Determine mu subscript x overbar and sigma subscript x overbar from the given parameters of the population and sample size. mu equals86, sigma equals24, nequals 64 mu subscript x overbar equals86 sigma subscript x overbar equalsnothing

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".  

Solution to the problem

Let X the random variable who represents the variable of interest. We know from the problem that the distribution for the random variable X is given by:

We select a sample of size n=64. That represent the sample size.  

From the central limit theorem we know that the distribution for the sample mean is given by:  

The mean for the sample distirbution would be given by:

And the deviation given by:

And then the distribution for the sample mean is:

he has now walked a total of what?

if i know what this is maybe i can

step-by-step explanation:

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west