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
0.27
inches? Round your answer to at least four decimal places.
you can play around with Z table stuff here:
http://davidmlane.com/hyperstat/z_table.html
To find the proportion of woodlice with antenna lengths that are more than 0.27 inches, we need to calculate the area under the normal distribution curve to the right of 0.27 inches.
Step 1: Standardize the value of 0.27 inches using the z-score formula:
z = (x - μ) / σ
Where:
x = 0.27 inches
μ = mean = 0.2 inches
σ = standard deviation = 0.05 inches
Substituting the values:
z = (0.27 - 0.2) / 0.05
z = 1.4
Step 2: Look up the corresponding cumulative probability from the z-table, which gives the proportion of values less than or equal to a certain z-score.
From the z-table, we find that the cumulative probability for a z-score of 1.4 is approximately 0.9192.
Step 3: Calculate the proportion of woodlice with antenna lengths greater than 0.27 inches. Since we want the proportion to the right of 0.27 inches, we subtract the cumulative probability from 1:
Proportion = 1 - 0.9192
Proportion = 0.0808
Therefore, approximately 0.0808, or 8.08% of woodlice, have antenna lengths that are more than 0.27 inches.