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.
Answer:
71: 19 72: 100 73:9 74:1 75:-2 76:4
Step-by-step explanation:
71. 2(12)-5 = 24-5 = 19
72. 2(5(12)-10) = 2(60-10) = 2(50) =100
73. 3+12/2 = 3+6 = 9
74. 4(5)/12+4(2)= 20/20 = 1
75. 2(2)^2-10 = 8-10 = -2
76. 12-10+2 =4
Answer:
the answer is A on Edge
A) The limit does not exist because the values of h(x) seem to oscillate between random values around x = 9.
Step-by-step explanation:
took the quiz :)
I got 23 u might want to check it tho
Answer:x+85=x+105
x+x=105+85
2x=190
x=190/2
x=95
Step-by-step explanation:
