Algebra II: Mean and Standard Deviation

Algebra II

California Standards Test: Algebra II
California Standards Test: Algebra II (Graphing Inequalities
CA Standards: Algebra II (Algebraic Division/Multiplication)
CA Standards: Algebra II
Algebra II: Simplifying Polynomials
Algebra II: Imaginary and Complex Numbers
Algebra II: Complex numbers and conjugates
Algebra II: Quadratics and Shifts
Examples: Graphing and interpreting quadratics
Algebra ||: Conic Sections
Algebra II: Circles and Logarithms
Algebra II: Logarithms Exponential Growth
Algebra II: Logarithms and more
Algebra II: Functions, Combinatorics
Algebra II: binomial Expansion and Combinatorics
Algebra II: Binomial Expansions, Geometric Series Sum
Algebra II: Functions and Probability
Algebra II: Probability and Statistics
Algebra II: Mean and Standard Deviation

Algebra II: Mean and Standard Deviation 79-80, mean and standard deviation

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Algebra II: Mean and Standard Deviation

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We're on problem 79.
A small business owner must hire seasonal workers as the
need arises.
The following list shows the number of employees hired
monthly for a five-month period.
OK, so we hired 4 in the first month, 13 in the second
month and so on.
If the mean of these data-- remember, data is plural.
I always say this data, but if they correctly say if the mean
of these data is approximately 7.
So the mean is the average, right?
So they're saying the average of these numbers is 7.
Fair enough.
If you add all of these up and divide by 5, you'll get
approximately 7.
What is the population standard
deviation for these data?
So, with population-- I'll make a statistical playlist
one day where I talk about the difference between a
population, standard deviation, and a sample
standard deviation and all that.
But the population standard deviation, just you have an
intuition, it's a measure of, on average, how far are each
of these numbers from the mean?
And it's not exactly the average of how far they are
from the mean, but I'll show you that in a second.
It's actually the square root of the average of the squared
distances of how far each of these are from the median.
And that might sound crazy for a second, but when you
actually work it all out the math isn't too bad.
So what we'll do first is figure out the variance.
That's the notation for the variance.
And the standard deviation actually is just the square
root of this.
So if you have the variance, it's very easy to figure out
the standard deviation.
But the variance-- and just to give you the intuition-- it's
equal to the average of the squared
distances from the mean.
So essentially I say, OK how far is 4 from 7?
Oh it's 3 away.
And I square it.
I say, OK, the squared distance is 9.
The squared distance here, let me figure it out.
So let's actually write them down.
Let me actually calculate it.
So I'll do it here.
4 minus 7 is minus 3.
If I were to square that, I'd get 9.
So the square of the distance from 4 to 7 is 9.
So that's a squared distance.
So the squared distance from 13 to 7.
See, they're 6 apart.
You square them.
And because we're squaring it doesn't matter if we do 7
minus 12 or 13 minus 7.
Because a negative times a negative is a positive.
Anyway 13 minus 7 is 6.
That squared is 36.
Then I have 5 minus 7.
That's minus 2 squared is equal to 4.
Then I have 6 minus 7.
That's negative 1.
Squared is equal to 1.
And then I have 9 minus 7.
That's 2.
Squared is equal to 4.
So the 9 is the square of how far 4 is from 7.
36 is a square of how far 13 is from 7.
So each of these are the squared distances of each the
data points from the mean.
That's where we got the 7 from, the
average of these numbers.
Now, we said the variance is the average
of the squared distances.
So it's just the average of these five numbers.
So what's the average of these five numbers?
It's going to be 9 plus 36 is 45.
Plus 4 is 49.
Plus 1 is 50.
Plus 1 is 54.
So it's going to be 54 divided by, 1, 2, 3, 4, 5.
Divided by 5.
So the variance is equal to the average
of the squared distances.
And that's-- what?
10 and 4/5 is equal to 10.8.
So that's the variance.
They don't want the variance.
They don't want the population variance.
They want the population standard deviation.
Well that's just the square of the variance.
And you can see, since you're taking the square of the
square distances, it kind of becomes close to the average
of the real distances.
But it's actually not that and I don't want to
confuse you right now.
For the sake of this problem, let's just take the square
root of this.
So the standard deviation is just equal to the
square root of the.
Variance Which is equal to the square root of 10.8.
And what is that?
Let's see, 3 squared is 9.
4 squared is 16.
So the square root of 10.8 is going to be pretty close to 3.
Between 3 and 4 but a lot closer to 3.
And if we look at all the choices, the only one that
even comes close is this one.
I think if you actually found the square root it's actually
very close to 3.3.
Next problem.
James found the mean and standard deviation of the set
of numbers given above.
If he adds 5 to each number, which of the
following will result?
All right.
If you add 5.
The mean will be multiplied by 5.
No, no, no.
That's not right.
Because if you add 5 to each of these numbers, the mean
will just become 5 bigger.
You're kind of just shifting over the
center to some degree.
Where you're measuring the center by the average over 5.
Actually I could even show it to you.
If I said 3 plus 5, and I'm going to add that to 6 plus 5
plus 2 plus 5.
I'm adding 5 to every number.
Plus 2 plus 5.
Plus 1 plus 5.
Plus 7 plus 5.
Plus 5 plus 5.
All of that divided by-- there's, 1,
2, 3, 4, 5, 6 numbers.
All of that divided by 6.
This is the same thing as-- let me write this-- if we, we
can get rid of parentheses.
All we're doing is adding.
So that's the same thing as 3 plus 6 plus 2 plus
1 plus 7 plus 5.
I just took the first terms in each of these.
Plus-- how many 5s do we have?
1, 2, 3, 4, 5, 6.
Plus 6 times 5.
All of that over 6.
Well, this is going to be equal to 3 plus 6 plus 2 plus
1 plus 7 plus 5, over 6.
Let me scroll down a little bit.
Plus-- what's 6 times 5 divided by 6?
Well that's just 5.
So this is the average of the original set of numbers.
And now, when you add 5 to all of them, the average of all of
them combined is just going to be the original
average plus 5.
So you're just shifting the average to the right.
So this is not right.
You're not multiplying the mean by 5.
You're just adding 5 to the mean.
That's not right.
The standard deviation will increase by 5.
Well, in the last video you saw that the standard
deviation is the square root of the variance.
But both of them are a measure of the average distance of the
numbers from the mean, or from the average.
So even if you shift everything by 5-- the mean
shifts by 5, all the numbers shift by 5-- their distances
from the mean aren't going to change, actually.
The standard deviation actually won't change.
So this isn't right.
The standard deviation won't increase by 5.
These numbers aren't going to become further
away from the average.
So that's not right.
The mean will not change.
Well, no.
We know that the mean will change.
I just showed you that the mean will
actually increase by 5.
When all the numbers have increased by 5.
That's not right.
So we already know that D is probably the answer,
but let's make sure.
The standard deviation will not change.
Right.
In general, the average dispersion, or how far on
average the numbers are away from the mean, won't change.
All the numbers are going to increase by 5 and the mean is
going to increase by 5.
So how far the numbers are from the mean, their variance
won't change.
The square of their distances won't change.
And if the variance doesn't change, then
the mean won't change.
So the answer's definitely D.
And we're all done.
See you in the next playlist.

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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...

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At 2:33, Sal says "double bonds" but should say "single bonds."

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