Question

use the following to answer the questions below.
use the normal distribution of fish lengths for which the mean is 11 inches and the standard deviation is 2 inches. assume the variable x is normally distributed.
what percent of the fish are longer than 12 inches? round your percent to two decimal places.
if 400 fish are randomly selected, about how many would you expect to be shorter than 8 inches?
round your answer to the nearest whole number.

Explanation:

Step1: Calculate the z - score for \(x = 12\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 11\) (mean) and \(\sigma=2\) (standard deviation).

$$z=\frac{12 - 11}{2}=\frac{1}{2}=0.5$$

We want to find \(P(X>12)\), which is equivalent to \(P(Z > 0.5)\). Using the property \(P(Z>z)=1 - P(Z\leq z)\). From the standard normal table, \(P(Z\leq0.5)=0.6915\). So \(P(Z > 0.5)=1 - 0.6915=0.3085\) or \(30.85\%\)

Step2: Calculate the z - score for \(x = 8\)

$$z=\frac{8 - 11}{2}=\frac{- 3}{2}=-1.5$$

We want to find \(P(X < 8)\), which is equivalent to \(P(Z<-1.5)\). From the standard normal table, \(P(Z<-1.5)=0.0668\)

If \(n = 400\) (number of fish selected), the expected number of fish shorter than 8 inches is \(n\times P(X < 8)\)

$$400\times0.0668 = 26.72\approx27$$

Answer:

The percent of fish longer than 12 inches is \(30.85\%\). The expected number of fish shorter than 8 inches out of 400 is \(27\)