Question

The average length of adult male american alligators is 3.4 meters. suppose the distribution of lengths is mound/bell shaped, and the standard deviation is 0.5 meters. what percent of adult male american alligators are between 2.9 and 3.9 meters long?

Answer 1

The percentage of adult male American alligators between 2.9 and 3.9 meters long is approximately 68.26%.

Step-by-step explanation:

To find the percent of adult male American alligators between 2.9 and 3.9 meters long, we need to find the area under the normal distribution curve between these two lengths. Since the distribution is bell-shaped, we can use the Empirical Rule to estimate this percentage.

1. The mean length of adult male American alligators is 3.4 meters and the standard deviation is 0.5 meters.
2. Calculate the z-scores for the lower and upper limits:
   For 2.9 meters: z = (2.9 - 3.4) / 0.5 = -1
   For 3.9 meters: z = (3.9 - 3.4) / 0.5 = 1
3. Look up the corresponding percentiles for the z-scores in the standard normal distribution table.
   For z = -1, the percentile is 0.1587 (15.87%)
   For z = 1, the percentile is 0.8413 (84.13%)
4. Subtract the lower percentile from the upper percentile to find the percentage between the two lengths: 84.13% - 15.87% = 68.26%

Answer 2

Approximately 68.26% of adult male American alligators are between 2.9 and 3.9 meters long. This was determined using the z-score formula and a standard normal distribution table.

Step-by-step explanation:

The question asks what percent of adult male American alligators are between 2.9 and 3.9 meters long, given the average length is 3.4 meters and the standard deviation is 0.5 meters. Since the distribution is mound/bell-shaped, this implies a normal distribution.

First, we standardize the lengths 2.9 and 3.9 meters using the z-score formula: z = (X - μ) / σ, where X is the length, μ is the mean, and σ is the standard deviation. For 2.9 meters, z = (2.9 - 3.4) / 0.5 = -1. For 3.9 meters, z = (3.9 - 3.4) / 0.5 = 1.

Using a standard normal distribution table, the area to the left of z = -1 is approximately 0.1587 (15.87%) and the area to the left of z = 1 is approximately 0.8413 (84.13%). To find the percent between, we subtract the smaller area from the larger area: 84.13% - 15.87% = 68.26%. Therefore, approximately 68.26% of adult male American alligators are between 2.9 and 3.9 meters long.