Solve for x (20 pts)
2 answers:
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Given Information:
Mean = μ = $282.45
Standard deviation = σ = $64.50
Sample size = n = 100
Confidence level = 95%
Required Information:
95% Confidence interval = ?
Answer:
95% Confidence interval = ($269.81, $295.09)
Step-by-step explanation:
We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.
What is Confidence Interval?
The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.
The confidence interval is given by
Where is the mean and MoE is the margin of error given by
Where σ is the standard deviation, n is the sample size and is the z-score corresponding to 95% confidence level.
Finally, the confidence interval is
Therefore, we are 95% sure that the true population mean amount spent per day by a family of four visiting Niagara Falls is within the interval of ($269.81, $295.09)
The unit disk can be parameterized by the function
where
and 
. The squared distance between any point in this region 
and the point (1, 1) is
The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.
![=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7D%5Br%5E2-2r%28%5Ccos%5Ctheta-%5Csin%5Ctheta%29%2B2%5Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D)
where
The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.