An operation manager at an electronic company wants to test their amplifiers. The design engineer claims they have a mean output of 305 watts with a variance of 121.
What is the probability that the mean amplifier output would be greater than 303 watts in a sample of 46 amplifiers if the claim is true?
An operation manager at an electronic company wants to test their amplifiers. (STAT QUESTION)?
2016-05-09 2:45 am
回答 (2)
2016-05-09 7:56 am
✔ 最佳答案
s = √121 = 11P( xbar > 303 )
= P( Z > (303-305)/(11/√46) )
= P( Z > - 1.23 )
= 0.8907 ..... Ans
2016-05-10 10:28 am
The sample mean Xbar should have a mean of 305 watts
and variance 121/46. With a sample of 46, Xbar will have an
approximate normal distribution.
P(Xbar>303)=P(Z>(303-305)/√(121/46))
=P(Z>-1.23315)
=P(Z<1.23315) from symmetry of the Z curve
=0.89124
~0.89 (2SF)
and variance 121/46. With a sample of 46, Xbar will have an
approximate normal distribution.
P(Xbar>303)=P(Z>(303-305)/√(121/46))
=P(Z>-1.23315)
=P(Z<1.23315) from symmetry of the Z curve
=0.89124
~0.89 (2SF)
收錄日期: 2021-05-02 14:07:31
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https://hk.answers.yahoo.com/question/index?qid=20160508184516AA6J8dg