A population is distributed with a mean of 50 and a standard deviation of 5. Within what range will 90% of the population fall?
50 +/- 1.65*5%, only if it is normal
50 (+/-) 1.65*5 (41.75, 58.25)
Both wrong…this one is tricky
Since you don’t specify if it’s a normal distribution or not, I’m going with 3.16228 standard deviations based on chebyshev’s inequality.
> 50 +/- 1.65*5%, only if it is normal That’s why I said if it is normal, if that’s what you mean.
So the range is?
This is the solution…can anyone explain? Lower end of range = 50 - (3 x 5), upper end of range = 50 + (3 x 5) Range = 35 to 65 chebyshev’s inequality. is not clear to me… WTF!!?
Since you don’t specify it’s a normal distribution, you need to apply chebyshev’s inequality which gives the ‘worst case’ range, if you will, to include a set number of observations in the range. 1 - (1/k^2). I just back solved for a 90% range to get 3.16228 I would argue the correct answer is 50 +/- 5 x 3.16228 (rather than just 3)
mcf Wrote: ------------------------------------------------------- > Since you don’t specify it’s a normal > distribution, you need to apply chebyshev’s > inequality which gives the ‘worst case’ range, if > you will, to include a set number of observations > in the range. > > 1 - (1/k^2). I just back solved for a 90% range > to get 3.16228 > > I would argue the correct answer is 50 +/- 5 x > 3.16228 (rather than just 3) could you break this out – how do we know what K is ?
mcf Wrote: ------------------------------------------------------- > Since you don’t specify it’s a normal > distribution, you need to apply chebyshev’s > inequality which gives the ‘worst case’ range, if > you will, to include a set number of observations > in the range. > > 1 - (1/k^2). I just back solved for a 90% range > to get 3.16228 > > I would argue the correct answer is 50 +/- 5 x > 3.16228 (rather than just 3) How do you get the 3? Sorry if I ask but I am completely confused on this
the k in chebychev’s happens to be # of std. deviations from the mean. so (1/k)^2 = (1/.9)^2 = 1.2345 so 50 +/- 1.2345 (5) == ??? that does seem to be a far cry, in my mind. I would think, given it is population, you can safely assume Normal Dist, and go with the 1.645 # used above. CP
Just a little algebraic manipulation, I think. Desired range is 90%. So… 1-(1/k^2) = 0.90 1/k^2 = 0.10 1 = 0.10 * k^2 1/.1 = k^2 10 = k^2 sqrrt (10) = k 3.16228 = k
68% of the population will fall between 45 and 55, 95% will fall between 40 and 60 99% will fall between 35 and 65 90% should fall somewhere slightly greater than 40 and somewhat slightly less than 60. so arbitarily i’ll say 43, 57 (but you could do precisely by using the distribution tables.
thanks a lot mcf!
SO YOU ARE SAYING IF IT DOES NOT SAY NORMALLY DISTRIBUTED, WE USE CHEYBOKOVS FORMULAS?
k represents the # of sd’s in the equation so if you have 1 - 1/k^2 = 0.9, solve for k, you get 3.1622778 great catch mcf I’m getting this wrong on the exam for sure…
If it does not say normally distributed and you have such limited information as you do in this question, chebyshev’s inequality is the most conservative way to come up with the estimated range. I’ve only seen one other question like this in all my sample questions (and I’ve done the entire qbank at this point). In other cases, it will be clear they want you to use a t or z table based test as they’ll specify a sample size.
but it does say POPULATION, doesn’t it…!!! why are you arguing about it? Is this a framed question from somebody’s mind? if so, it is IRRELEVANT. If there is an answer, share it, let every one learn from it, and move on!!!.. you guys seem to want to beat things to death, so close to the exam. This is not the time, believe me. When we were doing Dec, we were more into exam and question mode, learning from our mistakes, as to what to do, and not. Not the way this batch seems to be doing things, which is go ad-infinitum, ad-nauseum on a topic, with no end in sight!!! I am sorry for the outburst, but more action on the question front, so you are sharing more info, rather than trying to prove who knows more, is what is needed when you are at a point less than two weeks before the exam. CP
I’m with you cpk on the intensive posting, but as for this question it is clear that we are not talking about a normal distribution, so we have to use 1-1/k^2.