✔ 最佳答案
Q1. mean = 3.5 , SD = 0.01
3.48 < X <3.51
3.48 = 3.50(mean) - 2 *[ 0.01 (SD) ]
3.51 =3.50(mean) + 1*[0.01(SD)]
according to the normal distribution
- 2 (SD) = 34.13% +13.59% ( left side)
+1 (SD) = 34.13% (right side)
so total % are 34.13% +13.59% +34.13% =81.85%
(II) OUTSIDES =(100 - 81.85)% = 18.15%
(b) new mean = 3.51 , SD= 0.01
3.48 = 3.51(mean) - 3 *[ 0.01 (SD) ]
3.51 =3.51(mean) + 0*[0.01(SD)]
according to the normal distribution
- 3 (SD) = 34.13% +13.59% +2.14% ( left side)
+0 (SD) = 0% (right side)
so total % are 34.13% +13.59% +2.14% =49.86%
OUTSIDES =(100 - 49.86)% = 50.14%
Q2. GIVEN:
SANDY 'S SD-SCORE = 1.6
CINDY 'S SD-SCORE = -0.4
HENRY 'S SD-SCORE= -0.2 , and his score = 60
Sandy's score = Cindy's score +30
(a) difference of SD score between sandy and cindy = 1.6- (-0.4) = 2
SO, 2 * (SD) = 30 , and one (SD) = 15
(b) ( score - mean )/ SD = SD score
so, refer to HENRY ,
(60 -mean) / 15 = -0.2
mean= 63
(c)Cindy's mark = (score - 63)/ 15 =-0.4
Score= 57
Sandy's score = (score- 63)/15 =1.6
Score= 87