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. Estimate the 60th percentile of first-class letter weights.

Answer 1

To estimate the 60th percentile of first-class letter weights, we can use the standard normal distribution and the formula x = z * σ + μ. By calculating the z-score for the 60th percentile and plugging in the values, the estimated 60th percentile of first-class letter weights is approximately 0.79648 ounce.

Step-by-step explanation:

To estimate the 60th percentile of first-class letter weights, we need to find the z-score corresponding to the 60th percentile using the standard normal distribution. The z-score can be calculated using the formula:

z = (x - μ) / σ

where x is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation. We can rearrange the formula to solve for x:

x = z * σ + μ

Now, we substitute the values into the formula:

x = z * 0.16 + 0.69

To find the z-score for the 60th percentile, we look up the corresponding value in the z-table or use a calculator. The z-score for the 60th percentile is approximately 0.253. Plugging in the values, we get:

x = 0.253 * 0.16 + 0.69

x ≈ 0.10648 + 0.69

x ≈ 0.79648

Therefore, the estimated 60th percentile of first-class letter weights is approximately 0.79648 ounce.