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1. Between z = 1.32 and z = 1.82

Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region . 1. Between z = 1.32 and z = 1.82. 1.32. 1.82. 0. When z = 1.32 Area = 0.4066. When z = 1.82 Area = 0.4656.

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1. Between z = 1.32 and z = 1.82

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  1. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 1. Between z = 1.32 and z = 1.82 1.32 1.82 0 When z = 1.32 Area = 0.4066 When z = 1.82 Area = 0.4656 Since its shaded between two z’s on the same side, subtract the biggest – smallest. 0.4656 – 0.4066 = 0.0590

  2. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 2. Between z = -1.56 and z = -1.96 -1.96 -1.56 0 When z = -1.56 Area = 0.4406 When z = -1.96 Area = 0.4750 Since its shaded between two z’s on the same side, subtract the biggest – smallest. 0.4750 – 0.4406 = 0.0344

  3. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 3. Between z = -0.03 and z = 0.53 -0.03 0.53 0 When z = -0.03 Area = 0.0120 When z = 0.53 Area = 0.2019 Since its shaded between two z’s on opposite sides of z=0, add the areas. 0.0120 – 0.2019 = 0.2139

  4. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 4. To the right of z = 1.99 1.99 0 When z = 1.99 Area = 0.4767 Since its to the right of a positive z; 0.5 – area. 0.5000 – 0.4767 = 0.0233

  5. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 5. To the right of z = -1.36 -1.36 0 When z = -1.36 Area = 0.4131 Since its to the right of a negative z; 0.5 + area. 0.5000 + 0.4131 = 0.9131

  6. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 6. To the left of 1.68 0 1.68 When z = 1.68 Area = 0.4535 Since its to the left of a positive z; 0.5 + area. 0.5000 + 0.4535 = 0.9535

  7. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 7. To the left of z = -0.76 and to the right of z = 1.73 0 -0.76 1.73 When z = -0.76 Area = 0.2764 When z = 1.73 Area = 0.4582 Since it is both tails, add the areas then subtract 1 – the sum. 0.2764 + 0.4582 = 0.7346 1 – 0.7346 = 0.2654

  8. 8) Americans spend on average $617 per year on their insured vehicles. Assume the variable is normally distributed. If the standard deviation is $52, find the probability that a randomly selected insured vehicle owner will spend the following amounts on his or her vehicle per year. a) Between $600 and $700 First, find the z-values for both of your X’s µ = $617 σ = $52 Area =0.1293 600 – 617 52 X - µ σ When X = 600, z = = = -0.33 700 – 617 52 X - µ σ Area =0.4452 When X = 700, z = = = 1.60 Since you are crossing z = 0, add the areas. 0.1293 + 0.4452 = 0.5745 0 -0.33 1.60

  9. 8) Americans spend on average $617 per year on their insured vehicles. Assume the variable is normally distributed. If the standard deviation is $52, find the probability that a randomly selected insured vehicle owner will spend the following amounts on his or her vehicle per year. b) Less than $575 First, find the z-values for both of your X’s µ = 617 σ = 52 Area =0.2910 575 – 617 52 X - µ σ When X = 575, z = = = -0.81 Since it is just a tail and does not cross over Z = 0, you need to subtract 0.5 – the area. 0.5 – 0.2910 = 0.2090 0 -0.81

  10. 8) Americans spend on average $617 per year on their insured vehicles. Assume the variable is normally distributed. If the standard deviation is $52, find the probability that a randomly selected insured vehicle owner will spend the following amounts on his or her vehicle per year. c) More than $424 First, find the z-values for both X µ = 617 σ = 52 Area =0.4999 424 – 617 52 X - µ σ When X = 424, z = = = -3.71 Since you cross over z = 0 and shade all the Way to the tail, add 0.5 to the area. 0.5 + 0.4999 = 0.9999 0 -3.71

  11. 9) The average diastolic blood pressure of a certain age group of people is 85 mmHg. The standard deviation is 6. If an individual from this age group is selected, find the probability that his or her blood pressure will be the following. Assume the variable is normally distributed. a) Greater than 90 First, find the z-values for your X µ = 85 σ = 6 Area =0.2967 90 – 85 6 X - µ σ When X = 90, z = = = = 0.83 Since you are only shading the tail and not Crossing over z = 0, do 0.5 – area. 0.5 – 0.2967 = 0.2033 0 0.83

  12. 9) The average diastolic blood pressure of a certain age group of people is 85 mmHg. The standard deviation is 6. If an individual from this age group is selected, find the probability that his or her blood pressure will be the following. Assume the variable is normally distributed. b) Below 80 First, find the z-values for you X µ = 85 σ = 6 Area =0.2967 80 – 85 6 X - µ σ When X = 80, z = = = = -0.83 Since you are only shading the tail and not Crossing over z = 0, do 0.5 – area. 0.5 – 0.2967 = 0.2033 0 0.83

  13. 9) The average diastolic blood pressure of a certain age group of people is 85 mmHg. The standard deviation is 6. If an individual from this age group is selected, find the probability that his or her blood pressure will be the following. Assume the variable is normally distributed. c) Between 85 and 95 First, find the z-values for both of your X’s µ = 85 σ = 6 Area =0.0000 85 – 85 6 X - µ σ When X = 85, z = = = = 0.00 Area =0.4525 95 – 85 6 X - µ σ When X = 95, z = = = = 1.67 Since you are shading between z = 0 and your Z-value, your z-value will be your answer. 0.4525 0 1.67

