Question #110110
The mean diameter of a population of newly settled periwinkle is 8.5 cm with a standard deviation of 1.10 cm. What is the probability of selecting a random sample of 100 wrinkles that has a mean diameter greater than 8.52.
1
Expert's answer
2020-04-16T18:52:11-0400

P(Xˉ>8.52)=P(Z>xˉ−μσn)=P(Z>8.52−8.51.10100)=P(Z>0.18)=0.4286.P(\bar X>8.52)=P(Z>\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}})=P(Z>\frac{8.52-8.5}{\frac{1.10}{\sqrt{100}}})=P(Z>0.18)=0.4286.


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17.04.20, 16:39

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17.04.20, 05:29

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