SOLUTIONS TO HW 10: Chapter 16

2. (a) 253m.

(b) 253m.

(c) 61m

6. (a) Q1=2038 pts

(b) Q3=2670 pts

14. (a) (128-110)/12 = 1.5

(b) (100-110)/12 = -5/6

(c) (110-110)/12 = 0

(d) (71-110)/12 = -3.25

19.

(84-50)/SD = 2

34=2(SD)

SD=17

35. (a)52 pts

(b) 50%

(c) 68% (1 SD on either side)

(d) 1/2(100%-68%)=16%

36. (a) 30 is 2 SD below the mean and 74 is 2 SD above. So, 95% of 2500 scored between 30 and 74.

(b) .025*250=62.5 (round to 63 students)

(c) .0015*2500=3.75 (4 students)

38. (a) 50th%

(b) a little above 84th%

(c) a little above 75th%

(d) 99.85th%

48. (a) 2.5th%

(b) 16th%

(c) 22.5 lb

55. (a) 68%

(b) 16%

(c) 84%

58.

(a) mean=1200*.75=900

SD= sqrt[1200*.75*(1-.075)]=15

(b) Q1=900-.675*15=890

Q3=900+.675*15=910

(c) .50 because 900 to 945 consists of 3 SD.

64. (a) 55+12*2.33=82.96 (round to 83 pts)

(b) 55-12*.52=48.76 (49pts)

74. (a) probability of losing on red is 20/38=0.53 . By dishonest coin principle, Y has mean=10000*0.53=5300 and SD=sqrt[10000*0.53*0.47]=50

(b) Since the center of the distribution is at Y=5300, the chances that we will lose 5300 times or more are 50%.

(c) 5150 to 5450 is 3 SD in either direction, so it is 99.7%

(d) To break even or win we must have Y less than or equal to 5000 (i.e. you have to be more than 4 SD to the left of the center) the chances of this are essentially 0.