Statistics Homework Help

A population variable has a distribution with mean 50 and standard deviation 15. From this population a simple random sample of n observations is to be selected and the mean of the sample values calculated.

How big must the sample size n be so that the standard deviation of the sample mean, bar-x, is equal to 1.4, i.e., standard deviation = 1.4?


A. n = 11
B. n = 161
C. n = 115
D. n = 36
E. n = 21

If the population variable is known to be Normally distributed and the sample size used is to be n = 16, what is the probability that the sample mean will be between 48.35 and 55.74, i.e., P(48.35 ≤ bar-x ≤ 55.74)?

A. 0.393
B. 0.607
C. 0.937
D. 0.330
E. 0.008

ans: [ C ] n = 115 [ (15/1.4)^2 ]
------

z1 = (48.35-50)/(15/sqrt 16) = -0.44 , z2 = (55.74-50)/(15/sqrt 16) = 1.53
P[-0.44 <= z <=1.53] = 0.6070 <-------

The first question is asking: sigma(pop) / rad n = 1.4

do you see how? Your given the value of sigma pop and also the value of the standard deviation you want. THis is appropriate to do because of the Central Limit theorem.

so fill it in!

15/radn = 1.4
answe: c

the second problem:

I assume this is based on the information in the 1st problem. I usually like to write down what I'm given

n=16
s=1.4
mean of pop=50
stan. deviation of pop = 15

Once again your going to have to use the Central limit theorem.

so X bar approx. normal =(50, 15/rad16)

then you get (50, 3.75) where 50= SAMPLE mean and 3.75 is SAMPLE stand. deviation

then you standardize the equation!

P(48.35 ≤ bar-x ≤ 55.74) --> p(48.35-50/3.75 ≤ bar-x-50/3.75 ≤ 55.74-50/3.75)

then you get (--.44≤ z ≤ 1.53)
(notice how the bar x is now a z score!)

then you look in the back of your book for the phi values:

phi(-1.53) - phi(-.44) and that should be your answer!