Finding ∑fx and ∑fx^2 from the standard deviation and mean

[albertrichardf]

Homework Statement

This is not homework. This is a question given in class with the answers already given. Neither the teacher nor the students have been able to figure out how to obtain the answers though, so my question is how to find them?
Here is a picture of the question.

The standard deviation, mean and variance are all for a population. None of them are for a sample, since it is not in the syllabus.

The answer given is 492 for ∑f . x and 6114.1 for ∑f . x^2

Homework Equations

The definition of the mean is:
u = ∑(f . x)/∑f
The definition of the standard deviation is:
√v
where is the variance, which is equal to:

(∑(f. (x^2))/42 - u^2)

Which is the mean of the square minus the square of the mean.

The Attempt at a Solution

According to the definition of the mean, ∑f .x would be equal to 42u, since ∑f = 42. u = 72.3. This gives about 3036, which is obviously not equal to 492

Then, according to the definition of the variance, it is the square of the mean. Thus the square of 34.1 should be equal to ∑(f . (x^2)) - u^2/42. Thus:

42 * (34.1^2 + u^2) = ∑f . (x^2)

Plugging in u = 72.3 gives 268384.2, which is again a far cry from 6114.1

So my question is, what is it that I'm doing wrong, and how to obtain the given answers?
Thanks for any answers.

 

[Mark44]

Homework Statement

This is not homework. This is a question given in class with the answers already given. Neither the teacher nor the students have been able to figure out how to obtain the answers though, so my question is how to find them?
Here is a picture of the question.
View attachment 106611

The standard deviation, mean and variance are all for a population. None of them are for a sample, since it is not in the syllabus.

Actually, you are given $\bar{x}$ not $\mu$. $\bar{x}$ is the sample mean, while $\mu$ is the population mean. The problem states that the sample standard deviation, s, is 34.1 which is generally somewhat larger than the population standard deviation, $\sigma$.

The answer given is 492 for ∑f . x and 6114.1 for ∑f . x^2

Homework Equations

The definition of the mean is:
u = ∑(f . x)/∑f
The definition of the standard deviation is:
√v
where is the variance, which is equal to:

(∑(f. (x^2))/42 - u^2)

Which is the mean of the square minus the square of the mean.

The Attempt at a Solution

According to the definition of the mean, ∑f .x would be equal to 42u, since ∑f = 42. u = 72.3.

This gives about 3036, which is obviously not equal to 492

I get 3036 as well. Since your teacher is stumped by this problem, too, I begin to have doubts about the answers given for this problem.

Then, according to the definition of the variance, it is the square of the mean. Thus the square of 34.1 should be equal to ∑(f . (x^2)) - u^2/42. Thus:

42 * (34.1^2 + u^2) = ∑f . (x^2)

Plugging in u = 72.3 gives 268384.2, which is again a far cry from 6114.1

So my question is, what is it that I'm doing wrong, and how to obtain the given answers?
Thanks for any answers.

 

[albertrichardf]

Alright. Thanks. Actually, I know that μ is supposed to the population mean, but the syllabus specifically states that xbar is the population mean. In fact, it mentions that there is no need to know about the sample calculations. The formulae given too are only for the population values.
Well, anyway reducing n by one does not make much of a difference. It does look impossible that a set of data such that the average value is 72.3 adds up to 429.

 

[RUber]

I agree with your method. As was noted above, you need a slight adjustment to your variance formula for sample data.
I built a data set that has the properties given, with 21 responses of 106, and 21 responses of 38.6. Doing tests on that clearly provide a counterargument to the so-called answers you posted.

 

[albertrichardf]

Yeah it does seem weird. Unless the variables aren't what they are assumed to mean, which is improbable. Thanks for the answers