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 \(\sigma = 1.3\).

$$z=\frac{13.1 - 14.2}{1.3}=\frac{- 1.1}{1.3}\approx - 0.85$$

Step2: Use the standard normal distribution table

Looking up the value of \(z=-0.85\) in the standard normal distribution table. The value corresponding to \(z =-0.85\) is \(0.1977\)

Answer:

Approximately \(19.77\%\) of the birds have a bone length of less than \(13.1\)mm.