 0.66 1.2 z 1.8 1.2 1.2
Solving for data point given the standardized value: z -value to X . Example #3: A normally distributed data set of recent test scores has a mean, μ = 85, and standard deviation, σ = 4.5. You are told that you have a z-score of 2.1. What is your actual test score?
2.1 Standard Deviations above mean of 85
85 + 2.1 (4.5) = 94.45 Example 4: A normally distributed data set of heights has a mean, μ = 68 inches, and standard deviation, σ = 3 inches. A student has a z-score of -1.5. Some material used was contributed by Tom Koesters – East Forsyth High School What is the student’s actual height?
68 – 1.5(3) = 63.5
Example #5: Consider a normally distributed data set with mean μ = 235.7 meters and standard deviation σ = 41.58 meters. Find the data value X that corresponds to the standardized value a) z = –3.45? b) z = 2.6? X 343.808m X 235.7 Above Mean by 2.6 c) z = 0.3? 3.45 standard deviations X 235.7 41.58 0.3 X 92.249m 235.7 2.6(41.58) 41.58 X 248.174m
Example #6: Consider a normally distributed data set with μ = 12.5 and σ = 0.35. Find data value X that corresponds to the standardized value? a) z = 1.5? b) z = -2.4? 1.5 standard deviation X 12.5 c) z = 3? 2.4 above Mean .35 3 standard deviation 12.5 1.5(.35) 13.025 X 11.66 above Mean 12.5 3(.35) 13.55
Finding Standard Deviation and Mean based on z-scores Example #7: In a normal distribution, the data value of 45 has a z-score of 2 and the date value of 27 has a z-score of -1. A. What is the difference in data values? B. What is the different in z-scores? 45 – 27 = 18 2 - -1 = 3 standard deviations
C. What is the standard deviation? 3(SDs) = 18 SD = 6
D. What is the mean? 27 + 6 = 33 OR 45 – 2(6) = 33 Example #8: In a normal distribution, the data value of 4.8 has a z-score of -3 and the date value of 17.2 has a z-score of 1. What is the mean and standard deviation? 4.8 + 12.4 = 17.2 4(SDs) = 12.4 SD = 3.1 Some material used was contributed by Tom Koesters – East Forsyth High School Mean: 4.8 + 3.1 (3) = 14.1 = 17.2 – 3.1 ( 1)
Example #9: In a normal distribution, the data value of 173 has a z-score of -2 and the date value of 238 has a z-score of 3. What is the mean and standard deviation? 172 + 66 = 238 5(SDs) = 66 SD = 13.2 Mean: 173 + 13.2 (2) = 198.4 = 238 – 13.2(3)
PERCENTAGES/PROBABILITIES WITH NORMAL CURVES NORMAL = SYMMETRICAL
1. Consider the normal distribution shown in the graph with 68% of data between 53 and 65. a. Find the mean and the standard deviation of the distribution.
μ = __59 = (65+53)/2__ σ = ____6____ b. What percent of the data is greater than X = 65? 16% = (100 – 68)/2
What percent of the data is c. between the mean, μ, and 53? 84% = 68% + 16% 34% = (68)/2 d. greater than X = 53? e. Suppose the size of the population is N = 114. Approximately how many data values are less than 53? 0.16(114) =18.24 approximately 18 data values 2. Consider normal distribution shown in the graph. Some material used was contributed by Tom Koesters – East Forsyth High School 2a. Find the mean and the standard deviation of the distribution.
μ = __59 = (66+52)/2__ σ = __7______
2b. What percent of the data is greater than X = 52? 84% = 100 – 16% or 68% +_ 16%
2c. Suppose the size of the population is N = 127. Approximately how many data values are there between 52 and the mean? .34(127) = 43.18 approximately 43 data values
3. Consider normal distribution shown in graph. a. Find the mean of the distribution.
μ = __76___ b. What percent of the data is less than X = 72.25?
25% = 100% - 75% What percent of the data is… c. Greater than 79.75? d. Between than 72.25 and 79.75? 25% = 100% - 75% 50% = 100% - 25% - 25 % e. Suppose the size of the population is N = 89. Approximately how many data values are there between 72.25 and 79.75? 0.5(89) = 44.5 approximately 45 data values
4. Consider the normal distribution with mean = 22: 40% of the data is between 12 and 32. a. What percent of data is above 32? b. What percent of data is below 12?
(100 – 40)/2 = 30% (100 – 40)/2 = 30% c. What percent of data is below 32? d. What percent of data is above 12?
