Question

Earthworms of the species Lumbricus terrestris often surface after a large rain storm. It is unclear why they surface in such abundance after large amounts

of rain. The lengths of this species of earthworm are normally distributed with an average length of 180 mm with a standard deviation of 15.5 mm. What
length would we expect to find at the 30th percentile?
A. 78.89
B. 188.13
C. 22.54
D. 171.87
E. 109.89

Answer 1

The 30th percentile length of an earthworm for the species Lumbricus terrestris is approximately 171.87 mm, calculated using the z-score method for a normal distribution.

Step-by-step explanation:

To determine the length of an earthworm that we would expect to find at the 30th percentile, we will have to use the properties of the normal distribution. Given that the average length is 180 mm with a standard deviation of 15.5 mm, we can use z-scores to find the percentile. A z-score corresponds to the number of standard deviations a point is from the mean of a normal distribution. The z-score for the 30th percentile can be found using statistical tables or a calculator that provides inverse normal probabilities.

Once we have the z-score for the 30th percentile, we apply the following formula to find the actual length:

L = μ + (Z × σ)

Where L is the length of the earthworm, μ is the mean length, Z is the z-score for the 30th percentile, and σ is the standard deviation.

Using statistical software or a z-table, we find that the z-score for the 30th percentile is approximately -0.52. We then calculate:

L = 180 + (-0.52 × 15.5) = 180 - 8.06 ≈ 171.87

Hence, the length we expect to find at the 30th percentile is approximately 171.87 mm, which corresponds to option D.