  14. 9) The average diastolic blood pressure of a certain age group of people is 85 mmHg. The standard deviation is 6. If an individual from this age group is selected, find the probability that his or her blood pressure will be the following. Assume the variable is normally distributed. d) Between 88 and 92 First, find the z-values for both of your X’s µ = 85 σ = 6 Area =0.1915 88 – 85 6 X - µ σ When X = 88, z = = = = 0.50 Area =0.3790 92 – 85 6 X - µ σ When X = 92, z = = = = 1.17 Since you are shading between two z-values That are on opposite sides of z = 0, subtract 0.3790 – 0.1915 = 0.1875 0 0.50 1.17

  15. 10) The average repair cost for automatic washers is $73. Assume the variable is normally distributed with a standard deviation of $8. If 9 washers are repaired, find the probability that the mean of the repair bills will be less than $70. First, find the z-values for your X µ = $73 σ = $8 n = 9 Area =0.3708 (70 – 73) (8/√9) X - µ σ When X = $70, z = = = -1.13 Since it is less than, shade to the left. Since you Shade all the way to the tail and do not cross Z = 0, 0.5 – area. -1.13 0 0.5 – 0.3708 = 0.1292

  16. 11) Americans spend an average of 12.2 minutes in the shower. If the standard deviation of the variable is 2.3 minutes and the variable is normally distributed, find the probability that the mean time of a sample of 12 Americans who shower will be less than 11 minutes. First, find the z-values for your X µ = 12.2 σ = 2.3 n = 12 Area =0.4649 (12.2 – 11) (2.3/√12) X - µ σ When X = 11, z = = = -1.81 Since it is less than, shade to the left. Since you Shade all the way to the tail and do not cross Z = 0, 0.5 – area. -1.81 0 0.5 – 0.4649 = 0.0351

  17. 12. The average weight of an airline passenger’s suitcase is 45 pounds. The standard deviation is 2 pounds. If 15% of the suitcases are overweight, find the maximum weight allowed by the airline. Assume the variable is normally distributed. Top 15% σ = 2 µ = 45 0.3500 Found by doing:0.5 – 0.1500 = 0.3500 0.1500 0 Find the z-value that corresponds with the area between the middle line and line you want: z-values corresponding with Area of 0.3500 is z = 1.04. X = z(σ) + µ = 1.04(2) + 45 = 47.08

  18. 13. An educational study to be conducted requires a test score in the middle 40% range. If the mean is 100 and the standard deviation is 15, find the highest and lowest acceptable scores that would enable a candidate to participate in the study. Assume the variable is normally distributed. Middle 40% σ = 15 µ = 100 0.2000 0.2000 Found by doing:0.4 / 2 = 0.2000 0 Find the z-value that corresponds with the area between the middle line and line you want:Because its middle you will have a positive and negative z-value.z-values corresponding with Area of 0.2000 will be z = 0.52 and z = -0.52. Upper: X = z(σ) + µ = 0.52(15) + 100 = 92.2 Lower: X = z(σ) + µ = -0.52(15) + 100 = 107.8 99.2 ≤ X ≤ 107.8

  19. 14. A certain high-end law firm receives many applications daily. All applicants must have taken an entrance exam that has a mean of 184 with a standard deviation of 17. In order to cut down on applications to sort through, they throw out all applications with a score in the bottom 45% of the test scores. What is the cut-off score to be considered for the job? Assume normally distributed. Bottom 45% σ = 17 µ = 184 0.0500 Found by doing:0.5 – 0.4500 = 0.0500 0.4500 0 Find the z-value that corresponds with the area between the middle line and line you want: z-value on the left side of mean (negative) corresponding with Area of 0.0500 is z = -0.13 X = z(σ) + µ = -0.13(17) + 184 = 181.79

  20. 15. The national average for a math SAT score is 515 with a standard deviation of 116. If a student was to receive a 610 on the math portion of the SAT, nationally what percentile would they rank? First, find the z-value for your X µ = 515 σ = 116 Area =0.2939 610 – 515 116 X - µ σ When X = 610, z = = = 0.82 Since it goes to a tail and crosses over Z = 0, you need to add: 0.5 + the area. 0.5 + 0.2939 = 0.7939 0 0.82 To find “percentile” you always shade all the way to the left. A percentile is expressed as a percent (multiply by 100) and is rounded to the Nearest whole number. 0.7939 means the 79th percentile

  21. 16. A certain high school has an average GPA of a 2.64. If the data is normally distributed with a standard deviation of 0.79, what percentile would someone with a GPA of a 2.02 rank? First, find the z-value for your X µ = 2.64 σ = 0.79 Area =0.2823 2.02 – 2.64 0.79 X - µ σ When X = 610, z = = = -0.78 Since it its only a tail and does not cross over Z = 0, you need to subtract: 0.5 – the area. 0.5 – 0.2823 = 0.2177 0 -0.78 To find “percentile” you always shade all the way to the left. A percentile is expressed as a percent (multiply by 100) and is rounded to the Nearest whole number. 0.2177 means the 22nd percentile

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