# Find the standard deviation of the values of ##y##

• chwala
In summary, the conversation discusses finding the standard deviation of a variable, y, given a set of discrete variables, x, and their corresponding frequencies. The solution involves using the fact that the frequencies are the same and finding the ratio of a:b:c to determine the variance of y. The conversation also addresses a typo in one of the formulas and clarifies that the constant 1 does not affect the calculation of the variance.
chwala
Gold Member
Homework Statement
See attached- sent as received both question and solution.
Relevant Equations
stats
This is the question;

This is the solution as received;

I am not familiar with the approach used in the solution...my thinking was as follows

The frequencies are the same...the only thing changing are the discrete variables thus;

Let ##[x= 2,4,6]## and ##[y=7,13,19]## form a sequence...then the nth term is given by;

##x_n=2n## and ##y_n=(3⋅2n)+1=6n+1=3(x_n)+1##

##y_n=3(x_n)+1##

I will need to think on this...your input is welcome though as we already have the solution given...not unless it is wrong. Cheers.

I will look at this later...should be achievable! Since they've given us standard deviation for ##x## ... and both frequencies have same values then I would need to find the ratio of ##a:b:c## then use that to determine the standard deviation of ##y##.

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chwala said:
The frequencies are the same...the only thing changing are the discrete variables thus;
Let ##[x= 2,4,6]## and ##[y=7,13,19]## form a sequence...
The part about "form a sequence" doesn't make sense to me. Really, what you're talking about is a map (i.e., function) from the x values to the y values.
chwala said:
Let ##[x= 2,4,6]## and ##[y=7,13,19]## form a sequence...then the nth term is given by;
##x_n=2n## and ##y_n=(3⋅2n)+1=6n+1=3(x_n)+1##
##y_n=3(x_n)+1##
The last line gives the function. That is, Y = 3X + 1.

There are several properties of the variance (see https://en.wikipedia.org/wiki/Variance, in the section titled Addition and multiplication by a constant.
Among them are
• Var(X + a) = Var(X)
• Var(aX) = a2Var(X)
From these, you should be able to work out the value of Var(Y) = Var(3X + 1), and use the given value of ##\sigma_X## to find ##\sigma_Y## AKA SD(Y). Note that Var(X) = ##(\sigma_X)^2##.

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chwala
Mark44 said:
The part about "form a sequence" doesn't make sense to me. Really, what you're talking about is a map (i.e., function) from the x values to the y values.

The last line gives the function. That is, Y = 3X + 1.

There are several properties of the variance (see https://en.wikipedia.org/wiki/Variance, in the section titled Addition and multiplication by a constant.
Among them are×
• Var(X + a) = Var(X)
• Var(aX) = a2Var(X)
From these, you should be able to work out the value of Var(Y) = Var(3X + 1), and use the given value of ##\sigma_X## to find ##\sigma_Y## AKA SD(Y). Note that Var(X) = ##(\sigma_X)^2##.

That is fine and clear ...but what happens to the ##1##? all values are scaled by a constant...and i think ##1## is also a part of that constant. Should we not have;

##Var (Y)=\sqrt {a^2×Var(3x)+1}=2.82## to two decimal places?

Last edited by a moderator:
Sorry, I made a typo (now corrected) in one of the formulas I wrote: It should be Var(X + a) = Var(X), not Var(a) as I originally wrote.

So if U = 3x, Var(U) = Var(3x), and so Var(U + 1) = Var(U).

chwala said:
Should we not have; ##Var (Y)=\sqrt {a^2×Var(3x)+1}=2.82## to two decimal places?
No, that's wrong on several counts.
##Var (Y) \ne \sqrt {a^2×Var(3x)+1}## -- you're mixing up variance and standard deviation.
Var(Y) = Var(3X) = 32 Var(X)
So SD(Y) = ##\sqrt{9 Var(X)} = 3 \sqrt{Var(X)}##

The answer given in the image you posted, 2.64, is correct. 2.82 is not.

chwala said:
but what happens to the 1?
It doesn't affect the calculation of the variance. A vertical translation in the data makes no difference. For example, the sets {1, 2, 3} and {3, 4, 5} have different means but the same variance, and hence the same standard deviations.

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chwala

## 1. What is the formula for finding the standard deviation of a set of values?

The formula for finding the standard deviation of a set of values is √(∑(x-μ)^2 / N), where x is each individual value, μ is the mean of the values, and N is the total number of values.

## 2. How do I calculate the standard deviation using a calculator?

Most scientific calculators have a built-in function for calculating the standard deviation. Look for a button labeled "SD" or "σ" and follow the instructions in the calculator's manual to enter your values and find the standard deviation.

## 3. What does the standard deviation tell us about a set of values?

The standard deviation measures the amount of variation or spread in a set of values. A smaller standard deviation indicates that the values are closer to the mean, while a larger standard deviation indicates that the values are more spread out.

## 4. Can the standard deviation be negative?

No, the standard deviation cannot be negative. It is always a positive value or zero, depending on the spread of the values in the set.

## 5. How does the standard deviation differ from the variance?

The standard deviation and variance are both measures of the spread of a set of values, but the standard deviation is the square root of the variance. This means that the standard deviation is in the same units as the original values, while the variance is in squared units.

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