s need to include all required steps and answer(s) for full credit. All answers need to be reduced to lowest terms where possible. Answer the following problems showing your work and explaining (or analyzing) your results. Submit your work in a typed Microsoft Word document. 1. The final exam scores listed below are from one section of MATH 200. How many scores were within one standard deviation of the mean? How many scores were within two standard deviations of the mean? 99 34 86 57 73 85 91 93 46 96 88 79 68 85 89 2. The scores for math test #3 were normally distributed. If 15 students had a mean score of 74.8% and a standard deviation of 7.57, how many students scored above an 85%? 3. If you know the standard deviation, how do you find the variance? 4. To get the best deal on a stereo system, Louis called 8 out of 20 appliance stores in his neighborhood and asked for the cost of a specific model. The prices he was quoted are listed below: $216 $135 $281 $189 $218 $193 $299 $235 Find the standard deviation. 5. A company has 70 employees whose salaries are summarized in the frequency distribution below. Salary Number of Employees 5,001-10,000 8 10,001- 15,000 12 15,001- 20,000 20 20,001- 25,000 17 25,001- 30,000 13 a. Find the standard deviation. b. Find the variance. 6. Calculate the mean and variance of the sample data set provided below. Show and explain your steps. Round to the nearest tenth. 14, 16, 7, 9, 11, 13, 8, 10 7. Create a frequency distribution table for the number of times a number was rolled on a die. (It may be helpful to print or write out all of the numbers so none are excluded.) 3, 5, 1, 6, 1, 2, 2, 6, 3, 4, 5, 1, 1, 3, 4, 2, 1, 6, 5, 3, 4, 2, 1, 3, 2, 4, 6, 5, 3, 1 8. Answer the following questions using the frequency distribution table you created in No. 7. a. Which number(s) had the highest frequency? b. How many times did a number of 4 or greater get thrown? c. How many times was an odd number thrown? d. How many times did a number greater than or equal to 2 and less than or equal to 5 get thrown? 9. The wait times (in seconds) for fast food service at two burger companies were recorded for quality assurance. Using the data below, find the following for each sample. a. Range b. Standard deviation c. Variance Lastly, compare the two sets of results. Company Wait times in seconds Big Burger Company 10 5 67 78 12 0 17 5 11 5 120 59 The Cheesy Burger 13 3 12 4 200 79 10 1 14 7 118 125 10. What does it mean if a graph is normally distributed? What percent of values fall within 1, 2, and 3, standard deviations from the mean? Problems need to include all required steps and answer(s) for full credit. All answers need to be reduced to lowest terms where possible. Answer the following problems showing your work and explaining (or analyzing) your results. Submit your work in a typed Microsoft Word document. 1. The final exam scores listed below are from one section of MATH 200. How many scores were within one standard deviation of the mean? How many scores were within two standard deviations of the mean? 99 34 86 57 73 85 91 93 46 96 88 79 68 85 89 the sample :n=1 5 n 15 xi x i Mean : x = i=1 = i=1 n 15 x = 99+34+ 86+57+73+ 85+ 91+93+ 46+96+ 88+79+68+85+ 89 1,169 = 15 15 x =77.93333333 n ( x ix )2 Variance :s 2= i=1 n1 s 2=[ ( 9977.93333333 )2 + ( 3477.93333333 )2+ ( 8677.93333333 )2 + ( 5777.93333333 )2+ ( 7377.93333333 s 2=[ ( 21 . 