The mean attention span for adults in a certain village is 15 minutes with a standard deviation of 6.4 What is the probability of randomly selecting a sample of 30 scores from this population that has a mean attention span equal to or greater than 17 minutes?

Question

amistre64 · Answer

what ways do you have to approach this?  stats program, ti83, tables,... ???

amistre64 · Answer

out and out integrating?

slaaibak · Answer

I'd say just use the normal Z-statistic approach.  

P(X(BAR) > 17)

Then Z = X(BAR) - 15 / (6.4/sqrt(30))

then just calculating P(Z> (17-15)/(6.4/sqrt(30)) using standard tables

slaaibak · Answer

Z ~ N(0,1) btw

anonymous · Answer

hmm i don't think I completely understand the Z~N(0.1) part ? and thanks for helping btw!

slaaibak · Answer

the central limit theorem states that the sample mean is normally distributed with mean u and variance sigma^2/n

It follows that:

[x-bar minus - mean] / sigma/sqrt(n) is a standard normal distribution.