一道英语概率的题,非常急!It has been found that 72.3% of all unsolicited

一道英语概率的题,非常急!
It has been found that 72.3% of all unsolicited mail delivered to households goes unread.Over the course of a month,a household receives 80 pieces of unsolicited mail.
1、What is the mean of the sample proportion of pieces of unread mail?
2、What is the variance of the sample proportion?
3、What is the standard error of the sample proportion?
4、What is the probability that the sample proportion is greater than 0.
我爱宝宝_aa 1年前 已收到1个回答 举报
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himely 幼苗

共回答了13个问题采纳率：84.6% 举报

p = 72.3% = 0.723,n = 80
mean of the sample proportion = np = 0.723*80 = 57.84
variance s^2 = p(1-p)/(n-1) = (0.723)*(1-0.723)/(79) ≈ 0.002535
standard error s = √0.00254 ≈ 0.05040

p(p>0.7) = p(z>(0.7-p)/s) ≈ p(z>-0.46) = 0.6772

1年前

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