Suppose that the effects of copper on a second species (say, species B) of fish show the variance of ln(LC50) measurements to be .8. If the population means of ln(LC50) for the two species are equal, find the probability that, with random samples of ten measurements from each species, the sample mean for species A exceeds the sample mean for species B by at least 1 unit.

Hello the reference related to your question is missing attached below is the reference

answer : p ( x - y ≥ 1 ) = 0.0019

Step-by-step explanation:

Assuming 'X' and 'Y' are independent random samples

X : mean value = , variance = 0.4 /10 = 0.04

Y : mean value = , variance = 0.8/10 = 0.08

since the values are independent

V [ X - Y ] = V [ X ] + V [Y ] = 0.04 + 0.08 = 0.12

Determine the probability that the sample mean for species A exceeds the sample mean for species B by at least 1 unit

P ( x - y ≥ 1 ) = 0.0019

attached below is the detailed solution

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )

answer: i don't understand it

step-by-step explanation: sorry