30 + 40 = 70% 40+ 30 = 70%
5. Consider the normal distribution with mean = 45: 80% of the data is below 56. a. What percent of data is above 56? b. What percent of data is between 45 and 56? (100 – 80)/2 = 10% 80/2 = 40% Some material used was contributed by Tom Koesters – East Forsyth High School Section 16.4 68-95-99.7 Rule
99.7% 95% 68%
(99.7– 95)/2 (95 – 68)/2 68/2
In a set of data with a normal distribution, approximately 68% of the data fall within one standard deviation of the mean (between z = –1 and 1)
68% of data
95% of the data fall within two standard deviations of the mean (between z = –2 and 2) 95% of data
99.7% of the data fall within three standard deviations of the mean (between z = –3 and 3) 99.7% of data
Data Values μ – 3σ μ – 2σ μ – σ μ μ + σ μ + 2σ μ + 3σ
0.15% 2.35% 13.5% 34% 34% 13.5% 2.35% 0.15%
Z-values z = -3 z = -2 z = -1 z = 0 z = 1 z = 2 z = 3 4. What z-value has 16% of data above? 6. What z-value has 84% of data above?
Z = 1 Z = -1
5. What z-value has 2.5% of data below? 7. What z-value has 0.15% of data below? Z = - 2 Z = - 3 Some material used was contributed by Tom Koesters – East Forsyth High School 8. What percent of data is… a. between z = 1 and z = -1? c. between z = -2 and z = 1? Z = 68% 95/2 + 68/2 = 81.5%
b. between z = 1 and z = 2? d. between z = - 3 and z = - 2? 95/2 - 68/2 = 13.5% 99.7/2 – 95/2 = 2.35%
9. In a normal distribution with mean μ and standard deviation σ, what percent of the data a. fall below the value μ + σ? b. between μ + 3σ and μ + 2σ? c. above μ – 3σ? Z = 1 100 - (100 – 68)/2 Z = 3 and 2 99.7/2 – 95/2 Z = -3 100 –(100 – 99.7)/2 __84%_ _2.35%_ _99.85%
10. A normal distribution has a standard deviation σ = 6.1 cm and 84% of the data is above 47.1 cm. a. What z-value does 47.1 represent? b. Find the mean. 16% below X = 47.1 84% above = 47.1 + 6.1 _ 47.1 is one standard deviation below mean μ = _ 53.2 z = -1
11. A normal distribution has a standard deviation σ = 9.6 cm and 84% of the data is below 49.7 cm. a. What z-value does 49.7 represent? b. Find the mean. 16% below X = 49.7 84% above = 49.7 – 9.6 _ 49.7 is one standard deviation above mean μ = _40.1 z = 1
12. A normal distribution has a standard deviation σ = 7.4 and 2.5% of the data is above 89.3. a. What z-value does 89.3 represent? b. Find the mean. 97.5% below X = 89.3 2.5% above = 89.3 – 2(7.4)_ 83.3 is TWO standard deviation above mean μ = 74.5 _ z = 2 Some material used was contributed by Tom Koesters – East Forsyth High School 13. A normal distribution has mean μ = 11.8 and standard deviation σ = 3.4. Approximately what percent of the data fall between 1.6 and 22.0? a. Find z-values of data 1.6 = 11.8 – 3.4(3) = three standard deviations below mean 22.0 = 11.8 + 3.4(3) = three standard deviations above mean
b. What is the approximate percent? Between 3 Standard Deviations = 99.7%
14. A normal distribution has mean μ = 12.5 and standard deviation σ = 3.1. Approximately what percent of the data fall between 6.3 and 18.7? a. Find z-values of data 6.3 = 12.5 – 3.1(2) = two standard deviations below mean 18.7 = 12.5 + 3.1(2) = two standard deviations above mean
b. What is the approximate percent? Between 2 Standard Deviations = 95%
Chapter 16 – Normal Curve Practice Problems 1. Find the z-value for a given data point X in a normal distribution of data with mean μ = 100 and standard deviation σ = 10 (round your answers to the nearest tenth).
1a. X = 140 z = __4______1b. X = 112 z = __1. 2_____
1c. X = 95 z = __-0.5_____ 1d. X = 83 z = __-1.7_____
2. Find the data point X that corresponds to the given z-value in a normal distribution of data with μ = 180.3 ft. and standard deviation σ = 30.9 ft. (round your answers to the nearest tenth).
2a. z = –4 X = ___56.7__ 2b. z = 0.8 X = ___205.02___
2c. z = 4.5 X = ___319.35_ 2d. z = 0 X = ___180.3____
3. In a normal distribution, the data value X1 = 30 has the standardized value z1 = –2 and the data value X2 = 130 has the standardized value z2 = 3. Find the mean and standard deviation. 130 30 = 5 standard deviations … σ = 20 and μ = 70 Some material used was contributed by Tom Koesters – East Forsyth High School 4. In a normal distribution, the data value X1 = 30 has the standardized value z1 = –2 and the data value X2 = 75 has the standardized value z2 = 1. Find the mean and standard deviation. 75 30 = 3 standard deviations … σ = 15 and μ = 60
5. In a normal distribution, the data value X1 = 20 has the standardized value z1 = –3 and the data value X2 = 115 has the standardized value z2 = 2. Find the mean and standard deviation. 115 20 = 5 standard deviations … σ = 19 and μ = 77
6. Consider the normal distribution shown in the graph. 6a. Find the mean of the distribution. μ = ___84____
6b. What percent of the data is less than X = 95.50? ___75%__
7) The number of problems missed on a quiz follows a normal distribution with a mean of 15 and a standard deviation of 4.
