Answer to Question #180495 in Statistics and Probability for Silver

Question #180495

A study discussed that the percent purity of oxygen from a certain supplier yields a mean of 99.61% and a standard deviation of 0.08%. If the distribution of percent purity is said to be in normal distribution, what purity value exceed exactly 5% of the population?


1
Expert's answer
2021-04-14T07:39:16-0400

We need to calculate the value of "y" when the "P(Y\\geq y)=0.05." This is computed as follows


"P(Y\\geq y)=0.05, \\\\ 1-P(Y\\leq y) =0.05,\\\\ P(Y\\leq y)=0.95 ,\\\\P\\left(\\frac{Y-\\mu}{\\sigma}\\leq \\frac{y-\\mu}{\\sigma}\\right)=0.95, \\\\ P\\left(Z\\leq\\frac{y-\\mu}{\\sigma}\\right)=0.95,\\\\P(Z\\leq 1.645)=0.95."


From the above we know that


"\\frac{y-\\mu}{\\sigma}=1.645".


Solving for "y" we have


"y=(1.645)(\\sigma)+\\mu".


Substituting the values of "\\mu" and

"\\sigma" we have the value of "y" to be


"y=(1.645)(0.08)+99.61 \\Longrightarrow y=99.7416."


Thus, the purity value that exceeds exactly 5% of the population is "99.7416\\%."


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