06666666 )2 + (43 .93333333 )2+ ( 8.06666666 7 )2+ (20 .93333333 )2+ (4 .93333333 3 )2+ (7. 0 666666 2 s =[ 443.8044444 +1,930.137777+65.07111111+438.2044444+ 24.33777777+ 49.93777777+170.737777 s 2= 5,068.933333 s 2=362.0666666 14 Standard deviation: s= s 2= 362 .0666666 s=19.02804947 s 19.0 3 How many scores were within one standard deviation of the mean? x 77.93 ; s 19.0 3 x s x i x +s 77.9319.03 x i 77.93+19.03 58.90 x i 96.96 86 73 85 91 93 88 79 68 85 89 Answer: 10 scores were within one standard deviation of the mean How many scores were within two standard deviations of the mean? x 77.93 ; s 19.0 3 x 2 s x i x + 2 s 77.932 (19.03 ) x i 77.93+ 2 ( 19.03 ) 77.9338.06 x i 77.93+ 38.06 39.87 xi 115.99 99 86 57 73 85 91 93 46 96 88 79 68 85 89 Answer: 14 scores were within two standard deviations of the mean 2. The scores for math test #3 were normally distributed. If 15 students had a mean score of 74.8% and a standard deviation of 7.57, how many students scored above an 85%? Score : X the sample :n=1 5 Mean : =74.8 Standard deviation : =7.57 Probability that the student scored above an85 =P ( X > 85 ) ( P ( X > x )=P Z> x ; x=85 ( P ( X >85 )=P Z> ) 85 74.8 10.2 =P Z > =P ( Z >1.35 )=1P ( Z <1.35 ) 7.57 7.57 ) ( ) P ( Z <1.35 )=0.91149 ( Normal distribution table ) P ( X >85 )=10.91149 P ( X >85 )=0.08851 How many students scored above an 85%? 15 ( 0.08851 )=1.32765 1 Answer: One student scored above an 85% 3. If you know the standard deviation, how do you find the variance? Answer: The variance is the square of the standard deviation. 4. To get the best deal on a stereo system, Louis called 8 out of 20 appliance stores in his neighborhood and asked for the cost of a specific model. The prices he was quoted are listed below: $216 $135 $281 $189 $218 $193 $299 $235 Find the standard deviation. the sample :n=8 n 8 xi x i Mean : x = i=1 = i=1 n 8 x = 216+135+281+189+218+193+299+ 235 1, 766 = x =220.75 8 8 n ( x ix )2 Variance :s 2= i=1 n1 s 2=[ ( 216220.75 )2 + ( 135220.75 )2+ (281220.75 )2 + ( 189220.75 )2 + ( 218220.75 )2+ ( 193220.75 )2 + ( 299 (4.75 )2 + (85.75 )2+ ( 60.25 )2+ (31.75 )2 + (2.75 )2+ (27.75 )2 + ( 78.25 )2 + ( 14.25 )2 s= 7 2 s= 22.5625+7,353.0625+3,630.0625+1,008.0625+7.5625+770.0625+6,123.0625+203.0625 7 s 2= 19 ,117.5 s 2=2,731.071428 7 2 Standard deviation: s= s 2= 2,731.071428 s=$ 52 .25965392 s $ 52. 26 Answer: The standard deviation is $52.26. 5. A company has 70 employees whose salaries are summarized in the frequency distribution below. Salary Number of Employees 5,001-10,000 8 10,001- 15,000 12 15,001- 20,000 20 20,001- 25,000 17 25,001- 30,000 13 a. Find the standard deviation. Salary Midpoint (M) 5,001-10,000 10,001-15,000 15,001-20,000 20,001-25,000 25,001-30,000 7,500 12,500 17,500 22,500 27,500 Totals: Mean : x = Number of Employee s (n) 8 12 20 17 13 70 M*n M x ( M x )2 ( M x )2n 60,000 150,000 350,000 382,500 357,500 1,300,000 -11,071.42857 -6,071.428571 -1,071.428571 3,928.571428 8,928.571428 122,576,530.6 36,862,244.89 1,147,959.183 15,433,673.46 79,719,387.75 980,612,244.8 442,346,938.7 22,959,183.67 262,372,448.9 1,036,352,040 2,744,642,857 Mn = 1,300,000 x =18,571.42857 70 n Standard deviation : s= ( M x )2n = n 2,744,642,857 = 39,209,183.67 70 s=6,261.723698 s 6,261.72 The standard deviation is approximately 6,261.72 b. Find the variance. 