7a. What percent of students 7b. What percent of students 7c. What percent of students missed between 11 – 19 missed more than 23 less than 23 problems? problems? problems? 100 – (100-95)/2 = 68% (z = -1, 1) (100 – 95)/2 = 2.5% (z = 2) 97.5% (z = 2)
8) The final exam scores normal distribution with a mean of 80 and a standard deviation of 6.
8a. What percent of students 8b. What percent of students 8c. What percent of students scored between 80 - 92? scored between 68 -86? scored higher than 74? 47.5% (z = 0, 2) 81.5% (z = -2, 1) 84% (z = -1)
9) The class mean of height is 60 inches. We know that 68% of the students are between 56 and 64 inches. 9a. What is the standard deviation? 60 in (z = 0) 4 in z = 1 64, z = -1 56
9b. 50% of the students are shorter than? Some material used was contributed by Tom Koesters – East Forsyth High School 9c. 16% of students are taller than …? 9d. 84% of students are taller than …? 64 in 100 – 2*16 = 68% z = 1 56 inches (z = -1) Honors Discrete Practice Multiple Choice CHAPTER 16: NORMAL CURVE Pick the MOST ACCURATE Answer Choice For #1 – 11: 250 students in a math class take the final exam. The scores on the exam have an approximately normal distribution with center μ = 75 and standard deviation σ = 10. 1) The number of students scoring 75 or more is approximately A. 75 B. 83 C. 125 D. 158 250(.50) 2) The average score on the exam was approximately A. 10 B. 75 C. 85 D. 95
3) Approximately 95% of the class scored between A. 55 and 95 B. 0 and 95 C. 45 and 100 D. 65 and 85 (z = -2 and 2) 4) Assuming there were no outliers, the lowest score on the exam was around A. 0 B. 10 C. 45 D. 75 Z = - 3 5) Approximately what percent of the students scored between 65 and 75? A. 10% B. 34% C. 68% D. 95% (z = -1 to 0) 6) Peter’s score on the exam places him in the 16th percentile of the class. Peter’s score on the exam is approximately? A. 16 B. 45 C. 55 D. 65 (z = - 1) 7) Carol scored 85 points on the exam. In approximately what percentile does this score place her? A. 34th B. 68th C. 84th D. 95th (z = 1) 8) A z-value of 1.8 corresponds to a test score of A. 18 B. 57 C. 76.8 D. 93
9) A score of 95 corresponds to a standardized value (z-value) of A. – 2 B. 2 C. 8.5 D. 20
10) A score of 60 corresponds to a standardized value (z-value) of A. – 5 B. – 1.5 C. 1.5 D. 5 11) Approximately what percentage of test scores had standardized values between -3 and 3? A. 50% B. 68% C. 95% D. 99%
For #12 – 14: 95% of the data for a normal curve is between data values of 24 and 84. 12) Find the mean, μ, of the normal curve A. 12 B. 39 C. 48 D. 54 Z = -2 and 2 … mean is halfway 13) Find the standard deviation, σ, of the normal curve A. 10 B. 15 C. 30 D. 60 24 to 84 covers a distance of 60 or 4 standard deviations 14) What z-value represents the data value 84? A. 2.0 B. 3.0 C. 15 D. 30
For #15 – 17: 16% of the data for a normal curve is above the data value 56 and another 16% of the data for the normal curve is below the data value of 44. 15) Find the mean, μ, of the normal curve A. 38 B. 48 C. 50 D. 62 Z = 1 and -1 … mean is halfway between 16) Find the standard deviation, σ, of the normal curve A. 3 B. 6 C. 12 D. 24 44 to 56 covers a distance of 12 or two standard deviations 17) What z-value represents the data value 44? A. – 1.0 B. 1.0 C. – 2.0 D. 2.0
For # 18 - 21: As part of a study, 800 college football players are randomly chosen and their weights taken. The distribution of weights is approximately normal. The average weight is 235 pounds and the standard deviation is 25 pounds. 18) Approximately how many players weighed 210 pounds or less? A. 64 B. 128 C. 200 D. 544 800(.16) … z = -1 19) Assuming there were no outliers, the range of player weights was approximately A. 100 lbs B. 150 lbs C. 200 lbs D. 400 lbs (z = -3 to 3 6 standard deviations) 20) Approximately what percentage of players weighed over 285 pounds? (z = 2) A. 95% B. 97.5% C. 2.5% D. 5%
21) Approximately how many players weighed between 210 and 260 pounds?
A. 760 B. 128 C. 256 D. 544 = 800•(.68) Honors Discrete Practice Multiple Choice CHAPTER 16: NORMAL CURVE – Solutions 1) C 12) D 2) B 13)D 3) A 14)B 4) C 15)A 5) B 16)C 6) D 17)B 7) C 18)A 8) D 19)B 9) B 20)B 10)B 21)C 22) D 11)