Question

a biologist is studying rainbow trout that live in a certain river and she estimates their mean length to be 620 millimeters. assume that the lengths of these rainbow trout are normally distributed, with a standard deviation of 40 millimeters. use this table or the aleks calculator to find the percentage of rainbow trout in the river that are shorter than 576 millimeters. for your intermediate computations, use four or more decimal places. give your final answer to two decimal places (for example 98.23%).

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 576$ (the value we are interested in), $\mu=620$ (the mean), and $\sigma = 40$ (the standard deviation).
$z=\frac{576 - 620}{40}=\frac{-44}{40}=-1.1000$

Step2: Find the cumulative probability

We use the standard normal distribution table to find the cumulative probability $P(Z < - 1.10)$. Looking up the value in the standard - normal table, we find that $P(Z < - 1.10)=0.135767$.

Step3: Convert to percentage

To convert the probability to a percentage, we multiply by 100.
$0.135767\times100 = 13.58\%$ (rounded to two decimal places)

Answer:

$13.58\%$