Mean, Variance, Standard Variation

Question

I submitted this question the other week to my tutor, but it was wrong. the only thing correct was the mean. He did not tell me why it was wrong or the correct answers.

Can anybody tell me why this is wrong please? And where I went wrong?

Question and my answer below:

Why haven´t used more decimal places? For instance: $x_1-\overline x=22.36-22.395=-0.035$ And for the sample variance you have to divide it by 100 I think. As far as I´ve understood the question it is not asked for an unbiased estimator for the variance (of the population). — callculus42, May 05 '19 at 19:21
all the steps look fine, but your teacher could be expecting the proper rounding (show at least four digits after point), notations like $F$ must be $f$, etc. If you don't round and carry out all calculations in MS Excel, the SD will be $0.0128468...$ and the variance will be $0.000165...$. See the difference? — farruhota, May 05 '19 at 19:39

score 1 · Answer 1 · answered May 05 '19 at 18:59

1

I think that your method is correct. However, one place you might have messed up is rounding $\bar{x}=22.3981$ to $22.39$, when it is actually closer to $22.40$, as this might've messed up your answer. You should keep the value as $22.3981$ during calculations.

answered May 05 '19 at 18:59

Saketh Malyala

13,637

So recalculate with 22.3981 then this would of been correct? – Kyle Anderson May 05 '19 at 19:28
yup it should be right then – Saketh Malyala May 05 '19 at 19:30
I don't think that this causes the instructor to say the answer is wrong. – NoChance May 05 '19 at 19:32
but it had to have created a large enough error, and maybe the instructor was simply scanning the answers @NoChance – Saketh Malyala May 05 '19 at 19:33
He did say tiny errors. So maybe this could be why. If so that is so frustrating. I thought they was a big mistake, hence why I asked on this forum. – Kyle Anderson May 05 '19 at 19:35
@NoChance can you see any other errors that would cause me to fail? – Kyle Anderson May 05 '19 at 19:36
@KyleAnderson, pls check my answer. – NoChance May 05 '19 at 19:45

score 0 · Answer 2 · answered May 08 '19 at 12:29

0

Using the formulas from This Question, I noticed that the formulas are different from what you are using, specially in the denominator of the Variance.

I obtained the following values:

answered May 08 '19 at 12:29

NoChance

6,427

Mean, Variance, Standard Variation

2 Answers2