2 2 2 2 Variance :s =( s ) =( 6,261.723698 ) s =39,209,183.67 The variance is approximately 39,209,183.67 6. Calculate the mean and variance of the sample data set provided below. Show and explain your steps. Round to the nearest tenth. 14, 16, 7, 9, 11, 13, 8, 10 the sample :n=8 n 8 xi x i Mean : x = i=1 = i=1 = n 8 14 +16+7+ 9+11+13+8+ 10 88 = x =11 8 8 n ( x ix )2 Variance :s 2= i=1 n1 s 2= ( 1411 )2 + ( 1611 )2+ (711 )2+ ( 911 )2 + ( 1111 )2 + ( 1311 )2 + ( 811 )2+ (1011 )2 81 s 2= ( 3 )2 + ( 5 )2+ (4 )2 + (2 )2+ ( 0 )2 + ( 2 )2+ (3 )2 + (1 )2 7 s 2= 9+25+16+ 4+ 0+4 +9+1 68 s2= =9.714285714 s2 9.7 7 7 Answer: The mean is 11 and the variance is 9.7. 7. Create a frequency distribution table for the number of times a number was rolled on a die. (It may be helpful to print or write out all of the numbers so none are excluded.) 3, 5, 1, 6, 1, 2, 2, 6, 3, 4, 5, 1, 1, 3, 4, 2, 1, 6, 5, 3, 4, 2, 1, 3, 2, 4, 6, 5, 3, 1 Frequency distribution table Value 1 2 3 4 5 6 Frequenc y 7 5 6 4 4 4 8. Answer the following questions using the frequency distribution table you created in No. 7. a. Which number(s) had the highest frequency? The number 1 had the highest frequency. b. How many times did a number of 4 or greater get thrown? 4+4+4=12 12 times a number of 4 or greater got thrown. c. How many times was an odd number thrown? 7+6+4=17 17 times was thrown an odd number. d. How many times did a number greater than or equal to 2 and less than or equal to 5 get thrown? 5+6+4+4=19 19 times a number greater than or equal to 2 and less than or equal to 5 got thrown. 9. The wait times (in seconds) for fast food service at two burger companies were recorded for quality assurance. Using the data below, find the following for each sample. Company Wait times in seconds Big Burger Company 10 5 67 78 12 0 17 5 11 5 120 59 The Cheesy Burger 13 3 12 4 200 79 10 1 14 7 118 125 a. Range Range: R=V max V min Maximum value: V max Minimum value : V min For Big Burger Company: V max =175 V min =59 R=V maxV min=17559 R=116 seconds The Range for Big Burger Company is 116 seconds. For The Cheesy Burger: V max =200 V min =79 R=V maxV min=20079 R=121 seconds The Range for The Cheesy Burger is 121 seconds. b. Standard deviation the sample :n=8 n xi Mean : x = i=1 n 8 xi x = i=1 8 n ( x ix )2 Variance :s 2= i=1 n1 8 ( x i x )2 = i=1 81 8 ( x ix )2 s2= i=1 7 Standard deviation : s= s 2 For Big Burger Company: x = 105+ 67+78+120+175+115 +120+59 839 = x =104 . 8 75 8 8 s 2=[ ( 105104 . 8 75 )2+ ( 67104 .8 75 )2 + ( 7 8104 . 8 75 )2+ ( 120104 . 8 75 )2+ (175104 . 8 75 )2 + ( 11510 s 2= ( 0 .12 5 )2 + (37 .8 75 )2 + (26.875 )2 + ( 15 . 125 )2 + ( 70 . 125 )2 + ( 10.12 5 )2+ ( 15. 1 25 )2+ (45.875 )2 7 s 2= 0 . 015625+1 , 434 . 515625+722 .265625+228.765625+ 4,917.515625+102. 515625+228.765625+2 7 s 2= 9, 738. 87 5 s2=1,391 . 267857 7 Standard deviation: s= s 2= 1, 391 .267857 s=37.29970317 s 37.3 seconds Answer: The standard deviation for the Big Burger Company is approximately 37.3 seconds. For The Cheesy Burger: x = 133+124 +200+79+101+147+118 +125 1,027 = x =128 . 375 8 8 s 2=[ ( 1 331 28 . 375 )2+ ( 1241 28 .3 75 )2 + ( 200128 . 3 75 )2+ ( 791 28 .3 75 )2 + ( 1 011 28 .3 75 )2 + ( 1 47 s 2= ( 4 . 6 2 5 )2 + (4 . 3 75 )2+ (71 . 6 75 )2+ (49 .375 )2 + (27.375 )2 + ( 18 . 