a particular bird in the amazon rainforest has an average bone length of 14.2mm. they have a standard…

Explanation:

Step1: Calculate the z - score

The formula for the z - score is (z=\frac{x - \mu}{\sigma}), where (x = 13.1), (\mu=14.2), and (\sigma = 1.3). [ \begin{align*} z&=\frac{13.1-14.2}{1.3}\ &=\frac{-1.1}{1.3}\approx - 0.85 \end{align*} ]

Step2: Find the percentage using the standard normal distribution table

Looking up the z - score of (-0.85) in the standard normal distribution table. The value corresponding to (z=-0.85) is (0.1977). But we want the percentage of values less than (x = 13.1). Since the total area under the normal curve is (1), the percentage of birds with bone length less than (13.1)mm is (1-0.1977 = 0.8023\approx80.23%) (using more accurate table values or a calculator, if we use the empirical rule approximations more carefully: The mean (\mu = 14.2), (\mu-\sigma=14.2 - 1.3=12.9), (\mu - 2\sigma=14.2-2\times1.3 = 11.6), (\mu-3\sigma=14.2-3\times1.3 = 10.3), (\mu+\sigma=14.2 + 1.3=15.5), (\mu + 2\sigma=14.2+2\times1.3=16.8), (\mu+3\sigma=14.2+3\times1.3 = 18.1). The area to the left of (\mu-\sigma = 12.9) is (16%) (approximate from the empirical rule: about (68%) within (\mu\pm\sigma), so (34%) on each side of the mean within one - standard - deviation). The value (x = 13.1) is (14.2-13.1 = 1.1) away from the mean. Since (\sigma=1.3), (1.1\approx0.85\sigma). Using a more accurate normal distribution calculation (either via a calculator with normalcdf function (\text{normalcdf}(-\infty,13.1,14.2,1.3)) or a more detailed z - table), we get that the percentage is approximately (100\times(1 - 0.185)=81.5%))

Answer:

(81.5%)