Question

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?

Answer 1

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 $99.7416\%.$