62 5 )2+ (1 0 .37 5 )2+ (3.375 )2 7 s= 21.390625+19 .140625+5,130 . 140625+2,437 . 890625+ 749.390625+ 346 .890625+107.640625+11 7 s 2= 8,823.875 s 2=1, 260 .553571 7 2 Standard deviation: s= s 2= 1,260 . 553571 s=3 5 .50427539 s 3 5 . 5 seconds Answer: The standard deviation for The Cheesy Burger is approximately 35.5 seconds. c. Variance The Variance for the Big Burger Company is approximately 1,391.27. The Variance for The Cheesy Burger is approximately 1,260.55. Lastly, compare the two sets of results. The mean wait times in Big Burger Company (104.875 seconds) is less than in The Cheesy Burger (128.375 seconds), but the standard deviation of wait times in Big Burger Company (37.3 seconds) is more than in The Cheesy Burger (35.5 seconds). 10.What does it mean if a graph is normally distributed? What percent of values fall within 1, 2, and 3, standard deviations from the mean? It means that the graph has the shape of a symmetrical bell, where the midpoint is the mean of the data, and 50% of the data is to the left of this midpoint and 50% of the data is above of this midpoint. Approximately 68% of values fall within 1 standard deviation from the mean, 95% of values fall within 2 standard deviations from the mean, and 99.7% of values fall within 3 standard deviations from the mean. Problems need to include all required steps and answer(s) for full credit. All answers need to be reduced to lowest terms where possible. Answer the following problems showing your work and explaining (or analyzing) your results. Submit your work in a typed Microsoft Word document. 1. The final exam scores listed below are from one section of MATH 200. How many scores were within one standard deviation of the mean? How many scores were within two standard deviations of the mean? 99 34 86 57 73 85 91 93 46 96 88 79 68 85 89 the sample :n=1 5 n 15 xi x i Mean : x = i=1 = i=1 n 15 x = 99+34+ 86+57+73+ 85+ 91+93+ 46+96+ 88+79+68+85+ 89 1,169 = 15 15 x =77.93333333 n ( x ix )2 Variance :s 2= i=1 n1 s 2=[ ( 9977.93333333 )2 + ( 3477.93333333 )2+ ( 8677.93333333 )2 + ( 5777.93333333 )2+ ( 7377.93333333 s 2=[ ( 21 . 06666666 )2 + (43 .93333333 )2+ ( 8.06666666 7 )2+ (20 .93333333 )2+ (4 .93333333 3 )2+ (7. 0 666666 2 s =[ 443.8044444 +1,930.137777+65.07111111+438.2044444+ 24.33777777+ 49.93777777+170.737777 s 2= 5,068.933333 s 2=362.0666666 14 Standard deviation: s= s 2= 362 .0666666 s=19.02804947 s 19.0 3 How many scores were within one standard deviation of the mean? x 77.93 ; s 19.0 3 x s x i x +s 77.9319.03 x i 77.93+19.03 58.90 x i 96.96 86 73 85 91 93 88 79 68 85 89 Answer: 10 scores were within one standard deviation of the mean How many scores were within two standard deviations of the mean? x 77.93 ; s 19.0 3 x 2 s x i x + 2 s 77.932 (19.03 ) x i 77.93+ 2 ( 19.03 ) 77.9338.06 x i 77.93+ 38.06 39.87 xi 115.99 99 86 57 73 85 91 93 46 96 88 79 68 85 89 Answer: 14 scores were within two standard deviations of the mean 2. The scores for math test #3 were normally distributed. If 15 students had a mean score of 74.8% and a standard deviation of 7.57, how many students scored above an 85%? Score : X the sample :n=1 5 Mean : =74.8 Standard deviation : =7.57 Probability that the student scored above an85 =P ( X > 85 ) ( P ( X > x )=P Z> x ; x=85 ( P ( X >85 )=P Z> ) 85 74.8 10.2 =P Z > =P ( Z >1.35 )=1P ( Z <1.35 ) 7.57 7.57 ) ( ) P ( Z <1.35 )=0.91149 ( Normal distribution table ) P ( X >85 )=10.91149 P ( X >85 )=0.08851 How many students scored above an 85%? 