Help Solve Variance Question with 80 Students

[Yankel]

Hello,

Any help with this one would be appreciated.

In a college 80 students are taking the exam in the spring semester. Their grades has a mean and standard deviation.
In the summer semester, k additional students are being tested. All k students get grades which are equal to the average grade of the 80 students from the spring semester. After combining both samples, we get a standard deviation which is half of the standard deviation of the first 80 students. Find k.

I did something, and got k=80, but I am not sure it's correct, would appreciate if anyone can validate my answer.

what I did was to take S1 (the standard deviation before any addition) and I said it is equal to 2*S2 (which is the standard deviation of the two samples together).

Then I realized, that since all observations in the second sample are equal to the mean, than the numerator is equal in both parts of the equations (apart from the 2 of course). Moreover, the sums of deviance are equal.

Then I solved the equation to get k=80

P.S In this context, when I say variance I mean the formula in which we divide by n, not by n-1.

 

[I like Serena]

Hello,

Any help with this one would be appreciated.

In a college 80 students are taking the exam in the spring semester. Their grades has a mean and standard deviation.
In the summer semester, k additional students are being tested. All k students get grades which are equal to the average grade of the 80 students from the spring semester. After combining both samples, we get a standard deviation which is half of the standard deviation of the first 80 students. Find k.

I did something, and got k=80, but I am not sure it's correct, would appreciate if anyone can validate my answer.

what I did was to take S1 (the standard deviation before any addition) and I said it is equal to 2*S2 (which is the standard deviation of the two samples together).

Then I realized, that since all observations in the second sample are equal to the mean, than the numerator is equal in both parts of the equations (apart from the 2 of course). Moreover, the sums of deviance are equal.

Then I solved the equation to get k=80

P.S In this context, when I say variance I mean the formula in which we divide by n, not by n-1.

Hi Yankel! :)

What you have is pretty close!

The one oversight is that when you say that the numerator is equal, you are talking about the variance instead of the standard deviation.
In formula form:

$\sigma_1^2 = \dfrac{{SS}_1}{80}$

$\sigma_2^2 = \dfrac{{SS}_2}{80 + k} = \dfrac{{SS}_1 + k \cdot 0}{80 + k}$

where ${SS}_1$ is the sum of the squared deviations for the first sample, and ${SS}_2$ is the sum for the complete sample.

P.S.: A variance in which we divide by n is called a population variance, as opposed to a sample variance.