Final answer:
To determine the minimum wingspan for the largest 20% of robins, we use the z-score corresponding to the top 20% and the formula X = μ + zσ. With a mean of 14 inches and standard deviation of 0.7 inches, and a z-score of 0.84, we calculate that the value is 14.59 inches.
Step-by-step explanation:
The student is tasked with determining the minimum wingspan that the largest 20% of robins have, assuming the wingspan is normally distributed. Since we are dealing with a normal distribution, this is a problem that involves using z-scores associated with standard normal distribution. The z-score that corresponds to the largest 20% can be found in a z-table or calculated using statistical software or a calculator.
To find the wingspan value, we use the formula: X = μ + zσ, where X is the wingspan we are solving for, μ (mu) is the mean of the distribution, z is the z-score, and σ (sigma) is the standard deviation.
The mean wingspan, μ, is given as 14 inches, and the standard deviation, σ, is 0.7 inches. The z-score that corresponds to the top 20% is approximately 0.84. Plugging these values into the formula, we get: X = 14 + 0.84(0.7) = 14.588. Therefore, the largest 20% of robins have wingspans larger than 14.59 inches, when rounded to two decimal places.