Statistics-Mean & Standard Deviation of Absences.

[mesh]

Could someone guide me on the approach I need in resolving this question.
I have seen some online examples of the same question, but what is required in this is different. I'll very much appreciate your support.

Problem:
For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6.In a nationwide survey, suppose 100 boys and 50 girls are sampled.
Required:
What are the mean and standard deviation of the absences of the entire class

 

[HallsofIvy]

Re: Statistics-Mean & Standard Deviation of Absences.

Do you know the definitions of 'mean' and 'standard deviation'? To find the mean of a collection of data, add all of the values and divide by the number of values. If the mean number of absences of 100 boys is 15 then the total number of absences must have been 100*15= 1500. If the number of absences of 50 girls is 10, then the total number of absences must have been 50*10= 500. So there were a total of 1500+ 500= 2000 absences for all 150 students.

"Standard deviation" is a little more complicated. It is the square root of an average. Specifically, if the number of absences for the 100 boys was \(\displaystyle x_1\), \(\displaystyle x_2\), ..., \(\displaystyle x_{100}\), then first find the average of the squares of the difference of \(\displaystyle x_i\) and the mean, \(\displaystyle \mu\): \(\displaystyle \frac{1}{100}\sum (x_i- \mu)^2= \frac{1}{100}\left(\sum x_i^2- 2\sum \mu x_i+ \sum \mu^2\right)\).

Since \(\displaystyle \mu\) is constant, independent of I, we can take out of each sum: \(\displaystyle \frac{1}{100}\sum x_i^2- 2\mu \sum x_i+ \mu\sum 1\).

Of course, \(\displaystyle \mu\sum 1= 100\mu\) while \(\displaystyle \sum x_i= 100\mu\) so that becomes \(\displaystyle \sigma^2= \frac{1}{100}\sum x_i^2- \mu^2\) so that the standard deviation is \(\displaystyle \sigma= \sqrt{\frac{1}{100}\sum x_i^2- \mu^2}\). (That formula is probably in your textbook.)

If the standard deviation in the number of absences for 100 boy was 7 with mean 15, then \(\displaystyle 7= \sqrt{\frac{1}{100}\sum x_i^2- 225}\) so that \(\displaystyle 49= \frac{1}{100}\sum x_i^2- 225\) and \(\displaystyle \sum x_i^2= 100(49+ 225)= 27400\).

Similarly, for the 50 girls, the standard deviation in the number of absences was 6 with mean 10 so that \(\displaystyle 6= \sqrt{\frac{1}{50}\sum x_i^2- 100}\) so that \(\displaystyle 36= \frac{1}{50}\sum x_i^2- 100\) and \(\displaystyle \sum x_i^2= 50(36+ 100)= 6800.

So the "sum of squares" for both boys and girls is 27400+ 6800= 34200. Divide that by the total 100+ 50= 150 boys and girls. Subtract the mean that you found above, and take the square root.\)