Question

Assume the population of fully loaded delux SUVs has a mean cost of $62,000 with a...

Assume the population of fully loaded delux SUVs has a mean cost of $62,000 with a standard deviation of $5,100. We dont know the exact distribution of the costs, just the population mean and standard deviation.

A. We plan to take a ranodm sample of fully loaded deluxe SUVs and calculate the mean cost for analysis. Approximately, how big should our sample be so we dont care about the exact distribution of cost.

We take a random sample of 100 loaded deluxe SUVs and caluclate the mean cost.

B. What is the standard error of the mean

C. State the probability distribution (sampling distribution) of x bar

D. What is the probability x bar will fall between $61,000-$62,500?

E. What is the probability x bar will fall within $1,500 of the mean?

F. What will happen to the sampling distribution of x bar if we increase our sample size

G. What values of x bar would indicate we have an outlier?

Homework Answers

Answer #1

a)from central limit theorum minimum sample size required is 30 for sample mean to be approximately normal

b)

standard error of the mean =std deviation/sqrt(n)=5100/sqrt(100)=510

c)

sampling distribution of mean(xbar) is approximately normal with estimaed mean =62000 and std error of mean =510

d)

for normal distribution z score =(X-μ)/σx
here mean=       μ= 62000
std deviation   =σ= 5100.000
sample size       =n= 100
std error=σ=σ/√n= 510.0000
probability = P(61000<X<62500) = P(-1.96<Z<0.98)= 0.8365-0.025= 0.8115

e)

probability x bar will fall within $1,500 of the mean:

probability = P(60500<X<63500) = P(-2.94<Z<2.94)= 0.9984-0.0016= 0.9968

f)

as we increase  our sample size standard error of mean decreases .

g)

as 3 std deviation away values are outlier ;

therefore outlier fences =mean -/+3*std error=6200-/+3*5100=60470 to 63530

hence values below 60470 and above 63530 should be considered outlier.

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