MAT 102 - Introduction to Statistics [PDF]

Answer Key

MAT 102 - Introduction to Statistics Test 2 1. a.

(3 pts) min x  12, Q1  18, med  26, Q3  32, max x  55

b. (3 pts) Box & Whisker Plot for Students Online Hours per Week

12

18

26

32

55

c. (3 pts) Skewed right d. (3 pts) Yes a potential outlier is 55 hours from using the box-and-whisker plot which identifies the outlier by separating it from the whisker. or No 55 is not an outlier if you calculate the z-score since it is not greater than 3.

x  26.69 and s  11.38 and z 

55  26.69  2.49 11.38

or No 55 is not an outlier if you found the inner and outer fences. 2. a. (3 pts) 2 cars b. (2 pts) The probability is 0.10 c. (5 pts)   1.55 cars per household d. (5 pts)   0.80

*NOTE: for questions 3, 4, and 5 your answers may be slightly different if you added in one step in your calculator. 3. a. (5 pts) n  6 and p  4 5  0.80 s2

P  s  2

binompdf  6, 4 5 , 2  0.0154 b. (5 pts) n  6 and p  4 5  0.80

P  s  5  P  s  6 

binompdf  6, 4 5 ,5  binompdf  6, 4 5 , 6  0.3932  0.2621  0.6553

4. a. (5 pts) n  8 and p  0.40

P  s  0   P  s  1 binompdf  8, 0.40, 0   binompdf 8, 0.40,1 0.0168  0.0896  0.1064 b. (5 pts) n  8 and p  0.40

P  s  6   P  s  7   P  s  8 binompdf  8, 0.40, 6   binompdf  8, 0.40, 7   binompdf 8, 0.40,8 0.0413  0.0079  0.0007  0.0499

5. a. (3 pts) binompdf 10, 1 6 , 2   0.2907 b. (3 pts) binompdf 10, 1 6 , 0   binompdf 10, 1 6 ,1  binompdf 10, 1 6 , 2  0.1615  0.3230  0.2907  0.7752

6. a. (3 pts) normalcdf   E99,1.20  0.8849 b. (3 pts) normalcdf  1.64,0.55  0.6583 c. (3 pts) normalcdf 1.08, E99   0.1401

7. a.

(3 pts) invnorm  0.15  1.04  z b. (3 pts) invnorm  0.95  1.64  z c. (3 pts) invnorm  0.10  1.28  z and z  1.28 8. a. (4 pts) normalcdf 170, E99,150, 20  0.1587 b. (4 pts) normalcdf 140,160,150, 20  0.3829 c. (4 pts) normalcdf   E99,155,150, 20  0.5987 9. a.

(4 pts) normalcdf b. (4 pts) normalcdf c. (4 pts) normalcdf d. (4 pts) normalcdf

 375, E99,350,30  0.2023  E99,300,350,30  0.0478  340,360,350,30  0.2611  26.11%   E 99,335,350,30   0.3085  30.85%

31 is the percintile rank 335  P31