1. Between z = 0 and z = 1.97

1. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 1. Between z = 0 and z = 1.97 1.97 0 When z = 1.97 Area = 0.4756 Since its shaded between z = 0 and z = 1.97, 0.4756 is the answer.

2. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 2. Between z = -2.04 and z = 0 0 -2.04 When z = -2.04 Area = 0.4793 Since its shaded between z = 0 and z = -2.04, 0.4793 is the answer.

3. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 3. To the right of z = 3.00 3.00 0 When z = 3.00 Area = 0.4987 Since its shaded between z = 3.00 to the end of the tail, you must subtract from 0.5 0.5 – 0.4987 = 0.0013

4. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 4. Between z = -0.54 and z = 0.88 -0.54 0.88 0 When z = -0.54 Area = 0.2054 When z = 0.88 Area = 0.3106 Since its shaded between two z-values and crosses overz = 0, add. 0.2054 + 0.3106 = 0.5160

5. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 5. To the left of z = 1.23 1.23 0 When z = 1.23 Area = 0.3907 Since its shaded to the left of a positive z-value (crosses overz = 0) you add to 0.5. 0.5 + 0.3907 = 0.8907

6. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 6. Between z = -2.45 and z = -1.13 -2.45 -1.13 0 When z = -2.45 Area = 0.4929When z = -1.13 Area = 0.3708 Since its between two on the same side of the mean, subtract. 0.4929 – 0.3708 = 0.1221

7. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. • 7. To the right of z = -2.77 0 -2.77 When z = -2.77 Area = 0.4972 Since its shaded to the right of a negative z-value, (crossesover 0) add 0.5. 0.5 + 0.4972 = 0.9972

8. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 8. To the right of z = 1.49 and to the left of z = -2.04 -2.04 1.49 0 When z = 1.49 Area = 0.4319When z = -2.04 Area = 0.4793 Since its shaded area is both tails, add to get all of the notshaded area, then do 1 – (not shaded) to get shaded area Not shaded = 0.4319 + 0.4793 = 0.9112 1 – 0.9112 = 0.0888

9. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 9. To the left of z = -0.76 -0.76 0 When z = -0.76 Area = 0.2764 Since its shaded between z = -0.76 to the end of the tail, you must subtract from 0.5 0.5 – 0.2764 = 0.2236

10. Section 1: Draw a picture, shade the appropriate region, and find the probability of the shaded region. 10. Between z = 1.36 and z = 3.04 1.36 3.04 0 When z = 1.36 Area = 0.4131When z = 3.04 Area = 0.4988 Since its between two on the same side of the mean, subtract. 0.4988 – 0.4131 = 0.0857

11. 11. A particular fast food restaurant has an average cost of $11.81 per order with a standard deviation of $3.97. Assuming the data is normally distributed, find the probability that an order will be: a) Between $10 and $15 µ = 11.81 σ = 3.97 Area =0.1772 10 – 11.81 3.97 X - µ σ When X = 10, z = = = -0.46 15 – 11.81 3.97 X - µ σ Area =0.2881 When X = 15, z = = = 0.80 0.2881 + 0.1772 = 0.4653 0 -0.46 0.80

12. 11. A particular fast food restaurant has an average cost of $11.81 per order with a standard deviation of $3.97. Assuming the data is normally distributed, find the probability that an order will be: b) less than $5.25 µ = 11.81 σ = 3.97 Area =0.4505 5.25 – 11.81 3.97 X - µ σ When X = 5.25, z = = = -1.65 0.5 – 0.4505 = 0.0495 0 -1.65

13. 11. A particular fast food restaurant has an average cost of $11.81 per order with a standard deviation of $3.97. Assuming the data is normally distributed, find the probability that an order will be: d) In a random selection of 50 orders, how many do you expect to cost more than $17.32? µ = 11.81 σ = 3.97 Area =0.4177 17.32 – 11.81 3.97 X - µ σ When X = 17.32, z = = = 1.39 0.5 – 0.4177 = 0.0823 0 1.39 How many? = n(p) = 50 (0.0823) = 4.115

