Question

the wingspans of adult 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.
b. the largest 10 percent herons have wingspans of ____ cm or more. round your answer to 1 decimal place.
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.
d. determine the proportion of herons which have wingspans between 112 and 143 cm. round your answer to 4 decimal places.

Explanation:

Step1: Calculate the z - score for part a

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 127\), \(\sigma=13\), and \(x = 96\).

$$z=\frac{96 - 127}{13}=\frac{-31}{13}\approx - 2.3846$$

Using the standard normal table (or a calculator with a normal - distribution function, e.g., \(P(Z<z)\) for \(z=-2.3846\)), \(P(X < 96)=P(Z<-2.3846)\approx0.0086\)

Step2: Find the z - score for part b

If the largest \(10\%\) of herons are considered, then the area to the left of the z - score is \(A = 1 - 0.10=0.90\).
Looking up the z - score in the standard normal table (or using the inverse - normal function on a calculator), the z - score \(z\) such that \(P(Z < z)=0.90\) is approximately \(z = 1.28\)
Using the formula \(x=\mu+z\sigma\), we substitute \(\mu = 127\), \(\sigma = 13\), and \(z = 1.28\)

$$x=127+1.28\times13=127 + 16.64=143.6$$

Step3: Find the z - scores for part c

For the middle \(90\%\) of the data, the area in the two tails is \(1 - 0.90 = 0.10\), so the area in each tail is \(\frac{0.10}{2}=0.05\)
The z - score for the left - tail \(z_1\) (where \(P(Z<z_1)=0.05\)) is approximately \(z_1=-1.645\), and the z - score for the right - tail \(z_2\) (where \(P(Z < z_2)=0.95\)) is approximately \(z_2 = 1.645\)
Using \(x=\mu+z\sigma\)
For \(z=-1.645\): \(x_1=127-1.645\times13=127-21.385 = 105.6\)
For \(z = 1.645\): \(x_2=127 + 1.645\times13=127+21.385=148.4\)

Step4: Calculate the z - scores for part d

For \(x_1 = 112\): \(z_1=\frac{112 - 127}{13}=\frac{-15}{13}\approx - 1.1538\)
For \(x_2 = 143\): \(z_2=\frac{143 - 127}{13}=\frac{16}{13}\approx1.2308\)
\(P(112<X<143)=P(-1.1538<Z<1.2308)\)
\(P(-1.1538<Z<1.2308)=P(Z < 1.2308)-P(Z<-1.1538)\)
\(P(Z < 1.2308)\approx0.8907\), \(P(Z<-1.1538)\approx0.1241\)
\(P(-1.1538<Z<1.2308)=0.8907 - 0.1241=0.7666\)

Answer:

a. \(0.0086\)
b. \(143.6\)
c. \(105.6\), \(148.4\)
d. \(0.7666\)