一道英语概率的题,在线等答案,非常急!!!!

Ithasbeenfoundthat72.3%ofallunsolicitedmaildeliveredtohouseholdsgoesunread.Overthecou... 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 0.7?
马上就要啊啊啊,求答案啊,重谢!!!!
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p = 72.3% = 0.723, n = 80

  1. mean of the sample proportion = np = 0.723*80 = 57.84

  2. variance s^2 = p(1-p)/(n-1) = (0.723)*(1-0.723)/(79) ≈ 0.002535

  3. standard error s = √0.00254 ≈ 0.05040

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

追问
后三个一样,主要是第一问,为什么是np啊,我找书了看写的样本比例的μp=p啊?
追答
不好意思是我看错了。。mean of the sample proportion 和总体成功概率P是一样的,所以是0.723。
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