Question

a particular bird in the amazon rainforest has an average bone length of 14.2mm. they have a standard deviation of 1.3 mm. what percent of birds have a bone length of less than 13.1mm?

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 \(\sigma = 1.3\).

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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:

\(81.5\%\)