Answer:
Look below
Step-by-step explanation:
The mean of the sampling distribution always equals the mean of the population.
μxˉ=μ
The standard deviation of the sampling distribution is σ/√n, where n is the sample size
σxˉ=σ/n
When a variable in a population is normally distributed, the sampling distribution of for all possible samples of size n is also normally distributed.
If the population is N ( µ, σ) then the sample means distribution is N ( µ, σ/ √ n).
Central Limit Theorem: When randomly sampling from any population with mean µ and standard deviation σ, when n is large enough, the sampling distribution of is approximately normal: ~ N ( µ, σ/ √ n ).
How large a sample size?
It depends on the population distribution. More observations are required if the population distribution is far from normal.
A sample size of 25 is generally enough to obtain a normal sampling distribution from a strong skewness or even mild outliers.
A sample size of 40 will typically be good enough to overcome extreme skewness and outliers.
In many cases, n = 25 isn’t a huge sample. Thus, even for strange population distributions we can assume a normal sampling distribution of the mean and work with it to solve problems.
0.27 is the answer, hope it helps
Answer:
(0, -4)
(2,0)
Step-by-step explanation:
Basically x intercept is (0, -4)
Y intercept is (2, 0)
Hope I helped!
