Answer:
By the Central Limit Theorem, the sampling distribution of the sample mean amount of money in a savings account is approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Average of 1,200 dollars and a standard deviation of 900 dollars.
This means that ![\mu = 1200, \sigma = 900](https://tex.z-dn.net/?f=%5Cmu%20%3D%201200%2C%20%5Csigma%20%3D%20900)
Sample of 10.
This means that ![n = 10, s = \frac{900}{\sqrt{10}} = 284.6](https://tex.z-dn.net/?f=n%20%3D%2010%2C%20s%20%3D%20%5Cfrac%7B900%7D%7B%5Csqrt%7B10%7D%7D%20%3D%20284.6)
The sampling distribution of the sample mean amount of money in a savings account is
By the Central Limit Theorem, approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.
'justify your answer using complete sentences' means re-write your answer in a complete sentence. In other words, say your answer in a full and clear sentence.
Answer:
N$12 576. 11
Step-by-step explanation:
Loss%=10%
S.P= N$11 480.50
C.P =?
Loss%=100(C.P - S.P)/C.P
10=100(C.P-11 480.50)/C.P
10C.P=100C.P - 1 148 050
1 148 050=100C.P - 10C.P
1 148 050=90C.P
C.P=N$12 576.11
F(x)=467(<span>5^<span>1/4</span></span><span>)^<span>4<span>x is the closest I can get.</span></span></span>