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Answer to Question #110110 in Statistics and Probability for Mateus Michael Ndinomwene
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.
Expert's answer
"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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