15 ( 0.08851 )=1.32765 1 Answer: One student scored above an 85% 3. If you know the standard deviation, how do you find the variance? Answer: The variance is the square of the standard deviation. 4. To get the best deal on a stereo system, Louis called 8 out of 20 appliance stores in his neighborhood and asked for the cost of a specific model. The prices he was quoted are listed below: $216 $135 $281 $189 $218 $193 $299 $235 Find the standard deviation. the sample :n=8 n 8 xi x i Mean : x = i=1 = i=1 n 8 x = 216+135+281+189+218+193+299+ 235 1, 766 = x =220.75 8 8 n ( x ix )2 Variance :s 2= i=1 n1 s 2=[ ( 216220.75 )2 + ( 135220.75 )2+ (281220.75 )2 + ( 189220.75 )2 + ( 218220.75 )2+ ( 193220.75 )2 + ( 299 (4.75 )2 + (85.75 )2+ ( 60.25 )2+ (31.75 )2 + (2.75 )2+ (27.75 )2 + ( 78.25 )2 + ( 14.25 )2 s= 7 2 s= 22.5625+7,353.0625+3,630.0625+1,008.0625+7.5625+770.0625+6,123.0625+203.0625 7 s 2= 19 ,117.5 s 2=2,731.071428 7 2 Standard deviation: s= s 2= 2,731.071428 s=$ 52 .25965392 s $ 52. 26 Answer: The standard deviation is $52.26. 5. A company has 70 employees whose salaries are summarized in the frequency distribution below. Salary Number of Employees 5,001-10,000 8 10,001- 15,000 12 15,001- 20,000 20 20,001- 25,000 17 25,001- 30,000 13 a. Find the standard deviation. Salary Midpoint (M) 5,001-10,000 10,001-15,000 15,001-20,000 20,001-25,000 25,001-30,000 7,500 12,500 17,500 22,500 27,500 Totals: Mean : x = Number of Employee s (n) 8 12 20 17 13 70 M*n M x ( M x )2 ( M x )2n 60,000 150,000 350,000 382,500 357,500 1,300,000 -11,071.42857 -6,071.428571 -1,071.428571 3,928.571428 8,928.571428 122,576,530.6 36,862,244.89 1,147,959.183 15,433,673.46 79,719,387.75 980,612,244.8 442,346,938.7 22,959,183.67 262,372,448.9 1,036,352,040 2,744,642,857 Mn = 1,300,000 x =18,571.42857 70 n Standard deviation : s= ( M x )2n = n 2,744,642,857 = 39,209,183.67 70 s=6,261.723698 s 6,261.72 The standard deviation is approximately 6,261.72 b. Find the variance. 2 2 2 2 Variance :s =( s ) =( 6,261.723698 ) s =39,209,183.67 The variance is approximately 39,209,183.67 6. Calculate the mean and variance of the sample data set provided below. Show and explain your steps. Round to the nearest tenth. 14, 16, 7, 9, 11, 13, 8, 10 the sample :n=8 n 8 xi x i Mean : x = i=1 = i=1 = n 8 14 +16+7+ 9+11+13+8+ 10 88 = x =11 8 8 n ( x ix )2 Variance :s 2= i=1 n1 s 2= ( 1411 )2 + ( 1611 )2+ (711 )2+ ( 911 )2 + ( 1111 )2 + ( 1311 )2 + ( 811 )2+ (1011 )2 81 s 2= ( 3 )2 + ( 5 )2+ (4 )2 + (2 )2+ ( 0 )2 + ( 2 )2+ (3 )2 + (1 )2 7 s 2= 9+25+16+ 4+ 0+4 +9+1 68 s2= =9.714285714 s2 9.7 7 7 Answer: The mean is 11 and the variance is 9.7. 7. Create a frequency distribution table for the number of times a number was rolled on a die. (It may be helpful to print or write out all of the numbers so none are excluded.) 3, 5, 1, 6, 1, 2, 2, 6, 3, 4, 5, 1, 1, 3, 4, 2, 1, 6, 5, 3, 4, 2, 1, 3, 2, 4, 6, 5, 3, 1 Frequency distribution table Value 1 2 3 4 5 6 Frequenc y 7 5 6 4 4 4 8. Answer the following questions using the frequency distribution table you created in No. 7. a. Which number(s) had the highest frequency? The number 1 had the highest frequency. b. How many times did a number of 4 or greater get thrown? 