Could someone please help with this question

Question

A population of values has a normal distribution with μ=73 and σ=42.8. You intend to draw a random sample of size n=194.

Find the probability that a sample of size n=194 is randomly selected with a mean between 74.8 and 76.7.

P(74.8 < ¯¯¯X < 76.7) = Round to 4 decimal places.

Answer 1

This problem is an application of the central limit theorem (CLT)

The CLT says that for a normally distributed population, the distribution of the sampling means of sample size n from that population is normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation, divided by the square root of n.

So if we draw a random sample (X) of 192, its mean will be as if it was drawn from a sampling distributions means which has a normal distribution with a mean = 73, and a standard deviation of 42.8/√(192)=3.0888. Accordingly, the P(74.8<X<76.7)

= P([[74.8-73]/3.0888] < Z <[[76.7-73]/3.0888]) = P(.583 < Z <1.198).

The rest is looking up those values on a z-table (or using a calculator) to compute the probability. If you have questions on how to do that let me know, and I can walk you through that procedure as well.

Answer 2

Hi Love,

First you calculate the standard error (sample SD)=42.8/√194=3.0728

Want to calculate the Z scores for 74.8 and 76.7

Z1=(74.8-73)/3.0729=.5957

Z2=(76.7-73)/3.0729=1.2040

From your table of area under the normal curve

p(Z1<.5957)=.7257 and p(Z2<1.204)=.8925

So the P(74.8 < X < 76.7) = .8925-.7257=.1668

Regards

Jim