SOLUTION: Hello in need assistance with solving this question, i also want to know/see what steps im missing : Given a population mean IQ =100 and a standard deviation of 16, for a random

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Question 1037910: Hello in need assistance with solving this question, i also want to know/see what steps im missing :
Given a population mean IQ =100 and a standard deviation of 16, for a random sample of size 64, what is the probability that the sample mean IQ will be greater than 104?
would be less than 96
would be less than 98
Im not sure if im doing my calculations right,
z=104-100\16=0.25

Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Hello in need assistance with solving this question, i also want to
know/see what steps im missing :
Given a population mean IQ =100 and a standard deviation of 16, for
a random sample of size 64, what is the probability that the sample
mean IQ
will be greater than 104?
would be less than 96
would be less than 98
I'm not sure if im doing my calculations right,
z=104-100\16=0.25

You're getting two things mixed up.

1.  When you are selecting just ONE THING at random and looking 
    for the probability that that ONE THING's VALUE is within a
    certain range.

and 

2. When you are selecting A WHOLE LOT OF THINGS at random and looking 
   for the probability that THEIR AVERAGE (MEAN) value is within a
   certain range.

This is case 2, not case 1.  When you select a whole lot of things, 
the whole lot of things is called "a sample" and the number of things 
in the sample is called the "sample size", n.

In case 1 you're looking for a value of x. In case 2 you are looking
for a value of x-bar, written this way: x

In case 1 you use this formula for z

    x - m
z = —————— 
      s

In case 2 you use this formula for z

    x - m
z = —————— 
      

The big difference is that when you select many things (a sample) 
you must divide the standard deviation by the square root of the 
sample size.

You must use the second formula in this problem because you have a sample:


    104 - 100
z = —————————
     

Don't forget to divide the standard deviation by the square root
of the sample size (the number of things in the sample).  If you
need further help, tell me about it in the thank-you note form
below, and I'll get back to you by email.  BTW, I don't charge any
money.  I'm a retired math prof and do this as a hobby.

Edwin

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