Math Is Fun Forum

DWC256

Member

Registered: 2011-02-26

Posts: 1

Help please

Question:
The weight in kilograms, w, of the 15 players in a rugby team was recorded and the results summarised as follows.
Sum of w = 1145.3
Sum of w^2 = 88042.14

...

Due to injury, one of the players who weighed 79.2kg was replaced with another player who weighed 63.5kg.
-- without further calculation state the effect of this change on the mean and and variance of the weight of the players in the team. Explain your answers.

Problem:
This is a 4 mark question. So how am I suppose to answer and score all marks -- rather than just saying that the mean and standard deviation would be lower as a result.

Many thanks in advance

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help please

Hi DWC256;

One thing, I am not sure you are able to say that the standard deviation goes down...

Supposing we remove the 79.2 kg player we now have a mean of 76.15 kg. Adding the 63.5 kg seems to me that it will increase the variance as it is further from the mean.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

byronjordan

Member

Registered: 2011-01-30

Posts: 14

Re: Help please

Strange problem. What does "4 mark" mean? I can't do it "without further calculation" either and I'm not sure how the sum of w^2 helps. Maybe the sum of w^2 is an intentional red herring.

The change in the mean is obvious without further calculation since taking out a heavier and putting in a lower weight will always reduce the mean. The mean is the sum divided by the quantity of 15. The quantity doesn't change and the sum would obviously go down by taking out a heavier and putting in a lighter weight.

The change in standard deviation isn't at all obvious to me without further calculations. The orginal mean is 1145.3/15=76.35, but I get (1145.3-79.2+63.5)/15=75.31 for the reduced mean.

Still, since 63.5 is considerably farther from the reduced mean of 75.31, than 79.2 was from the original mean of 76.35 ((75.31-63.5)>(79.2-76.35)), the standard deviation would go up. Standard deviation is basically a measure of how far spread out weights are from the mean. Standard deviation is the square root of the mean of the squares of the difference between the weights and the mean.

The conclusion is clearly the same as bobbym's.