4+4+4=12 12 times a number of 4 or greater got thrown. c. How many times was an odd number thrown? 7+6+4=17 17 times was thrown an odd number. d. How many times did a number greater than or equal to 2 and less than or equal to 5 get thrown? 5+6+4+4=19 19 times a number greater than or equal to 2 and less than or equal to 5 got thrown. 9. The wait times (in seconds) for fast food service at two burger companies were recorded for quality assurance. Using the data below, find the following for each sample. Company Wait times in seconds Big Burger Company 10 5 67 78 12 0 17 5 11 5 120 59 The Cheesy Burger 13 3 12 4 200 79 10 1 14 7 118 125 a. Range Range: R=V max V min Maximum value: V max Minimum value : V min For Big Burger Company: V max =175 V min =59 R=V maxV min=17559 R=116 seconds The Range for Big Burger Company is 116 seconds. For The Cheesy Burger: V max =200 V min =79 R=V maxV min=20079 R=121 seconds The Range for The Cheesy Burger is 121 seconds. b. Standard deviation the sample :n=8 n xi Mean : x = i=1 n 8 xi x = i=1 8 n ( x ix )2 Variance :s 2= i=1 n1 8 ( x i x )2 = i=1 81 8 ( x ix )2 s2= i=1 7 Standard deviation : s= s 2 For Big Burger Company: x = 105+ 67+78+120+175+115 +120+59 839 = x =104 . 8 75 8 8 s 2=[ ( 105104 . 8 75 )2+ ( 67104 .8 75 )2 + ( 7 8104 . 8 75 )2+ ( 120104 . 8 75 )2+ (175104 . 8 75 )2 + ( 11510 s 2= ( 0 .12 5 )2 + (37 .8 75 )2 + (26.875 )2 + ( 15 . 125 )2 + ( 70 . 125 )2 + ( 10.12 5 )2+ ( 15. 1 25 )2+ (45.875 )2 7 s 2= 0 . 015625+1 , 434 . 515625+722 .265625+228.765625+ 4,917.515625+102. 515625+228.765625+2 7 s 2= 9, 738. 87 5 s2=1,391 . 267857 7 Standard deviation: s= s 2= 1, 391 .267857 s=37.29970317 s 37.3 seconds Answer: The standard deviation for the Big Burger Company is approximately 37.3 seconds. For The Cheesy Burger: x = 133+124 +200+79+101+147+118 +125 1,027 = x =128 . 375 8 8 s 2=[ ( 1 331 28 . 375 )2+ ( 1241 28 .3 75 )2 + ( 200128 . 3 75 )2+ ( 791 28 .3 75 )2 + ( 1 011 28 .3 75 )2 + ( 1 47 s 2= ( 4 . 6 2 5 )2 + (4 . 3 75 )2+ (71 . 6 75 )2+ (49 .375 )2 + (27.375 )2 + ( 18 . 62 5 )2+ (1 0 .37 5 )2+ (3.375 )2 7 s= 21.390625+19 .140625+5,130 . 140625+2,437 . 890625+ 749.390625+ 346 .890625+107.640625+11 7 s 2= 8,823.875 s 2=1, 260 .553571 7 2 Standard deviation: s= s 2= 1,260 . 553571 s=3 5 .50427539 s 3 5 . 5 seconds Answer: The standard deviation for The Cheesy Burger is approximately 35.5 seconds. c. Variance The Variance for the Big Burger Company is approximately 1,391.27. The Variance for The Cheesy Burger is approximately 1,260.55. Lastly, compare the two sets of results. The mean wait times in Big Burger Company (104.875 seconds) is less than in The Cheesy Burger (128.375 seconds), but the standard deviation of wait times in Big Burger Company (37.3 seconds) is more than in The Cheesy Burger (35.5 seconds). 10.What does it mean if a graph is normally distributed? What percent of values fall within 1, 2, and 3, standard deviations from the mean? It means that the graph has the shape of a symmetrical bell, where the midpoint is the mean of the data, and 50% of the data is to the left of this midpoint and 50% of the data is above of this midpoint. Approximately 68% of values fall within 1 standard deviation from the mean, 95% of values fall within 2 standard deviations from the mean, and 99.7% of values fall within 3 standard deviations from the mean