QM計probability 問題

n = 200

p = 0.1

np = 20 > 10.

Hence, by Central limit theorem, it is well enough to approximate by normal distribution.

Let X be the number of widowed members selected

X ~ N(np , npq)

X ~ N(20 , 20(1 - 0.1))

X ~ N(20 , 18)

P(X >= 10)

= P[z >= (10 - 0.5 - 20) / √18]

= P(z >= -2.47487)

= 0.5 + 0.4933

= 0.9933