Ch 2 HW Solutions



Ch 2 HW Solutions

CH2H1

1. Eleanor: [pic] Gerald: [pic] Eleanor’s score is higher.

2. Cobb: [pic] Williams: [pic] Brett: [pic]

3. a) Judy’s bone density score is about one and a half standard deviations below the average score for all women her age

b) [pic] [pic]

5. a) (Graph)

b) Mean= 4.896%, Standard deviation= .976%.

5 Number Summary: 2.7%, 4.1%, 4.8%, 5.5%, 7.1%

The distribution is symmetric with a center at 4.896%, a range of 4.4%, and no gaps or outliers.

c) [pic]

84th percentile; Illinois has one of the higher unemployment rates in the country

d) Minnesota; z= -.61

e) [pic](at least 0%). [pic](at least 75%). [pic](at least 89%). [pic](at least 93.75%). [pic](at least 96%)

9. a) (Graph) b) (Graph)

10. a) The area under the curve is a rectangle with height 1 and width 1.

b) 20% c) 60% d) 50% e) 0.5

11. (Graph)

12. a) mean C, median B

b) mean A, median A

c) mean A, median B

13. a) The curve satisfies the two conditions of a density curve: curve is on or above horizontal axis, and the total area under the curve = area of triangle + area of 2 rectangles = 1

b) 0.2

c) 0.6

d) 0.35

e) The area between 0 and 0.2 is 0.35. The area between 0 and 0.4 is 0.6. Therefore the “equal areas point” must be between 0.2 and 0.4.

15. Women: [pic] Men: [pic]

The z-scores tell us that 6 feet is quite tall for a woman but not at all extraordinary for a man.

16. a) (Graph)

b) Five-number summary: 316000, 775000, 2875000, 7250000, 22000000

The mean is 4410897 and s is 4837406. Skewed to the right. $22 million is an outlier.

c) [pic]

14th percentile

19. a) Erik had a relatively good race compared with those of the other athletes who completed the state meet, but he had a poor race by his own standards.

b) Erica was only a bit slower than usual by her own standards, but she was relatively slow compared to the other swimmers at the state meet.

20. a) (Graph)

b) 0.5 the quartiles are approximately at 0.3 and 0.7

c) 25%

d) 50%

Ch2H2

23. Approximately 0.2 for the tall one and 0.5 for the short one.

24. (Graph)

25. a) Approximately 2.5%

b) Between 64 inches and 74 inches

c) Approximately 16%

d) 84th percentile

26. a) (Graph)

b) [pic]

16th percentile

c) [pic] [pic]

[pic] About 84%

CH2H3

29. a) .9978

b) 1-.09978= 0.0022

c) 1-.0485= 0.9515

d) .9978-.0485= 0.9493

31. a) [pic] b) [pic]

1-.7734= .2266 .9994-.7734= .2260

c) [pic] [pic]

1-.3085= .6915

32. a) -.67

b) .25

33. a) [pic]

About 5.2%

b) [pic]

.5987-.052=.5467

About 55%

c) [pic] x=279.6

The longest 20% of pregnancies last approximately 279 or more days.

34. a) [pic]

1-.3446=.6554 About 65.54%

b) [pic]

1-.9452=.0548 About 5.5%

c) [pic] x=161.25 so 162 or higher on IQ test

35. a) [pic]

b) [pic] standard deviations

255.2 and 276.8

CH2H4

39. a) The graph has a gap between 20 and 22 and a peak between 16 and 18. The graph has somewhat of a normal shape except for the gap. The center appears to be around 15. There is a dramatic change from 11 to 13. The range is 15.6.

b) The mean is 15.586, and the median is 15.75. These two measures of center are very close to one another, as excepted for a symmetric distribution.

c) Yes. 68.2% of the lengths fall within one standard deviation of the mean, 95.5% of the lengths fall within two standard deviations of the mean, and 100% of the lengths fall within 3 standard deviations of the mean.

d) (Graph)

e) Yes, because the normal probability plot shows a linear pattern. There is reason to believe the data is normal.

43. a) 0.8997

b) 1-0.3372= 0.6628

c) 1.8997-0.3372= 0.5625

d) 0.3372-0.1003= 0.2369

44. a) 2.06

b) .77

46. a) -21% and 45%

b) [pic]

c) [pic]

1-.7852= .2148

47. a) The deciles are -1.28 and 1.28. Look up .1000 in the middle of the chart and go up and over to find your z-score of -1.28

b) The deciles are 61.3inches and 67.7 inches (Solve [pic] to get 61.3”

48. Find that Q1 and Q3 are -.67 and .67 respectively for a standard normal curve. Plug these numbers into our outlier formula: 1.5IQR = 1.5(.67 - -.67) = 2.01. This is our “cloud number” that we need to add and subtract to Q3 and Q1. When you do this you get

-.67 – 2.01 = -2.68 and .67 + 2.01 = 2.68/ This means that any data point that is more than 2.68 standard deviation above or below the mean are outliers. To find this percentage, look -2.68 up in the table to get about .0035. Double this since you are looking for above and below to get .70%

CH2H5

51. The percentage is 1% which you multiply by how many students there are (1300) to get about 13.

52. Sketches will vary but the peak at 0 should be tall and skinny while the curve at 1 should be short & fat

53. 89.8% of scores should fall below 27 using z-scores. In the actual data, it was 1052490/1171460 = 89.84%...pretty close to what it is supposed to be if normal so the normal distribution should fit it well.

54. a) Joey did as well or better than 97% of the student that took the reading test and did as well or better than 72% of the students that took the math test.

b) We cannot determine the z-score unless we know that the data is normally distributed or are given the mean and standard deviation.

55. Custom made helmets will be needed for soldiers with head circumferences less than approximately 21” or greater than approximately 24.6” (Look up .0500 in the middle of the chart to find a z-score of -1.64 and use this in the z-score formula)

56. a) The endpoint of the density curve should be ([pic].

b) The median is 1. Q1 = 0.707 and Q3 = 1.225 (You will need to set up equations where you know the area of the triangles (.5 for the median, .25 for Q1 and .75 for Q3)

c) The mean will be slightly below the median since our triangle is skewed left.

d) 12.5% are below 0.5 and 0% are above 1.5

57. a) The mean ($17,776) is greater than the median ($15, 532). Q3 is further from the median than Q1.

b) In the Normal distribution, Q3 has a P = .75 which is a z = .67. Plug this into z-score formula to get a Q3 = $25,889 which is above the actual value.

58. a) About 0.6% b) About 31%

62. Use distr/normalcdf(-1000, 27, 20.9, 4.8)

63. Use distr/invnorm(.05) = -1.64 (this is your z-score)

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