Answer:
Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.
A. The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.
C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.
E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake
The answer is E.
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Starting population = 4000
Addition per month = 170
decline on population per month = 70
Increase rate in population per month (dt) :
Starting population = 4000
Number of births per month = 170
However, the population declines by 70 individuals each month
Hence,
Number of births - number of deaths(d) = 70
170 - d = - 70 ( decline?
170 + 70 = d
240 = d
d = number of deaths
Per capita death :
Total number of deaths per. Month / starting population
= 240 / 4000
= 0.06
