Question

a local post office weighs outgoing mail and finds that the weights of first - class letters are approximately normally distributed with a mean of 0.69 ounce and a standard deviation of 0.16 ounce.
find the 60th percentile of this distribution. ounces (round to 2 decimal places.)

Explanation:

Step1: Find the z - score corresponding to the 60th percentile

We know that if \(X\sim N(\mu,\sigma^{2})\), and we want to find the \(x\) value such that \(P(X\leq x)=0.6\). Using the standard normal distribution \(Z = \frac{X-\mu}{\sigma}\sim N(0,1)\). We look up the \(z\) - value in the standard normal table (or use a calculator with a normal - distribution function). Using a calculator (e.g., in R: qnorm(0.6) or in Excel: NORM.S.INV(0.6)), the \(z\) - score \(z\approx0.253\).

Step2: Use the formula \(x=\mu + z\sigma\)

Given \(\mu = 0.69\) (mean) and \(\sigma=0.16\) (standard deviation). Substitute the values into the formula: \(x=0.69+0.253\times0.16\).
First, calculate \(0.253\times0.16 = 0.04048\).
Then, \(x=0.69 + 0.04048=0.73048\).

Answer:

\(0.73\)