急! More about Dispersion 13Q7

Question 20


圖片參考：http://faculty.virginia.edu/PullenLab/WJIIIDRBModule/standard%20normal%20distribution.jpg


(a)

(i) Between 3.48 cm and 3.51 cm is
between 3.50 cm - 0.02 cm and 3.50 cm + 0.01 cm,
that is between mean - 2*(s.d.) and mean + 1*(s.d.)
That would cover
95% - (95% - 68%)/2
= 95% - (27%)/2
= 95% - 13.5%
= 81.5%

(ii) Percentage outside the acceptable range
= 100% - Percentage within the acceptable range
= 100% - Percentage covering (mean ± 2*(s.d.))
= 100% - 95%
= 5%

(b) Recall that the acceptable range is from 3.50-0.02 cm to 3.50+0.02 cm,
that is, from 3.48 cm to 3.52 cm.
Now, with mean = 3.51 cm and s.d. = 0.01 cm.
That acceptable range is from mean - 3*(s.d.) to mean + 1*(s.d.)
Therefore, percentage outside the acceptable range
= 100% - [99.7% - (99.7%-68%)/2 ]
= 100% - [99.7% - (31.7%)/2 ]
= 100% - [99.7% - 15.85%]
= 100% - 83.85%
= 16.15%


Question 21

Let m be the mean mark and s be the standard deviation.
Let S be Sandy's mark and C be Cindy's mark.

It is given that
(S - m)/s = 1.6 ...(1)
(C - m)/s = -0.4 ...(2)
(60 - m)/s = -0.2 ...(3)
S = C+ 30 ...(4)

From (3), 60 - m = -0.2s and then m = 60 + 0.2s ...(5)

Put (4) and (5) into (1) and (2):
{ (C + 30 - 60 - 0.2s)/s = 1.6
{ (C - 60 - 0.2s)/s = -0.4

{ C + 30 - 60 - 0.2s = 1.6s
{ C - 60 - 0.2s = -0.4s

{ C - 30 - 0.2s = 1.6s
{ C - 60 - 0.2s = -0.4s

Subtracting gives 30 = 2s, s = 15.

C - 30 = 1.8s = 27
C = 57

S = 57 + 30 = 87

m = 60 + 0.2*15 = 63

Therefore,

(a) The s.d. is s = 15

(b) The mean mark is m = 63

(c) Sandy's mark is S = 87, Cindy's mark is C = 57