Apopulation has a mean, μ = 89 and a standard deviation,σ = 24.
find the mean and standard deviation of a sampling distribution of sample means with sample size n = 64.

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".  

Solution to the problem

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

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

The sample mean is defined as:

And if we find the expected value and variance for the sample mean we got:

Var(\bar X) = \frac{\sigma^2}{n}[/tex]

The distribution for the sample mean is given by:

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7