Thread: Mean & Median

Hi, I need some help with this please:

On a math test for a class of 32 students, the mean mark was 68, and the median mark was 64. Sue recieved 84, but her test was incorrectly marked. If her teacher giver her 10 more marks, what is:

a) The new class mean?
b) The new class median?

Thanks...

Originally Posted by StackFryer

Hi, I need some help with this please:

On a math test for a class of 32 students, the mean mark was 68, and the median mark was 64. Sue recieved 84, but her test was incorrectly marked. If her teacher giver her 10 more marks, what is:

a) The new class mean?
b) The new class median?

Thanks...

Let me show you an example problem.

If two people averaged 15 years old, how old are they added together?

remember, the average is: average=(the sum of all tries)/(the number of tries)

substitute: 15=(The sum of all tries)/2

multiply both sides by two to get: 30=The sum of all tries

but what if someone messed up and one of the people's ages were 4 years older, what would be the mean then?

We know that the sum is already 30, so all we need to do is add 4 on and find the new mean, so:

Average=(The sum of all tries)/(the number of tries)

substitute: Average=34/2

divide: Average=17

Can you do your problem now?

Originally Posted by StackFryer

Hi, I need some help with this please:

On a math test for a class of 32 students, the mean mark was 68, and the median mark was 64. Sue recieved 84, but her test was incorrectly marked. If her teacher giver her 10 more marks, what is:

a) The new class mean?
b) The new class median?

Thanks...

b) As Sue's mark was in the top 50% (above the median) increasing
that mark leaves the meadian unchanged.

RonL

I got it, thanks guys.