统计作业代做 STAT 301 REVIEW FOR EXAM 1 – ANSWERS:

STAT 301 REVIEW FOR EXAM 1 - ANSWERS:

统计作业代做 A sophomore from Harrison High School.   b.All 800 sophomores at Harrison High School.   c.The 150 selected sophomores from

1.a. C

2.A

3.B

4.a. 2 | 5 7 8 8 8 9 9 9

3 | 0 1 2 4 6 9

4 | 2 5

5 | 2  统计作业代做

1. Skewed right and unimodal


2. 25, 28, 30, 37.5, 52


3. Yes, 52 is an outlier. Because it is more than Q3+1.5*IQR=37.5+1.5*(37.5- 28)=51.75

e.

f.Median and IQR since they are resistant to outliers and work best with skewed

distributions.

3.a. sports cars (smallest time needed to get to speed)  统计作业代做

b.small cars

4.a. A sophomore from Harrison High School.

b.All 800 sophomores at Harrison High School.

c.The 150 selected sophomores from Harrison High School.

d.GPA (quantitative), whether the student took the SAT as a sophomore (categorical).

e.SRS

5. c


6. e

7.a. All preschool children  统计作业代做

b. The 68 preschool children

c.The amount of improvement (Integers ranging from -4 to 9). Quantitative

d.Experiment – a treatment (6 months of piano lessons or 6 months of computer

e.lessons) was imposed on the subjects.

8. a. 1

b.3

9.c

10.a. If z~2 :(12.1 to 20.5). If z=1.96 :(12.2, 20.4)

b.Almost zero: z = (14-16.3)/(2.1/sqrt(25)) = -5.48, P  0

c.3.92%: z = (20-16.3)/2.1 = +1.76 P = 1 - .9608 = .0392

d.18.988 pounds: z for P  0.9 = 1.28 X = 1.28*2.1 + 16.3

11. 
    

    a. 11.51%: z = (1200 – 1500)/250 = -1.20 P (z<-1.20) = .1151  统计作业代做

    
    

b. 1887.5 hours: z (for P  0.94) = 1.55 (or 1.56) X = 1.55*250 + 1500 = 1887.5 (or 1890 if z=1.56))

c.0.9948: z = (1400 -1500)/(250/sqrt(49)) = -2.8 P = .0026 1 – 2P = .9948

12. A


13. C


14. D 15. E 16. C 17. F

18.A

19.C 20. E 21. B 22. E 23. C 24. F

25. B: z = (205 – 202)/4 = .75 P = 1 - .7734 = .2266


26. B 27. C

28.F 29.G 30. H

35.A

36. B 37.P(X>3.54)

38.P(Z>2.00)=0.0228 - An answer of 0.025 would also be considered correct if  calculated as P(X>3.50+2*0.02)=(1-0.95)/2=0.025.

39.a:3.5,b:0.9772,c:0.0228  统计作业代做

40.B 41.A 42.P(X_bar>3.52) 43.z=(3.52-3.5)/(0.02/sqrt(20)) = 4.47 P = 1 – 1  0

44.A

45. C,F

46.A 47.C 48.G

52.IQR=13 Q1-1.5IQR=21.5 No low outliers Q3+1.5IQR=73.5 No high outliers

53. P(X<x)=0.09=P(Z<z) z=-1.34 x=63.96

54. P(X>85)=P(Z>2.17)=0.015


55. Mean=72 at center. 85 to the right of the mean at the tail. Area to the right of 85


56. P(X_bar<68)=P(Z<-1.49)=0.068157.B 58.B 59.E 60.B 61.A 62.B

63. 
    

    Brown: (36,000-28,000)/5,500=1.45 Andrews:(35,000-30,000)/4,000=1.25 Browns make more.

    
    

64.B

65.B 66.C 67.D 68.A 69.F

70.P(X>700)=P(Z>0.5077)=0.3050

71.mean 667 at center,700 to the right, area to the right of 700 is P(X>700).

72.P(X_bar<655)=P(Z<-0.58)=0.2810

73.P(X>x)=0.1=P(Z>z) z=1.28 x=750.2  统计作业代做

74.B

75. Lurking variable (for example): lifestyle.


76. P(X<15)


77. z=-3.00


78. 0.0013 (or 0.0015 from 99.7% rule)


79. 1) Value at the center: 0

2) Value of the z-score for X=15: -3.00

3) Value of shaded area to the left of the z-score: 0.0013 (or 0.0015)

4) Value of area to the right of the z-score: 0.9987 (or 0.9985)

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