Suppose that the antenna lengths of woodlice are approximately normally distributed with a mean of 0.2 inches and a standard deviation of 0.05 inches. What proportion of woodlice have antenna lengths that are more than 0.23 inches? Round your answer to at least four decimal places.

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

.7257

The distribution of heights of American women aged to is approximately normally distributed with a mean of inches and a standard deviation of inches. Calculate the -score for a woman six feet tall.

To find the proportion of woodlice with antenna lengths greater than 0.23 inches, we need to calculate the area under the normal distribution curve to the right of 0.23. Here's how you can do it step by step:

1. Standardize the value 0.23 using the formula for z-score:

z = (x - μ) / σ

Where:
x = 0.23 (value we want to find the proportion for)
μ = 0.2 (mean)
σ = 0.05 (standard deviation)

Plugging in the values:
z = (0.23 - 0.2) / 0.05
z = 0.03 / 0.05
z = 0.6

2. Look up the proportion corresponding to the z-score of 0.6 in the standard normal distribution table. The table provides the proportion of values to the left of a given z-score.

Based on the table, the proportion to the left of 0.6 is 0.7257.

3. Subtract the result from 1 to find the proportion to the right of 0.6:

Proportion to the right = 1 - 0.7257
Proportion to the right ≈ 0.2743

The proportion of woodlice with antenna lengths greater than 0.23 inches is approximately 0.2743.

Note: This answer is rounded to four decimal places as requested in the question.