Question

the wingspans of about herons are approximately normally distributed with a mean of 127 cm and a standard deviation of 13 cm.
a. determine the proportion of herons that have wing spans less than 96 cm. round your answer to 4 decimal places.
0.0086
b. the largest 10 percent herons have wingspans of ____ cm or more. round your answer to 1 decimal place.
143.6
c. the middle 90 percent of herons have wingspans between __ and __ cm. round your answers to 1 decimal place. put the smaller answer on the left and the larger answer on the right.
105.6 148.4
d. determine the proportion of herons which have wingspans between 112 and 143 cm. round your answer to 4 decimal places.
0.7666
e. determine the proportion of herons which have wingspans greater than 154 cm. round your answer to 4 decimal places

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 127\) (mean), \(\sigma=13\) (standard deviation), and \(x = 154\) (value).

$$z=\frac{154 - 127}{13}=\frac{27}{13}\approx2.0769$$

Step2: Find the proportion

We want \(P(X>154)\), which is equivalent to \(1 - P(X\leq154)\). Using the standard normal distribution table or a calculator with a normal - distribution function (e.g., in Excel: \(=1 - NORM.DIST(154,127,13,TRUE)\) or in R: \(=1 - pnorm(154,127,13)\)), we know that \(P(Z\leq2.0769)\approx0.9812\)

$$P(X > 154)=1 - P(X\leq154)=1-0.9812 = 0.0188$$

Answer:

\(0.0188\)