anonymous
  • anonymous
The accompanying table shows the scores on a classroom test. What is the population standard deviation x: 100, 95, 90, 80, 75, 72, 70 f: 7, 2, 10, 4, 2, 3, 4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
ooh statistics.....dont remember this but tiere's a formula for Standard Deviation...google it lol
anonymous
  • anonymous
f stands for frequency, so that's the number of x' of that particular value that occurred. So x = 100 occurred 7 times in the population. x = 95 occurred twice. And so on. Find the mean of your population is the first step. That means summing over all the scores of the class and dividing by the class size.
anonymous
  • anonymous
\[\sum_{n=1}^{3}i^(n-1)\]

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anonymous
  • anonymous
that should read i^(n-1)
anonymous
  • anonymous
what's that?
anonymous
  • anonymous
sum?
anonymous
  • anonymous
What is that sum for formula for?
anonymous
  • anonymous
the top part means you do the formula on the right 3 times using n=1 n=2 n=3 and then add up all those values
anonymous
  • anonymous
but the i is raised to the n-1
anonymous
  • anonymous
What's the formula represent? Is this a different problem? I thought you were asking about standard deviation.
anonymous
  • anonymous
yeah its a different problem i figured out the other one ill use my calculator
anonymous
  • anonymous
So what's the sum asking you to do? You want to find an explicit formula for the sum?
anonymous
  • anonymous
no you need to find the sum of all those terms
anonymous
  • anonymous
I have a feeling you wrote the problem wrong. As it is, it makes no sense. Unless the teacher just wants you to write: i^(0) + i^(1) + i^(2). That's the sum.
anonymous
  • anonymous
yeah u do i^1+i^2+i^3 and add them up
anonymous
  • anonymous
you start with 1 because thats what n=
anonymous
  • anonymous
You raise it to (n-1). So when n = 1, n-1 is 0. So the first power is 0. The next power increased by 1, so it's 1.
anonymous
  • anonymous
nev ermind its fine..can you look at my probability question
anonymous
  • anonymous
no i gotta go
anonymous
  • anonymous
if you post it on here, someone will be sure to help. there's a lot of people on right now
anonymous
  • anonymous
okk
anonymous
  • anonymous
good luck

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