14. 12. One business polled all of their employees to determine the number miles driven between work and home. The data proved an average of 16.2 miles driven between work and home with a standard deviation of 6.4 miles. If the data is normally distributed, find the probability that the distance one randomly selected worker drives will be : a) Less than 5 miles or more than 10 miles σ = 6.4 µ = 16.2 Area =0.4599 5 – 16.2 6.4 X - µ σ When X = 5, z = = = -1.75 15 – 16.2 6.4 X - µ σ Area =0.2224 When X = 15, z = = = 0.59 Not Shaded:0.4599 + 0.2224 = 0.6823 To get Shaded Portion:1 – 0.6823 = 0.3177 0 -1.75 0.59

15. 12. One business polled all of their employees to determine the number miles driven between work and home. The data proved an average of 16.2 miles driven between work and home with a standard deviation of 6.4 miles. If the data is normally distributed, find the probability that the distance one randomly selected worker drives will be : b) More than 10.2 miles σ = 6.4 µ = 16.2 Area =0.3264 10.2 – 16.2 6.4 X - µ σ When X = 10.2, z = = = -0.94 0.3264 + 0.5 = 0.8264 0 -0.94

16. 12. One business polled all of their employees to determine the number miles driven between work and home. The data proved an average of 16.2 miles driven between work and home with a standard deviation of 6.4 miles. If the data is normally distributed, find the probability that the distance one randomly selected worker drives will be : b) . If every 3rd worker in alphabetical order is questioned for a total of 25 employees, how many do you expect drive more than 17.2 miles between work and home? σ = 6.4 µ = 16.2 Area =0.0636 17.2 – 16.2 6.4 X - µ σ When X = 17.2, z = = = 0.16 0.5 – 0.0636 = 0.4364 0 0.16 How many? = n(p) = 25 (0.4364) = 10.91

17. 13. The average GPA for a specific graduating class is 2.34 with a standard deviation of 0.89. If only the top 20% of the graduates will be recognized as graduating with honors, what is the cutoff GPA for graduating with honors? Top 20% σ = 0.89 µ = 2.34 0.3000 Found by doing:0.5 – 0.2000 = 0.3000 0.2000 0 Find the z-value that corresponds with the area between the middle line and line you want: z-values corresponding with Area of 0.3000 is z = 0.84. X = z(σ) + µ = 0.84(0.89) + 2.34 = 3.0876

18. 14. A career fairs hosting various jobs around the area accumulates an average starting salary pay of $19,000 with a standard deviation of $7,037. Dustin was an average student and does not want to apply for jobs in which he is not qualified and not likely to get, yet he does not want a low paying job either. If she wants to target salaries in the middle 50% range at the career fair, she would be looking for jobs paying between what two salaries? Middle 50% σ = 7,037 µ = 19,000 0.2500 0.2500 Found by doing:0.5 / 2 = 0.2500 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 0.2500 will be z = 0.67 and z = -0.67. Upper: X = z(σ) + µ = 7037(0.67) + 19,000 = 23,714.79 Lower: X = z(σ) + µ = 7037(-0.67) + 19,000 = 14,285.21 14,285.21 ≤ X ≤ 23,714.79

19. 14. Danielle is looking at buying a new BMW Mini Cooper Convertible but does not want to pay for all of the expensive add-ons. The dealership explains that these cars brand new sell for an average of $24,783 with a standard deviation of $3,997. If she wants to look for one in the bottom 10% price range, what would be the most she is willing to pay? Bottom 10% σ = 3,997 µ = 24,783 0.4000 Found by doing:0.5 – 0.1000 = 0.4000 0.1000 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 0.4000 is z = -1.28. X = z(σ) + µ = -1.28(3,997) + 24,783 = 19,666.84