急 ! F5 about Dispersion 13q7

2013-03-23 6:14 am
請詳細步驟教我計以下二條 :


圖片參考:http://imgcld.yimg.com/8/n/HA05788109/o/20130322221358.jpg

回答 (1)

2013-03-23 8:31 am
✔ 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


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