有關統計數學問題!!

for 197563 samples
n = 197563
pA = 20 / 197563 = 0.0101% ==> qA = 1 - pA = 99.9899%
pB = 730 / 197563 = 0.3695% ==> qB = 1 - pC = 99.6305%
pC = 340 / 197563 = 0.1721% ==> qC = 1 - pC = 99.8279%

for 10000 samples
n = 10000
E(A) = n x pA = 1.0123
E(B) = n x pB = 36.9502
E(C) = n x pC = 17.2097
sA2 = n x pA x qA = 1.0122
sB2 = n x pB x qB = 36.8137
sC2 = n x pC x qC = 17.1801
for 5% significance
the no. of A in 10000 sample should not exceed E(A) + 1.96 x sA = 2.98 > 2
the no. of B in 10000 sample should not exceed E(B) + 1.96 x sB = 48.84 > 48
the no. of C in 10000 sample should not exceed E(C) + 1.96 x sC = 25.33 > 25

197563----A:20 B:730 C:340

10000--- A:???% B:???% C:???%

不良品A所佔既百分比= 20/197563 * 100% = 0.0101%
不良品B佔既百分比= 730/197563 * 100% = 0.3695%
不良品C佔既百分比= 340/197563 * 100% = 0.1721%