A sampling distribution problem?A random sample of size 81 is drawn from a population with a standA random sample of size 81 is drawn from a population with a standard deviation of 16.If only 18% of the time a sample mean greater than 300 is obtained
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A sampling distribution problem?A random sample of size 81 is drawn from a population with a standA random sample of size 81 is drawn from a population with a standard deviation of 16.If only 18% of the time a sample mean greater than 300 is obtained
A sampling distribution problem?A random sample of size 81 is drawn from a population with a stand
A random sample of size 81 is drawn from a population with a standard deviation of 16.If only 18% of the time a sample mean greater than 300 is obtained,what is the mean of the population?
A sampling distribution problem?A random sample of size 81 is drawn from a population with a standA random sample of size 81 is drawn from a population with a standard deviation of 16.If only 18% of the time a sample mean greater than 300 is obtained
一个抽样分布问题.从一个总体中抽取81个数据的随机样本.
已知总体的标准差为16,如果抽取的样本的平均值大于300的概率仅为18%,那么总体的均值为多少?
设总体均值为mu
不管总体服从什么分布,样本均值均服从正态分布N(mu,256/81)
P(x>300)=18%
18%的分位数为：NORMINV(0.84,0,1)=0.9945
(300-mu)/(16/9)=0.9945
mu=298
Assume K= the mean of the parameter
the mean of sample~N(K,256/81)
convert normal distribution into standard normal distribution
Z=(300-K)/(16/9)
P(z<(300-K)/(19/9))=1-18%
according to standard normal distribution table：(300-K)/(16/9)=0.915
K=298.4