The average length of an adult blue whate is 30 meters with a standard deviation of 5.7 meters. If a sample of 31 whales is taken, what is the probability that their average lenght differs from the mean by more than 2.0 meters
0.1635
To determine the probability that the average length differs from the mean by more than 2.0 meters, we can use the Central Limit Theorem (CLT) and the Z-score.
The Central Limit Theorem states that if we have a sample size that is large enough (ideally greater than 30) and the population is normally distributed, the distribution of sample means will be approximately normal regardless of the shape of the population distribution.
First, we calculate the standard error of the mean (SEM) using the formula SEM = standard deviation / square root of sample size:
SEM = 5.7 / √31 ≈ 1.02
Next, we calculate the Z-score using the formula Z = (X - μ) / SEM, where X is the desired length, μ is the mean length, and SEM is the standard error of the mean:
Z = (2.0 - 0) / 1.02 ≈ 1.96
The Z-score of 1.96 corresponds to the value at the 97.5th percentile of the standard normal distribution. Since we are interested in the probability that the average length differs from the mean by more than 2.0 meters (either in the positive or negative direction), we calculate the probability of being more extreme on both sides of the distribution.
Using a Z-table or a statistical software, we find that the area to the right of the Z-score (1.96) is approximately 0.025. Since we are looking for the area in the tails (both sides), we double this probability:
P(Z > 1.96) = 2 * 0.025 ≈ 0.05
Therefore, the probability that the average length of the sample differs from the mean by more than 2.0 meters is approximately 0.05 or 5%.